1. An object, constrained to move along the x-axis is acted upon by a force F(x) where = a = 5 N/m, b = -2N/m F(x) = ax + bx² The object is observed to proceed directly from x = 1m to x = 3.0m. How much work was done by the object by the force? Does the process of integration take into account the fact that the force F(x) changes sign in the interval.
Answers
The change in the object's kinetic energy is given by:
ΔK = Kf - Ki
where Kf is the final kinetic energy and Ki is the initial kinetic energy.
Since the object starts from rest, its initial kinetic energy is zero. The final kinetic energy can be found using the work-energy principle:
W = ΔK
where W is the work done by the force.
The work done by the force is equal to the integral of the force over the displacement:
W = ∫ F(x) dx
W = ∫ (ax + bx^2) dx
W = (1/2)ax^2 + (1/3)bx^3 + C
where C is the constant of integration.
We can evaluate the constant of integration by using the given initial and final values of x:
W = (1/2)a(3.0)^2 + (1/3)b(3.0)^3 - (1/2)a(1)^2 - (1/3)b(1)^3
W = 22.5 - 1.0
W = 21.5 J
Therefore, the work done by the force is 21.5 J.
When we integrate the force F(x) over the interval from x = 1m to x = 3.0m, the fact that the force changes sign in the interval is automatically taken into account. This is because the integral of the negative part of the force cancels out the integral of the positive part of the force, resulting in the net work done by the force.
Related Questions
A platinum resistance thermometer has resistances of 210.0 Ω when placed in a 0°C ice bath and 237.8 Ω when immersed in a crucible containing a melting substance. What is the melting point of this substance? (Hint: First determine the resistance of the platinum resistance thermometer at room temperature, 20.0°C.)
Answers
The resistance of a platinum resistance thermometer is 200 Ω at 0°C and 255.8 Ω at the melting point. Using the resistance-temperature relationship and calculations, the estimated melting point is approximately 19.93°C.
The resistance of a platinum resistance thermometer is 200 Ω when placed in a 0°C ice bath and 255.8 Ω when immersed in a crucible containing a melting substance. To determine the melting point of the substance, we need to calculate the temperature at which the resistance reaches 255.8 Ω.
First, we find the temperature coefficient of resistance (α) using the formula α = (R - R₀) / (R₀ * T), where R is the resistance at the melting point, R₀ is the resistance at 0°C, and T is the temperature at the melting point.
Substituting the given values, we have α = (255.8 - 200) / (200 * T₀), where T₀ is the known room temperature of 20°C.
Calculating α, we find α ≈ 0.014.
Next, we use the resistance-temperature relationship equation R = R₀(1 + αT) to solve for the melting point temperature (T). Substituting the known values, we have 255.8 = 200(1 + 0.014 * T).
Simplifying the equation, we find 1.279 = 1 + 0.014T.
Solving for T, we get T ≈ 19.93°C.
Therefore, based on the given data, the estimated melting point of the substance is approximately 19.93°C.
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A wave has a speed of 360 m/s. It has a frequency of 20hz what is its wavelength (include correct unit)
Answers
Answer:
18m
Explanation:
v=frequency × wavelenght
wavelength=v/f
wavelength=360÷20
=18m
Answer:
18m
Explanation:
wave speed / frequency = wavelength
360 / 20 = 18
Question 5 (1 point)
You push a 34.2 kg box with a force to 97.7 N. At what rate will the box accelerate?
Your Answer:
Answers
Answer:
2.86 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
\(a = \frac{force}{mass} \\ \)
From the question
force = 97.7 N
mass = 34.2 kg
We have
\(a = \frac{97.7}{34.2} = 2.856725 \\ \)
We have the final answer as
2.86 m/s²Hope this helps you
When exercising in the heat, you should __________.
A.
wear tight-fitting clothes to absorb your sweat
B.
dress in layers
C.
choose dark colored clothing
D.
vary the intensity and duration of the exercise
Please select the best answer from the choices provided.
Answers
When exercising in the heat, you should wear tight - fitting clothes to absorb your sweat, therefore the correct answer is option A.
What is exercise?Exercise is a physical activity that involves bodily movement that improves a person's health, fitness, and well-being while lowering their chance of contracting illnesses.
Dressing in layers is useful in extreme cold conditions and extremely cold weather.
Dark-colored clothing absorbs more heat as compared to light - colored clothing.
When exercising in the heat, you should wear tight-fitting clothes to absorb your sweat, therefore the correct answer is option A.
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A rifle can shoot a 4.00 g bullet at a speed of 998 m/s. Find the kinetic energy of the bullet. What work is done on the bullet if it starts from rest?
Answers
Answer:
1992.008J
Explanation:
a roller coaster start at a height of 40Meters and reached a height of 20meter. does mechanical energy change
Answers
Mechanical energy changes when a roller coaster starts at a height of 40 meters and reaches a height of 20 meters. The potential energy decreases, while the kinetic energy increases.
When a roller coaster starts at a height of 40 meters and reaches a height of 20 meters, mechanical energy changes. In physics, mechanical energy is the sum of potential and kinetic energy that is present in the objects. When an object is moved, it gains or loses energy, thus the mechanical energy changes. There are two forms of mechanical energy, namely kinetic energy and potential energy. Kinetic energy is the energy that a moving object possesses due to its motion, while potential energy is the energy that an object possesses due to its position or shape.
In the case of a roller coaster, when it starts at a height of 40 meters, it has potential energy that is equal to its mass multiplied by the acceleration due to gravity multiplied by its height. As it moves down the track, the potential energy gets converted into kinetic energy, which is the energy of motion. When the roller coaster reaches a height of 20 meters, it has a lower potential energy compared to when it started. The difference in potential energy is equal to the amount of work done by the force of gravity in bringing the roller coaster down from a height of 40 meters to a height of 20 meters. At the same time, the roller coaster has a higher kinetic energy than when it started, as it gained speed during the descent.
Therefore, in summary, mechanical energy changes when a roller coaster starts at a height of 40 meters and reaches a height of 20 meters. The potential energy decreases, while the kinetic energy increases.
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Trina launches a water balloon at a speed of 11.9 meters per second, in the positive x-direction, at an angle of 24.9° above horizontal. What are the components of the velocity vector right before the balloon strikes the ground?
Select one:
a. {v_{x}} = 10.8 m/s, {v_{y}} = -5.01 m/s
b.{v_{x}} = 5.27 m/s, {v_{y}} = -10.8 m/s
c. {v_{x}} = 10.8 m/s, {v_{y}} = 5.01 m/s
d.{v_{x}} = -5.27 m/s {v_{y}} = 10.8 m/s
Answers
Answer:
a. {v_{x}} = 10.8 m/s, {v_{y}} = -5.01 m/s
Explanation:
The initial velocity components are
Horizontal 11.9cos24.9 = 10.8 m/s
Vertical 11.9sin24.9 = 5.01 m/s (where up is assumed positive)
In the absence of air resistance, the horizontal velocity will remain unchanged and, assuming the ground is horizontal, the vertical velocity will have completely reversed directions.
Find the torque t due to the spring. Assume that theta is small enough that the spring remains effectively horizontal and you can approximate sin(theta) = theta and cos(theta) =1.
Express the torque as a function of theta and other parameters of the problem. In this context, the torque will be a 1D vector; therefore, your equation must correctly express the relationship between the direction of torque and the direction of ant other 1D vectors within your equation.
Hints:
Deflecting the rod will stretch or compress the spring by a length L. The spring will react with a restoring force given by Hooke's law: F=-kL. What is L? Remember that the angle theta is assumed to be so small that sin(theta) = theta. express in terms of L and theta.
The torque τ about a point is defined as the product of the force F acting on a body times the moment arm (perpendicular distance d from the line of action of the force to the center point): T = Fd. What is d for the given situation? Remember that the angle θ is assumed to be so small that cos(θ)≈1.
Answers
Therefore, the frequency of oscillation when the spring is connected 1/5 of the way from the pivot to the end of the rod is approximately 1.34 Hz.
How to solveSince the rod is thin and uniform, its moment of inertia about the pivot point can be approximated as:
I = (1/3)ML^2
When the spring is connected 1/5 of the way from the pivot to the end of the rod, the effective length of the rod becomes:
l_eff = l/5 + (4/5)(l/2) = 9l/10
So, the frequency of oscillation is: 8.42 rad (after calculations)
The frequency of oscillation when the spring is connected 1/5 of the way from the pivot to the end of the rod is approximately 1.34 Hz.
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A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
Answers
The final speed (in meters per second), given that the coefficient of friction between her skis and the snow is 0.38 is 14.65 m/s
How do I determine the final velocity?First, we shall obtain the force. This can be obatined as follow:
Mass (m) = 60 KgAcceleration due to gravity (g) = 9.8 m/s²Normal reaction (N) = mg = 60 × 9.8 = 588 NCoefficient of kinetic friction (μK) = 0.38Force (F) = ?F = μKN
F = 0.38 × 588
F = 223.44 N
Next, we shall obtain the acceleration of the skier. This can be obtained as follow:
Mass (m) = 60Force (F) = 223.44 NAcceleration (a) = ?a = F / m
a = 223.44 / 61
a = 3.724 m/s²
Finally, we shall determine the final velocity. This can be obtained as follow:
Initial speed (u) = 14 m/sDistance (s) = 2.5 mDeceleration (a) = 3.724 m/s²Final speed (v) =?v² = u² + 2as
v² = 14² + (2 × 3.724 × 50)
v² = 196 + 18.62
v² = 214.62
Take the square root of both sides
v = √214.62
v = 14.65 m/s
Thus, the the final speed is 14.65 m/s
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The final speed right at the top will be 14.65 m/s and the coefficient of friction between her skis and the snow is 0.38.
What is velocity?The ratio between displacement to time is referred to as the velocity of the object. It is a vector quantity with SI unit meter per second (m/s).
According to the question, the given information is :
Skier's mass, m = 60 kg
The acceleration due to gravity, g = 9.8 m/s²
Initial Speed, u = 14 m/s
Distance, s = 2.5 m
Deceleration is given as, a = 3.724 m/s².
Normal reaction force (N) will be equal to mg.
N = 60 × 9.8
N = 588 N
The coefficient of kinetic friction is given as, μ = 0.38
By using the formula,
F = μKN
F = 0.38 × 588
F = 223.44 N
Now, let's calculate the acceleration of the skier :
f = ma
a = f/m
a = 223.44/60
a = 3.724 m/s².
Now, calculate the velocity,
v² = u² + 2as
v² = 14² + 2 (3.724) (50)
v² = 196 + 18.62
v² = 214.62
v = 14.65 m/s.
Hence, the final speed right at the top will be 14.65 m/s.
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If 62.9 cm of copper wire (diameter = 1.15 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of 8.43 mT/s, at what rate is thermal energy generated in the loop?
Answers
Answer:
The answer is "\(\bold{7.30 \times 10^{-6}}\)"
Explanation:
length of the copper wire:
L= 62.9 cm
r is the radius of the loop then:
\(r=\frac{L}{2 \pi}\\\)
\(=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\\)
area of the loop Is:
\(A_L= \pi r^2\)
\(=100.2001\times 3.14\\\\=314.628\)
change in magnetic field is:
\(=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}\)
then the induced emf is: \(e = A_L \times \frac{dB}{dt}\)
\(=314.628 \times 0.01\\\\=3.14\times 10^{-5}V\)
resistivity of the copper wire is: \(\rho =\) 1.69 × 10-8Ω·m
diameter d = 1.15mm
radius (r) = 0.5mm
\(= 0.5 \times 10^{-3} \ m\)
hence the resistance of the wire is:
\(R=\frac{\rho L}{\pi r^2}\\\)
\(=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\\)
Power:
\(P=\frac{e^2}{R}\)
\(=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}\)
The final answer is: \(\boxed{7.30 \times 10^{-6} \ W}\)
Can someone help me please
Answers
Given,
The distance of the race, d=100 m
The extended distance of the race, s=200 m
The velocity of the red car throughout the race, u₁=10 m/s
The initial velocity of the blue car, u₂=0 m/s
The blue car gains a velocity of 2 m/s every second.The constant acceleration of the blue car is
hus t
\(\begin{gathered} a=\frac{2}{1} \\ =2\text{ m/s}^2 \end{gathered}\)As the red car maintains the same velocity, the speed when it reaches the finish line will be 10 m/s.The time it takes for the red car to reach the finish line when the race was 100 m
\((t_1)_{100}=\frac{d}{u_1}\)On substituting the known values,
\(\begin{gathered} (t_1)_{100}=\frac{100}{10} \\ =10\text{ s} \end{gathered}\)The time it takes for the red car to reach the finish line after the race is extended,
\((t_1)_{200}=\frac{s}{u_1}\)On substituting the known values,
\(\begin{gathered} (t_1)_{200}=\frac{200}{10} \\ =20\text{ s} \end{gathered}\)From the equation of motion,
The final velocity of the blue car, when it reaches the finish line of 100 m race is given by the equation of motion,
\(v^2_{100}=u^2_2+2ad\)Where v₁₀O is the final velocity of the blue car at the end of the 100 m race.n substituting the known values,₀
\(\begin{gathered} v^2_{^{}100}=0+2\times2\times100 \\ =400 \\ v=\sqrt[]{400} \\ =20\text{ m/s} \end{gathered}\)hus the speed of the blue car when it reaches the finish line of 100 m race is 20 m/s
he time it takes for the blue car to reach the end of the 100 m race is given by,
\(v_{100}=u_2+a(t_2)_{100}\)Where (t₂)₁₀₀ is the time it takes for the blue car to reach the end of the 100 m race.
On substituting the known values in the above equation,
\(\begin{gathered} 20=0+2(t_2)_{100} \\ \Rightarrow(t_2)_{100}=\frac{20}{2} \\ =10\text{ s} \end{gathered}\)hus both cars take 10 s to reach the end of the 100 m race. Thus they both reach the finitsh line together.
The time it takes for the blue car to reach the end of the 200 m race can be calculated using the equation,
\(s=u_2(t_2)_{200}+\frac{1}{2}a\lbrack(t_2)_{200}\rbrack^2\)Where (t₂)₂₀₀ is the time it takes for the blue car to reach the end of the 200 m race.
On substituting the known values,
\(\begin{gathered} 200=0+\frac{1}{2}\times2\times\lbrack(t_2)_{200}\rbrack^2 \\ \lbrack(t_2)_{200}\rbrack^2=200 \\ \Rightarrow(t_2)_{200}=\sqrt[]{200} \\ =14.14\text{ s} \end{gathered}\)hus while rthe ed car takes 200 s to reach the finish line of the 200 m race, the blue car takes 14.14 s.
Therefore, if the race was extended, the blue car will reach the finish line first.
B. Calculate the total resistance of the circuit below. (4 points)
c. In the circuit diagram above, meters 1 and 2 are connected as shown. Write 2 - 3 sentences identifying each type of meter and how it is connected with the 30.0 Ω resistor in the circuit. (4 points)
d. In the circuit diagram above, predict which resistors (if any) will stop working when the switch is opened. Write 2 - 3 sentences explaining your reasoning. (4 points)
Answers
B. The equivalent resistance of the two resistors is 20.0 ohms.
C. The voltmeter will measure the voltage across the 30.0 ohm resistor.
D. The 30.0 ohm resistor will stop working.
How to determine resistance?B. The total resistance of the circuit is 60.0 ohms. This is because the 30.0 ohm resistor and the 60.0 ohm resistor are in parallel, and the equivalent resistance of two resistors in parallel is equal to the product of the resistors divided by the sum of the resistors.
R_T = 1/(1/R_1 + 1/R_2 + ...)
In this case, the product of the resistors is:
30.0 ohms × 60.0 ohms = 1800 ohms,
and the sum of the resistors is:
30.0 ohms + 60.0 ohms = 90.0 ohms.
Therefore, the equivalent resistance of the two resistors is 1800 ohms / 90.0 ohms = 20.0 ohms.
C. Meter 1 is an ammeter, and it is connected in series with the 30.0 ohm resistor. This means that the ammeter will measure the current flowing through the 30.0 ohm resistor.
Meter 2 is a voltmeter, and it is connected in parallel with the 30.0 ohm resistor. This means that the voltmeter will measure the voltage across the 30.0 ohm resistor.
D. When the switch is opened, the 30.0 ohm resistor will stop working. This is because the switch is in series with the 30.0 ohm resistor, and when the switch is opened, the circuit is broken.
The 60.0 ohm resistor will continue to work, because it is in parallel with the switch, and the current will continue to flow through the 60.0 ohm resistor even when the switch is opened.
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A student uses a motion detector to study the kinematics of a block suspended on a vertical spring that obeys Hooke’s law. The student pulls the block a distance y from equilibrium, releases it from rest, and records the speed v of the block as it passes the equilibrium position. The student repeats this process several times for different values of y. Which variables should be plotted on the horizontal and vertical axes to yield a linear graph?
Answers
When the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
Hooke's lawHooke's law states that force or load applied to elastic material is directly proportional to the extension produced in the material. This law can be written as;
F = kx
where;
F is the applied forcek is spring constantx is extension of the materialVelocity of the blockThe speed of the block increases with decreasing displacement of the bock. This is because the kinetic energy of the block is maximum at zero displacement and minimum at maximum displacement of the block.
Thus, when the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
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consider the combination of two capacitor c1 and c2 with c2>c1,when connected in parallel,the equivalent capacitance is 8 times the equivalent capacitance of the same connected in series. calculate the ratio of capacitor c2/c1?
Answers
Explanation:
Current flowing through the circuit, I = 2A Applying Kirchoff’s Ist law at junction P, I = I1 + I2 I2 = I – I1 …(1) Applying Kirchoff’s Ind law at junction PQSP 5 I1 + 10 Ig – 15 I2 = 0 5 I1 + 10 Ig -15(I – I1) = 0 20 I1 + 10 Ig = 15 I 20 I1 + 10 Ig = 15 x 2 ÷ by 10 21 I1 + Ig = 3 … (2) Applying Kirchoff’s II law at junction QRSQ 10(I1 – Ig ) – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10g – 20(I – I1 – Ig ) – 10 Ig = 0 10 I1 – 10 – 20 I + 20 I1 -20Ig – 10 Ig = 0 30 I1 – 40 Ig = 20 I ÷ by 10 ⇒ 3 I1 – 4 Ig = 20 I 3 I1 – 4 Ig = 2 I 3 I1 – 4 Ig = 2 x 2
hope this helps :)
HELP PLEASE THIS IS URGENT!!!
Answers
The neutron number of an atom X, which undergoes alpha, and beta decay reduces the neutron number by 6.
Alpha decay is the nuclear process in which the parent nucleus emits an alpha or helium particle to form a daughter nucleus. When a particle emits an alpha nucleus, the nucleus loses its two protons and two neutrons. Beta decay is the nuclear process in which the parent nucleus undergoes the emission of electrons to produce a daughter nucleus.
Alpha decay decreases the atomic mass number decreases by 4 and the atomic number decreases by 2. In beta decay, the neutron is converted into a proton and the atomic number decreases by one. The neutron number is affected by alpha decay.
From the given,
X atom undergoes alpha decay. X -----> ₐ₋₂Xᵇ⁻⁴ + He₂⁴. The neutron number decreases by two. ₐ₋₂Xᵇ⁻⁴ -----> ₐ₋₂₋₂Xᵇ⁻⁴⁻⁴ + He₂⁴. The neutron number decreases by two.
When the X atom undergoes beta decay, ₐ₋₄Xᵇ⁻⁸---> ₐ₋₅Xᵇ⁻⁸ + ₋₁e⁰. The neutron number does not get affected. When the atom again undergoes alpha decay, ₐ₋₅Xᵇ⁻⁸ -----> ₐ₋₇Xᵇ⁻¹². Thus, the neutron number decreases by 6 when the atom undergoes three alpha decay.
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2 What distance will a sprinter travel in 4.0 s if his or her
acceleration is 2.5 m/s²? Assume the sprinter starts
from rest.
A 13 m
B 20 m
C 80 m
D 90 m
Answers
The distance travelled by sprinter in 4 sec is 20 m
It is given that
acceleration of sprinter = 2.5 m/s²
time taken by sprinter = 4.0 s
initial speed of sprinter = 0 m/s
We have to find distance travelled by sprinter
We will use the following equation of motion
S = ut + 1/2at^2
S = 1/2at^2
S = 1/2 x 2.5 x (4)^2
S = 1/2 x 2.5 x 16
S = 20 m
Hence, distance travelled by sprinter in 4 sec is 20 m
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A bicycle possesses 1000 units of momentum. what would be the bicycle's momentum if,
A.its velocity is doubled
B. its mass is tripled
Answers
what is the value of pi(8.104)^2 written with correct significant numbers
Answers
Answer:206.3
Explanation:
If a 93000 kg truck collides with a 60 kg car
Answers
Cs-124 has a half-life of 30.8 s
A) If we have 7.5 μg initially, how many Cs nuclei are present?
B) How many nuclei are present 2.6 min later?
Answers
A) There are 3.70 x 10¹⁶ Cs nuclei present initially and B) There are 2.27 x 10¹⁵ Cs nuclei present 2.6 min later.
A) To solve this problem, we can use the following equation,
\(N = N'2^{\frac{-t}{T_{1/2}}}\)), number of Cs nuclei present is N, initial number of Cs nuclei present is N', elapsed time is t, half-life of Cs-124 is T½. First, we need to convert the initial mass of Cs-124 to the number of nuclei present,
7.5 μg Cs-124(1g/10⁶μg)(6.022x10²³nuclei/1g)
= 4.52 x 10¹⁵ Cs-124 nuclei.
Using the equation above, we can find the number of Cs-124 nuclei present after 0 s,
\(N = 4.52 * 10^{15} * 2^{\frac{-0}{30.8}}}\)
= 4.52 x 10¹⁵ Cs-124 nuclei
Therefore, initially, there are 4.52 x 10¹⁵ Cs-124 nuclei present.
B) After 2.6 min (156 s), we can again use the same radioactivity equation to find the number of Cs-124 nuclei present,
\(N = 4.52 * 10^{15} * 2^{\frac{-156}{30.8}}}\)
N = 1.60 x 10¹⁴ Cs-124 nuclei
Therefore, 2.6 min later, there are 1.60 x 10¹⁴ Cs-124 nuclei present.
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The change in motion (acceleration) of an object depends on
The size of the force
The mass of the object
BOTH the size of the force AND the mass of the object
Answers
Answer:
BOTH the size of the force AND the mass of the object
Explanation:
Acceleration of an object is the rate of change of its velocity.
The relation between force, mass and acceleration is given by the formula as follows :
F = ma
m is mass
a is acceleration
It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).
Answer5 ms 3. A football player has a mass of 95 kg, and he is running with a velocity of 15 m/s. What is his momentum? Answer:
Answers
Answer:
1425kgm/s
Explanation:
Given parameters:
Mass = 95kg
Velocity = 15m/s
Unknown:
Momentum = ?
Solution:
The momentum of a body is the amount of motion it posses;
Momentum = mass x velocity
Insert the parameters and solve;
Momentum = 95 x 15 = 1425kgm/s
How would you describe the mass and size the players need to have to play the game?
Answers
Answer:
mass and size are the main important to any playing person for more the two extremely don't Matcha the person doesn't play any exercise
In Figure 4.27 four particles form a square of edge length
a=5cm and have charges q1=+10nc, q2=-20nc,q3 = +20nc, q4=-10.0nc. In unit-rector notation, what net
electric field do the particles produce at the centre of square?
Answers
Answer:
so sorry
don't know but please mark me as brainliest please
How far does the object travel between 5 and 10 s
Answers
The distance travelled by the object between 5 and 10 seconds is 25 m.
What is the distance travelled by the object?The area under velocity time graph gives the distance travelled by the object.
The distance travelled by the object between 5 and 10 seconds is calculated from the area of curve within this time range.
Area of the curve between 5 and 10 seconds = Area of triangle
Distance travelled between 5 and 10 seconds = Area of triangle formed between 5 and 10 seconds
Area of triangle = ¹/₂bh
where;
b is the base of the triangleh is the height of the triangleA = ¹/₂ x (10 - 5 ) x ( 10 )
A = 25 m
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100 POINTS AND BRAINLIEST! What were the Magdeburg Hemispheres?
Answers
Answer:
Magdeburg hemispheres are two half-spheres of equal size. Placing them together traps air between them. This air is merely trapped, and not compressed, so the pressure inside is the same as the pressure of the atmosphere outside the spheres. The spheres thus pull apart with nearly no resistance.
The Magdeburg hemispheres are a pair of large copper hemispheres, with mating rims. They were used to demonstrate the power of atmospheric pressure. When the rims were sealed with grease and the air was pumped out, the sphere contained a vacuum and could not be pulled apart by teams of horses.
Ebo throws the ball again at an initial velocity of 18 m/s but this time at an angle of 53 to the ground. a) Work out the velocity in component vector form. b) Why will the ball spend a longer time in the air than it did before. c) Calculate the range of the ball.
Answers
(a) The velocity in component vector form is ( 10.8 i + 14.4 j ) m/s.
(b) The ball spent longer time in air than it did before because of the angle of projection.
(c) The range of the ball is determined as 31.8 m.
What is the velocity in component vector form?
The velocity in component vector form is calculated as follows;
Vy = V sinθ
Vx = V cosθ
where;
V is the initial velocityθ is the angle of projectionVy = ( 18 m/s ) sin(53) = 14.4 j
Vx = ( 18 m/s ) cos(53) = 10.8 i
The time of motion of the ball is calculated as follows;
T = 2u sinθ / g
T = ( 2 x 18 sin53 ) / (9.8)
T = 2.93 seconds
The range of the projectile is calculated as follows;
R = u² sin (2θ) / g
R = (18² sin (2 x 53) ) / 9.8
R = 31.8 m
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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
. The figure illustrates the configuration.
A system of two blocks connected by a rope passing over a pulley. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle. Box M subscript 1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M sub 1 and the surface is mu subscript 1. Box M subscript 2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle. The coefficient of friction between box M sub 2 and the surface is mu subscript 2.
Answers
The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.
The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.
The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.
The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.
The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).
The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).
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Find the minimum mass required to prevent a meter stick from tipping when the overhang location is at 39 can and then mass of the weight is 150 grams. The mass of the meter stick is 80 grams and the center of mass location on the meter stick is 50.3 cm.
Answers
Answer:
1.03 kg.
Explanation:
To prevent the meter stick from tipping when the weight is placed at 39 cm, we need to ensure that the torque due to the weight is balanced by the torque due to the mass of the meter stick itself.
The torque due to the weight is:
τ_w = F_w * r_w
where F_w is the force due to the weight (which is equal to the weight of the mass, W = m*g), and r_w is the distance between the weight and the fulcrum (which is 39 cm).
Plugging in the given values, we get:
τ_w = (0.150 kg * 9.81 m/s^2) * 0.39 m
τ_w = 0.571 J
The torque due to the mass of the meter stick is:
τ_m = m_m * g * r_m
where m_m is the mass of the meter stick (which is 0.080 kg), g is the acceleration due to gravity (which is 9.81 m/s^2), and r_m is the distance between the center of mass of the meter stick and the fulcrum (which is 50.3 cm).
Plugging in the given values, we get:
τ_m = (0.080 kg * 9.81 m/s^2) * 0.503 m
τ_m = 0.397 J
To prevent the meter stick from tipping, the torques due to the weight and the meter stick must be equal, so we have:
τ_w = τ_m
Solving for the required mass, we get:
m = (τ_m / r_w) / g
Plugging in the values for τ_m, r_w, and g, we get:
m = (0.397 J / 0.39 m) / 9.81 m/s^2
m ≈ 1.03 kg
Therefore, the minimum mass required to prevent the meter stick from tipping when the weight is placed at 39 cm is approximately 1.03 kg.
What are the real life applications of Hooke's Law?
Answers
Answer:
1) a rubber band
2) the spring of retractable pen
3) a spring loaded toy gun
Explanation:
Hooke's law states that; provided the elastic limit of a material is not exceeded, the force exerted on an elastic material is directly proportional to its extension. This relationship was first captured by Robert Hooke in 1660 when he asserted that 'as the extension, so is the force!'.
Hooke's law generally deals with elastic or stretchable materials. These materials can be deformed, but returned to their original shapes when the deforming force is removed. This deforming force causes an extension in the material which is directly proportional to the deforming force. That is F= Kx where K is the called the force constant, F is the deforming force and x is the magnitude of extension brought about by the force.
Various real life applications of Hooke's law have been listed in the answer. Any material that makes use of a loaded spiral spring or indeed any kind of elastic material obeys Hooke's law.
Explanation:
☞Hooks law is applied in:
☆Rubber bands
☆Fridge covers
☆Springs