Answer:
160m/s
Explanation:
The speed of a wave is related to its frequency and wavelength, according to this equation:
v=f ×λ
A lighter particle moving with a speed of 19 m /s collides with an object of quadruple its mass moving in the opposite direction with a third of its speed. Assume that the collision is a one-dimensional elastic collision. What will be the speed of both particles after the collision?
Answer:
Speed of lighter particle = -21.53 m/s
Speed of other particle = 3.8 m/s
Explanation:
Let mass of the lighter object be m
Thus mass of heavier object = 4m
Speed of lighter particle = 19 m/s
Speed of second particle with opposite direction = - ⅓(19) m/s = -19/3 m/s
Now, from the formulas of momentum before collision = momentum after collision, and also kinetic energy before collision = Kinetic Energy after collision, we have;
v_bf = [2m_a/(m_a + m_b)]v_ai + [(m_b - m_a)/(m_a + m_b)]v_bi
Now, in this question;
m_a = m
m_b = 4m
v_ai = 19
v_bi = -19/3 m/s
Thus;
v_bf = [2m/(m + 4m)]19 + [(4m - m)/(m + 4m)](-19/3)
Simplifying to get;
v_bf = 19(2m/5m) - (19/3)(3m/5m)
>> v_bf = 38/5 - 19/5
>> v_bf = 19/5 m/s
>> v_bf = 3.8 m/s
Similarly;
v_af = [(m_a - m_b)/(m_a + m_b)]v_ai + [2m_b/(m_a + m_b)]v_bi
v_af = 19((m - 4m)/(m + 4m)) - (19/3)((2 × 4m)/(m + 4m))
This gives;
v_af = 19(-3m/5m) - (19/3)(8m/5m)
v_af = -(57/5) - (152/15)
v_af = -323/15
v_af = -21.53 m/s
Two 11.5-mg balls are connected by a 28.9-cm-long insulating rod of negligible mass. One ball has a charge of +q and the other has a charge of −q. The system is placed in a 592 N/C uniform electric field. Initially the system is at rest and makes an angle α= 58.9° with respect to the field. It is then released, and when it is momentarily aligned with the electric field, its kinetic energy is 4.86 mJ. What is the magnitude of q?
Answer:
58.8 μC
Explanation:
The two balls create a dipole of dipole moment , p = qd where q = charge and d = distance between the charges = 28.9 cm = 0.289 m. The potential energy change ΔU = -W where W is the work done by the electric field of magnitude E = 592 N/C.
Now U = -pdEcosФ = -qdEcosФ where Ф is the angle between p and E. Since Ф is initially 58.9° and then becomes 0° when p and E align, the potential energy change is thus.
ΔU = -qdEcos0 - (-qdEcos58.9°) = qdE(cos58.9° - 1) = qdE(0.5165 - 1) = -0.4835qdE.
W = - ΔU = -(-0.4835qdE) = 0.4835qdE
This work equals the kinetic energy of the systems. Since K.E = 4.86 mJ = 4.86 × 10⁻³ J.
So K.E = W
K.E = 0.4835qdE
making q subject of the formula
q = K.E/0.4835dE
q = 4.86 × 10⁻³ J/(0.4835 × 0.289 m × 592 N/C)
q = 4.86 × 10⁻³ J/82.72
q = 0.0588 × 10⁻³ C
q = 58.8 × 10⁻⁶ C
q = 58.8 μC
The picture shows two solutions of salt water. Which solution is more concentrated (has a higher concentration)?
Question 15 options:
The first solution is more concentrated
The second solution is more concentrated
The solutions have the same concentration.
In order to determine which of the two solutions of salt water is more concentrated, we need to first understand what concentration means and how it is measured. Concentration refers to the amount of solute dissolved in a given amount of solvent. It is typically measured in units of mass per volume, such as grams per liter (g/L) or milligrams per milliliter (mg/mL). so The second solution is more concentrated
When comparing the concentration of two solutions, the one with a higher concentration has more solute dissolved in the same amount of solvent. Therefore, in the picture provided, we can determine which solution is more concentrated by looking at the relative amounts of solute in each solution.If the solutions have the same concentration, then they must have the same amount of solute dissolved in the same amount of solvent. From the picture, we can see that both solutions are in the same size container and have the same amount of solvent (water) in them. Therefore, we can conclude that they have the same concentration of salt.The amount of solute dissolved in a solution can be increased by either adding more solute or by reducing the amount of solvent. If we were to add more salt to one of the solutions, we would increase the concentration of that solution. Alternatively, if we were to evaporate some of the water from one of the solutions, we would reduce the amount of solvent and increase the concentration of that solution.
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Which one of the following statements is not true.
A. Soil is the second layer of Earth's crust.
B. Metals can be extracted from minerals.
C. Oil is made from dead plants and animals.
D. Chemical weathering can break down rocks into fragments.
4. A golf ball resting on the ground is struck by a golf club and given an initial velocity of
50 m/s at an angle of 30° above the horizontal. The ball heads toward a fence 12 m high at
the end of the golf course, which is 200 m away from the point at which the golf ball was
struck. Neglect any air resistance that may be acting on the golf ball.
K
50 m/s
30⁰ go
200 m
Tout
a. Calculate the time it takes for the ball to reach the plane of the fence.
b. Will the ball hit the fence or pass over it?
Answer:
See below
Explanation:
Horizontal component of the velocity will carry the ball to the fence in how long?
Hor vel = 50 cos 30 = 43.3 m/s 200m / 43.3 m/s = 4.61 seconds
Vertical position given by
Df = vo t + 1/2 a t^2 with t = 4.61 secs vo = 50 sin 30
Df = height = 11 m <===== will not clear the fence
what is the pressure of a tank of uniform cross sectional area 4.0m2 when the tank is filled with water a depth of 6m when given that 1 atm=1.013 x 10^5pa density of water=1000kgm-3 g=9.8m/s2
The pressure of the tank, when filled with water at a depth of 6 m, is approximately 580.124 atmospheres (atm). To calculate the pressure of the tank, one can use the equation: Pressure (P) = Density (ρ) × g × Depth (h)
Pressure (P) = Density (ρ) × g × Depth (h)
Given: Density of water (ρ) = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Depth (h) = 6 m
Using the given values, one can calculate the pressure:
Pressure = 1000 kg/m³ × 9.8 m/s² × 6 m Pressure
= 58800 kg·m⁻¹·s⁻²
Now, let's convert the units to pascals (Pa) using the conversion 1 atm = 1.013 x \(10^5\) Pa:
Pressure = 58800 kg·m⁻¹·s⁻² × (1 atm / 1.013 x\(10^5\) Pa)
Pressure = 580.124 atm
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A ball thrown with an initial velocity of u(10i+15j)m/s when it reaches the top of it trajectly neglucating air resistance what is avelocity & acceleration ?
At the apex of the trajectory, the ball's velocity and acceleration are u (10i+15j) m/s and zero
The initial velocity of the ball is u (10i+15j) m/s.
Velocity at the top of the trajectory: Since the ball is thrown with an initial velocity, it will reach the top of its trajectory with the same velocity, since there is no air resistance. Therefore, the velocity of the ball at the top of the trajectory is u (10i+15j) m/s.
Acceleration: The acceleration of the ball at the top of the trajectory is zero, since the ball is not accelerating (there is no acceleration due to air resistance).
Therefore, the velocity and acceleration of the ball at the top of the trajectory are u (10i+15j) m/s and zero, respectively.
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A meteor falls from the sky to the Earth. The meteor already had an initial velocity downward when it
was spotted. If it hit the Earth at 335 m/s after being seen for 30 seconds, then what was the initial
velocity of the meteor?
a=9.8
Six vectors (a through f ) have the magnitudes and directions indicated in the figure. (Figure 1)
1.Rank the vector combinations on the basis of their magnitude.
Rank from largest to smallest. To rank items as equivalent, overlap them.
O a+c
O f+c
O d
O a+b
O a+e
O a+d
1. The rank of the vector combinations on the basis of their magnitude is - a+b = a +d > a+c > f +c = d > a + e
What is the value of the vector?A physical quantity that has both directions and magnitude is referred to as a vector quantity. A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.
Position, force, torque, velocity, and displacement are a few examples of vector quantities.
1. a + c
⇒ 2+1 unit
⇒ 3 unit
2. a + b
⇒ 2√2 unit
3. a + e
⇒ 2 -1
⇒ 1 unit
4. a +d
⇒ 2√2 unit
5. f+ c
⇒ 2 unit
6. a+d
⇒ 2√2 unit
7. f+c
⇒ 2 unit
8. a+b
⇒ 2√2 unit
9. a+e
⇒ 1 unit
10. a+c
⇒ 3 unit
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Complete Question -
Six vectors (a through f ) have the magnitudes and directions indicated in the figure. (Figure 1)
1.Rank the vector combinations on the basis of their magnitude.
Rank from largest to smallest. To rank items as equivalent, overlap them.
a+c
f+c
d
a+b
a+e
a+d
A speeding car is traveling at a constant 30.0 m/s when it passes a sationary police car. If the police car delays for 1.00s before starting, what must be the magnitude of the constant accleration of the police car. pdf
Answer:
30.0m/s²Explanation:
Given
final speed v = 30.0m/s
initial velocity u = 0m/s
Time taken t = 1.00s
Required
constant acceleration of the police car a
Substitute the given parameters into this equation of motion
v = u + at
30 = 0 +1a
30 = a
a = 30.0m/s²
Hence the magnitude of the constant acceleration of the police car is 30.0m/s²
Batman attempts to hide out in a Batbox.
The top is a mirrored surface, but the vertical
sides are made of clear Batplastic (refractive
index np = 1.333). It is located x = 2.9 m
horizontally from the edge of the pool.
What is the minimum depth (measured to
the top of the Batbox) that the Batbox can be
below the surface of the water for the Caped
Crusader to remain hidden from the Joker?
Assume: The Joker gets down close to the
water for the best view.
Answer in units of m
The arcsine of 1 is 90 degrees, the critical angle is 90 degrees.the minimum depth (measured to the top of the Batbox) is equal to the height of the Batbox.
To determine the minimum depth of the Batbox below the surface of the water for Batman to remain hidden from the Joker, we need to consider the principle of total internal reflection. When light travels from a medium with a higher refractive index to a medium with a lower refractive index, it can undergo total internal reflection if the angle of incidence exceeds the critical angle. In this case, the light traveling from water (refractive index nw = 1.333) to the Batplastic (refractive index np = 1.333) can undergo total internal reflection at the interface between the two.
To calculate the minimum depth, we need to find the critical angle. The critical angle can be determined using the formula:
Critical angle = arcsin(np/nw)
Given:
Refractive index of water, nw = 1.333
Refractive index of Batplastic, np = 1.333
Calculating the critical angle:
Critical angle = arcsin(1.333/1.333)
Critical angle = arcsin(1)
Since the arcsine of 1 is 90 degrees, the critical angle is 90 degrees.
To ensure that Batman remains hidden, the Batbox should be placed at a depth below the surface of the water such that the light undergoes total internal reflection. In this case, the minimum depth is determined by the height of the Batbox. Therefore, the minimum depth (measured to the top of the Batbox) is equal to the height of the Batbox. However, the specific height of the Batbox is not provided in the question, so it cannot be determined without additional information.
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1)Answer the following question in sort
a)Define pressure ?
b)What is the value of standard atmospheric pressure?
c)Mention any one application of liquid pressure in our daily life?
d)Mention in the name of the instruments used to measure the pressure of compressed air?
e)Which instrument is used to measure atmospheric pressure ?
f)What is the unit of compressed air?
g)Define standard atmospheric pressure?
h)Which property of liquid is applicable in water supply system in cities?
i)Which property of liquid supports to use in it in hydraulic machine?
2)Answer the following questions in detail a)Define atmospheric pressure? Prove the presence of atmospheric pressure with the help of an activity?
b)Derive that P=dgh?
c)Describe the structure and working method of mercury barometer briefly?
d)Enlist any three points to show the importance of atmospheric pressure?
e)Enlist any four application of liquid pressure?f)Mention any three events occurred in our daily life which are directly related with pressure?
1)a) Pressure is the force exerted per unit area, measured in units such as pascals (Pa) or pounds per square inch (psi).
b) The value of standard atmospheric pressure is approximately 101.3 kilopascals (kPa) or 1 atmosphere (atm).
c) One application of liquid pressure in our daily life is in hydraulic systems, like car brakes, where liquid pressure is used to transmit force and amplify it.
d) The instruments used to measure the pressure of compressed air include pressure gauges or manometers.
e) An instrument called a barometer is used to measure atmospheric pressure.
f) The unit of compressed air is typically measured in pounds per square inch (psi) or pascals (Pa).
g) Standard atmospheric pressure is the pressure exerted by the Earth's atmosphere at sea level. It is approximately equal to 1 atm or 101.3 kPa.
h) The property of liquid that is applicable in water supply systems in cities is its ability to flow and exert pressure, allowing water to be distributed through pipes and reach different levels in buildings.
i) The property of liquid that supports its use in hydraulic machines is its incompressibility, allowing it to transmit force and energy effectively.
2) a)Atmospheric pressure is the force exerted by the weight of the Earth's atmosphere on a surface.
b) The equation P = dgh. This equation can be derived by considering the weight of the fluid column and the force it exerts on a unit area at the base.
c) A mercury barometer consists of a glass tube filled with mercury, inverted into a dish of mercury. The mercury in the tube adjusts its height based on the atmospheric pressure.
d) The importance of atmospheric pressure can be seen in its role in weather patterns, maintaining the balance of gases in the atmosphere, and facilitating breathing for humans and animals.
e) Applications of liquid pressure include hydraulic systems in machinery, such as lifts and cranes, hydraulic brakes in vehicles, and water towers for maintaining water pressure in buildings.
f) Events in daily life directly related to pressure include inflating a balloon, using a bicycle pump to inflate tires, and squeezing toothpaste out of a tube.
1)a) Pressure is defined as the force per unit area. Its unit in the S.I system is newtons per square meter (N/m²) or Pascal (Pa).
b) The value of standard atmospheric pressure at sea level is 101.3 kPa (kilopascals) or 1 atm (atmosphere). c) Liquid pressure has numerous applications in our daily life, but one of the most common ones is the hydraulic braking system used in cars.
d) An instrument used to measure the pressure of compressed air is called a pressure gauge. e) An instrument used to measure atmospheric pressure is called a barometer.
f) The unit of compressed air is generally psi (pounds per square inch).
g) Standard atmospheric pressure is the pressure exerted by the atmosphere at sea level and is equal to 101.3 kPa or 1 atm.
h) The property of liquids that is applicable in water supply systems in cities is their incompressibility. i) The property of liquids that supports their use in hydraulic machines is their incompressibility.
2)a) Atmospheric pressure is defined as the force per unit area exerted by the weight of the atmosphere on the surface. It is proven with the help of the following activity: Take a glass full of water and place a cardboard over it. Hold the cardboard tight and invert the glass. The water will not spill out of the glass, which is because the atmospheric pressure is greater on the cardboard than the pressure inside the glass.
b) The pressure exerted by a fluid can be derived using P = dgh, where P is the pressure, d is the density, g is the acceleration due to gravity, and h is the height of the fluid column.
c) A mercury barometer is made up of a glass tube that is closed at one end and filled with mercury. The tube is inverted and placed in a container of mercury. The pressure of the atmosphere on the open surface of the container forces the mercury in the tube to rise to a height that is proportional to the atmospheric pressure.
d) The importance of atmospheric pressure can be explained by the following points: it enables breathing, regulates the weather, and causes the ocean tides.
e) Some applications of liquid pressure include hydraulic brakes in cars, hydraulic lifts, and hydraulic jacks.
f) Some events that are directly related to pressure include gas escaping from a pressurized container, balloons being inflated, and soda cans being opened.
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Which of the following scenarios shows an object with a negative velocity and a negative acceleration?
Question 6 options:
A pig moves toward the reference point speeding up.
A pig moves away from the reference point slowing down.
A pig moves toward the reference point slowing down.
A pig moves away from the reference point speeding up.
C. The following scenarios that shows an object with a negative velocity and a negative acceleration is "a pig moves toward the reference point slowing down."
What is negative velocity?A negative velocity occurs when an object moves in the negative direction.
Example of negative velocity;
If a car is moving away from you (reference point) then the velocity is positive. If a car is coming towards you (reference point) then the same vector reverses direction and hence the velocity is negative.What is negative acceleration?If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion. Thus, the acceleration of the object will be negative.
Thus, the following scenarios that shows an object with a negative velocity and a negative acceleration is "a pig moves toward the reference point slowing down."
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What does the difference between gauge pressure and absolute pressure equal?
a.
the pressure within the fluid
c.
the pressure at the bottom of the fluid
b.
the pressure at the surface of the fluid
d.
zero
The actual difference between the guage pressure of a system and an absolute pressure is the same as the pressure at the surface of the fluid.
The correct answer choice is option b.
Why the pressure at fluid surface equals absolute and guage pressure.It follows that the two pressure system mentioned in the given task above has a zero point system as their principle of operation.
When we talk of guage pressure, it uses atmospheric pressure as its own zero point system. However, the sum of both the guage pressure and atmospheric pressure gives what we know as the absolute pressure.
So therefore, we can now deduced that the pressure on a liquid surface is equivalent to the variation between the gauge pressure and absolute pressure.
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Two protons are held a distance d apart. The electrostatic force and the gravitational force that one proton exerts on the other are Fe and Fg, respectively. Which of the following correctly compares the magnitude and direction of these forces?
Fe > Fg, opposite direction
a. Fe > Fg, same direction
b. Fe < Fg opposite direction
c. Fe < Fg same direction
Answer:
Explanation:
charge on each proton = 1.6 x 10⁻¹⁹ C
mass of proton = 1.67 x 10⁻²⁷ kg
Electrostatic force of repulsion Fe = 9 x 10⁹ x ( 1.6 x 10⁻¹⁹ )² / d²
= 23.04 x 10⁻²⁹ / d²
Gravitational force of attraction = G Mm / d²
M = m = 1.67 x 10⁻²⁷ kg
Gravitational force of attraction Fg = 6.67 x 10⁻¹¹ x ( 1.67 x 10⁻²⁷ )² / d²
= 18.60 x 10⁻⁻⁶⁵ / d²
So Fg is far less than Fe and former is attractive , later is repulsive .
Fe > Fg, opposite direction , is the answer .
Frozen food will thaw more quickly when placed under running water than if it
is immersed in water. Why?
Answer:
google answers, i'm not 100% on how to type my own answer
Explanation:
Thaw food submerged under running water at a temperature of 70˚F (21˚C) or lower. The water flow must be strong enough to wash food particles into the overflow drain. In a microwave oven. You can safely thaw food in a microwave oven if the food will be cooked immediately.
The reason defrosting in water is preferable is because water conducts heat better than air. And the faster food is thawed, the better it tastes.
Answer:
The reason defrosting in water is preferable is because water conducts heat better than air. And the faster food is thawed, the better it tastes.
Calculate the quantity of heat energy which must be transferred to 2.25 kg of brass to raise its temperature from 20°C to 240°C if the specific heat of brass is 394 J/kgK.
The quantity of heat energy that must be transferred to 2.25 kg of brass to raise its temperature from 20 °C to 240 °C is 195030 J
How do i determine the quantity of heat energy?First, we shall list out the given parameters from the question. This is shown below:
Mass of brass (M) = 2.25 Kg Initial temperature of brass (T₁) = 20 °CFinal temperature of brass (T₂) = 240 °CChange in temperature of brass (ΔT) = 240 - 20 = 220 °CSpecific heat capacity of brass (C) = 394 J/kgKQuantity of heat energy (Q) =?The quantity of heat energy that must be transferred can be obtained as follow:
Q = MCΔT
= 2.25 × 394 × 220
= 195030 J
Thus, we can conclude quantity of heat energy that must be transferred is 195030 J
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Will the 79 kg skier in the figure below slide down if f the coefficient of static friction is 0.25?
Answer:
Man will not slide down
Explanation:
Given:
Coefficient of static friction = 0.25
Angle = 13°
Computation:
Man will slide down if
tan13° > Coefficient of static friction
Tan 13 = 0.23
So,
0.23 < 0.25
So,
Man will not slide down
the object that emit light when they are hot called.......
A motorboat is a lot heavier than a pebble. Why does the boat float?
Answer:
The boat has more buoyancy
Explanation:
A spiral spring extends by 6cm when a load of mass 200g is hung on it the mass is set into small vertical oscillations Assuming that hook'es law is obeyed calculate the force constant of spring by 6cmthe period of oscillation of the mass of oscillation of the mass
Answer: To calculate the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for Hooke's Law is:
F = -kx
Where:
F is the force exerted by the spring,
k is the force constant (also known as the spring constant), and
x is the displacement from the equilibrium position.
Given:
Extension of the spring (displacement), x = 6 cm = 0.06 m
Load mass, m = 200 g = 0.2 kg
We can calculate the force exerted by the spring using the formula:
F = m * g
Where g is the acceleration due to gravity, approximately 9.8 m/s².
F = 0.2 kg * 9.8 m/s²
F = 1.96 N
Now, we can rearrange Hooke's Law to solve for the force constant, k:
k = -F / x
k = -1.96 N / 0.06 m
k ≈ -32.67 N/m
Since the spring constant (force constant) is typically represented as a positive value, we can take the absolute value of k:
k ≈ 32.67 N/m
To calculate the period of oscillation, we can use the formula:
T = 2π * √(m / k)
T = 2π * √(0.2 kg / 32.67 N/m)
T ≈ 0.713 s
So, the force constant of the spring is approximately 32.67 N/m, and the period of oscillation of the mass is approximately 0.713 seconds.
The time period of oscillation of the mass is 1.29 seconds.
A spiral spring extends by 6cm when a load of mass 200g is hung on it the mass is set into small vertical oscillations. Assuming that Hook's Law is obeyed, the force constant of spring by 6cm and the period of oscillation of the mass is given as follows:Given that:Mass of the load, m = 200g = 0.2kgExtension of the spring, x = 6cm = 0.06m
Since Hook's Law is obeyed, the force constant, k can be calculated as follows:Force applied = k × extensionx = F / k0.06 = (0.2 × 9.8) / kk = (0.2 × 9.8) / 0.06 = 32.67 N/mTherefore, the force constant of the spring is 32.67 N/m.The time period of the spring can be calculated as follows:The time period, T = 2π√(m / k)T = 2π√(0.2 / 32.67)T = 2π(0.205)T = 1.29s
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what is one reason that john adams and benjamin franklin believed the articles of confederation should be revised.
Answer:
Explanation:
Their most intimate experience probably happened during an unsuccessful peace mission in September 1776. The British forces had recently raced across Long Island (New York) and almost destroyed the American Army. The British commander, Adm. Lord Richard Howe, then offered peace. Congress sent Adams, Franklin, and Edward Rutledge (South Carolina) to meet Howe on Staten Island.
Howe hoped to resolve the differences between what Great Britain still considered its colonies and the mother country. The Americans insisted on British recognition of independence, but Howe had no such authority, and Adams and Franklin had little of their own. Although cordial, the meeting broke up without success after just three hours.
During the mission, Adams and Franklin lodged together at crowded inn in a small room with only one window. Adams records an unforgettable and amusing story in his diary about that evening and hearing Franklin’s theory of colds.
The known values are given in the table below them. Find the missing values.
The value of the missing resistance, R₃ = 10.35 ohms.
The value of the missing voltages, V₁ = 6 V, V ₃ = 24 V.
The value of the missing currents, I₁ = 3 A, I₃ = 2.32 A.
What is the value of missing component of the circuit?The values of the missing component of the circuit is calculated by applying the following formula.
The total resistance of the circuit;
For R₂, R₃, 1/R = 1/R₂ + 1/R₃
1/R = 1/12 + 1/R₃
1/R = (R₃ + 1)/(12R₃)
R = 12R₃ / (R₃ + 1)
For, R₁, R₂ and R₃, total resistance;
R = 12R₃ / (R₃ + 1) + R₁
R = [12R₃ / (R₃ + 1)] + 2
R = (12R₃ + 2(R₃ + 1) ) / (R₃ + 1)
R = (12R₃ + 2R₃ + 2 ) / (R₃ + 1)
R = (14R₃ + 2 ) / (R₃ + 1)
The total current in circuit is calculated as;
I = V/R
I = 30 / R
I = ( 30 ) / (14R₃ + 2 ) / (R₃ + 1)
I = (30R₃ + 30) / (14R₃ + 2) ------- (1)
The voltage in parallel circuit is the same
V₂ = V₃ = 24 V
V₃ = IR₃
24 = IR₃
I = 24/R₃ --------- (2)
Solve (1) and (2) together as follows;
24/R₃ = (30R₃ + 30) / (14R₃ + 2)
30R₃² - 306R - 48 = 0
Solve the quadratic equation, using formula method.
R₃ = 10.35 ohms
I₃ = V₃/R₃
I₃ = 24 V / 10.35
I₃ = 2.32 A
If the voltage drop at R₂ and R₃ = 24 V, the voltage drop at R₁ = 30V - 24 V = 6 V
The current in R₁ = V₁/R₁ = 6 V / 2 V = 3 A
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A 62 kg bike and rider accelerate from 4 m/s to 14 m/s.How much work was done by the rider? Round answerto 4 significant figures.
Given,
The mass of the bike and the rider, m=62 kg
The initial velocity of the bike and the rider, u=4 m/s
The final velocity of the bike and the rider, v=14 m/s
The work done in accelerating an object from its initial velocity to its final velocity is given by the change in its initial and final kinetic energy.
Therefore the work done by the rider is,
\(\begin{gathered} W=K_f-K_i \\ =\frac{1}{2}mv^2-\frac{1}{2}mu^2 \\ =\frac{1}{2}m(v^2-u^2) \end{gathered}\)On substituting the known values,
\(\begin{gathered} W=\frac{1}{2}\times62\times(14^2-4^2) \\ =5580\text{ J} \end{gathered}\)Therefore the work done by the rider is 5580 J
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheric pressure is Pa = 1.01 × 105 Pa. The volume of the container is V0 = 2.45 × 10-4 m3. The maximum difference between the pressure inside and outside that this particular container can withstand before bursting or imploding is ΔPmax = 2.35 × 105 Pa.
For this problem, assume that the density of air maintains a constant value of ρa = 1.20 kg / m3 and that the density of seawater maintains a constant value of ρs = 1025 kg / m3.
A) The container is taken from sea level, where the pressure of air is Pa = 1.01 × 105 Pa, to a higher altitude. What is the maximum height h in meters above the ground that the container can be lifted before bursting? Neglect the changes in temperature and acceleration due to gravity with altitude.
B)What is the maximum depth dmax in meters below the surface of the ocean that the container can be taken before imploding?
The maximum height the container can be lifted before bursting is 970 meters above sea level, and the maximum depth the container can be taken before imploding is 35 meters below the surface of the ocean.
Gauge pressure is the pressure measured relative to the atmospheric pressure at a particular location. It does not take into account the atmospheric pressure and only represents the pressure above or below the atmospheric pressure.
A) To find the maximum height h in meters above the ground that the container can be lifted before bursting, we need to find the new gauge pressure at this higher altitude. We can use the relationship between pressure and altitude:
P = P0 + ρgh
where P is the gauge pressure at the new altitude, ρ is the density of air, g is the acceleration due to gravity (assumed constant), and h is the height above sea level. Solving for h, we get:
h = (P - P0) / (ρg)
We know that the maximum pressure difference the container can withstand is ΔPmax = 2.35 × 105 Pa, so the new gauge pressure at the higher altitude can be found by adding this to the sea level pressure:
P = Pa + ΔPmax = 1.01 × 105 Pa + 2.35 × 105 Pa = 3.36 × 105 Pa
Substituting this into the equation above, along with the given values for ρ and g, we get:
h = (3.36 × 105 Pa - 2.02 × 105 Pa) / (1.20 kg/m3 × 9.81 m/s2) ≈ 970 meters
So, the maximum height the container can be lifted before bursting is approximately 970 meters above sea level.
B) To find the maximum depth dmax in meters below the surface of the ocean that the container can be taken before imploding, we need to find the new gauge pressure at this depth. We can use a similar equation to the one used above, but with the density of seawater instead of the density of air:
P = P0 + ρsgd
where g is the acceleration due to gravity (assumed constant), d is the depth below the surface of the ocean, and ρs is the density of seawater. Solving for d, we get:
d = (P - P0) / (ρsg)
We know that the maximum pressure difference the container can withstand is ΔPmax = 2.35 × 105 Pa, so the new gauge pressure at the maximum depth can be found by subtracting this from the sea level pressure:
P = P0 - ΔPmax = 2.02 × 105 Pa - 2.35 × 105 Pa = -0.33 × 105 Pa
(Note that this gives a negative value for pressure, which means the container will implode rather than burst.)
Substituting this into the equation above, along with the given values for ρs and g, we get:
d = (-0.33 × 105 Pa - 1.01 × 105 Pa) / (1025 kg/m3 × 9.81 m/s2) ≈ -35 meters
So, the maximum depth the container can be taken before imploding is approximately 35 meters below the surface of the ocean.
Therefore, The container can be lifted to a maximum height of 970 meters above sea level without bursting, and it can be submerged to a maximum depth of 35 meters without imploding.
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paleomagnetism shows at equal distances from the mor, the polarity of the earth's magnetism in the rock is the same. T/F
False, Paleomagnetism, is the study of glamorous fields recorded in jewels, deposition, or archeological accoutrements . Geophysicists who specialize in paleomagnetism are called paleomagnetists.
Certain glamorous minerals in jewels can record the direction and intensity of Earth's glamorous field at the time they formed. This record provides information on the once geste of the geomagnetic field and the once position of monumental plates. The record of geomagnetic reversals saved in stormy and sedimentary gemstone sequences provides a time- scale that's used as a geochronologic tool.
substantiation from paleomagnetism led to the reanimation of the international drift thesis and its metamorphosis into the ultramodern proposition of plate tectonics. Apparent polar wander paths handed the first clear geophysical substantiation for international drift, while marine glamorous anomalies did the same for seafloor spreading. Paleomagnetic data continues to extend the history of plate tectonics back in time, constraining the ancient position and movement of mainland and international fractions( topographies).
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Find the y-component of this
vector:
65.0⁰
15.3 m
Remember! Angles are measured from
the +x axis!
y-component (m)
Enter
The y-component of the vector of magnitude 15.3 m is 6.47 m.
What is y-component of a vector?The y-component tells us how far the vector travels in the vertical direction.
To calculate the y-component of the vector, we use the formula below.
Formula:
Dy = Dcos∅.................. Equation 1Where:
Dy = Y-component of the vectorD = Value of the given vector∅ = AngleFrom the question,
Given:
D = 15.3 m∅ = 65°Substitute these values into equation 1
Dy = 15.3×cos65°Dy = 6.47 mHence, the y-component is 6.47 m.
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Suppose an object starts from rest and achieves an operating speed of 33,500 rev/min. If it requires 1.6 s for the tool to reach operating speed and it is held at that speed for 50.0 s, how many rotations has the bit made? Suppose it requires another 10.0 s for the tool to return to rest. (Express your answer to two significant figures.)1. How many rotations does the tool complete from rest to finish?
From the information given, the object starts from rest. This means that
initial angular velocity, wo = 0
It achieves an angular speed of 33,500 rev/min. This means that
final angular velocity, wf = 33,500 rev/min
We would convert 33,500 rev/min to rev/s by dividing by 60. It becomes
558.33 rev/s
time required = 1.6
We would calculate the angular displacement, θ during this period this period by applying the formula,
θ = 1/2(wo + wf)t
θ = 1/2(0 + 558.33)1.6
θ = 446.664 rev
The object continued at this speed for 50s. This means that the angular acceleration is zero since the speed was constant. During this period,
wo = 558.33 rev/s
a = 0
t = 50
We would find the number of revolutions by applying the formula,
θ = wot + 1/2at^2
θ = 558.33 * 50 + 1/2 x 0 x 50^2
θ = 27916.5 rev
Total number of revolutions at this point is
446.664 + 27916.5 = 28363.16 revs
For the final part,
It returns to rest after 10 s. Thus,
t = 10
wf = 0
wo = 558.33
We would find the number of revolutions in this part by applying the formula
θ = 1/2(wo + wf)t
θ = 1/2(558.33 + 0)10
θ = 2791.65 rev
Total revolutions completed is
28363.164 + 2791.65 = 31154.81 revolutions
a question was asked by a teacher to a student. She gave the student a jumbled word and told him to make words out of it. The jumbled word is gzeysktqix. Now you know what to do. see ya!
When the teacher asked the student to make words out of the jumbled word gzeysktqix, the student was being tested on his ability to unscramble words. Unscrambling words is the process of taking a word or series of letters that are out of order and rearranging them to form a word that makes sense.
When trying to unscramble a word, it is important to look for any patterns that can help identify smaller words within the jumbled letters. This can help make the process easier and quicker. For example, in the jumbled word gzeysktqix, one might notice that the letters "sktqix" appear together.
This could indicate that these letters could potentially form a word. By looking at the remaining letters, one could notice that the letters "g", "z", "e", and "y" could also form smaller words. After some rearranging, the letters can be unscrambled to form the words "sky", "zig", "sex", and "yet". These are just a few examples, as there are likely many other words that can be formed from this jumbled word.
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Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .