Which one is not a fundamental quantity ?Mass Weight LengthTime

(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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To find the mass in kilograms, you need to know the object's weight in newtons and the acceleration due to gravity. The formula for finding mass is mass = weight / acceleration due to gravity. So if you have an object with a weight of 100 N and the acceleration due to gravity is 9.8 m/s^2, the mass would be 10.204 kg.

The mass of the block is 0.025 kg or 25 g, when the spring has k = 28 N/m, and compresses 0.11 m before bringing the block to rest.

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Answer:

Explanation:

We shall apply Wien's displacement law which is as follows .

λ T = b where λ is wavelength of light that is coming out of hot body to maximum extent .

Putting the value of temperature given and b

λ x 500 = 2898 μmK

λ = 5.796 μm = 5796 nm

For temp 1050 K

λ = 2760 nm

For temp 1800 K

λ = 1610  nm

For temp 2500 K

λ = 1159.2  nm

The visible range starts from 740 nm .

Hence we can expect that some amount of visible light may emerge at the temperature of 2500K because the wavelength that we have calculated above gives the value of peak wavelength of a spectrum of light coming out of hot body .  

Answer:

gang

Explanation:

yupppp

ANSWER

False.

EXPLANATION

Nuclear binding energy is the energy that is required to split the nucleus (of an atom) into nucleons: protons and neutrons.

The binding energy of a nucleus is always positive because nuclei require energy to separate them and it is impossible for nuclei to gain energy by being separated.

Therefore, the answer is false.

To solve the given problems, we'll use the principles of work-energy and conservation of energy. Let's address each question one by one:

(1) What is the work done by Chris on the box when the speed of the box reaches 2.1 m/s?

The work done by Chris on the box is equal to the change in the box's kinetic energy. Since the box starts from rest, the initial kinetic energy is zero. The final kinetic energy can be calculated using the formula:

Kinetic energy = (1/2) * mass * velocity^2

Plugging in the values:

Mass of the box (m) = 53 kg

Final velocity (v) = 2.1 m/s

Kinetic energy = (1/2) * 53 kg * (2.1 m/s)^2

Calculate the value of the kinetic energy, which represents the work done by Chris on the box.

(2) What is the speed of the box when it reaches the bottom of the slope (Point B in the diagram)?

To determine the speed at the bottom of the slope, we'll use the principle of conservation of energy. The total mechanical energy of the box is conserved as it moves from the top to the bottom of the slope.

The initial potential energy at the top of the slope is converted into kinetic energy at the bottom of the slope, neglecting any energy losses due to friction.

Potential energy at the top = m * g * h1

Where:

Mass of the box (m) = 53 kg

Acceleration due to gravity (g) = 10 m/s^2

Height difference between floors (h1) = 3.2 m

Calculate the initial potential energy.

The final kinetic energy at the bottom is given by:

Kinetic energy at the bottom = (1/2) * m * v^2

Where:

Mass of the box (m) = 53 kg

Velocity at the bottom (v) = ?

Equating the initial potential energy to the final kinetic energy, solve for v to find the speed of the box at the bottom of the slope.

(3) To what speed does the box slow down when it reaches the bottom of the ramp to the pickup truck?

Since the ramp connecting the first floor to the bed of the pickup truck is frictionless, there is no external force doing work on the box. Thus, the mechanical energy of the box is conserved as it moves from the bottom of the slope to the bottom of the ramp.

Using the same principle of conservation of energy, equate the final kinetic energy at the bottom of the slope to the initial potential energy at the bottom of the ramp.

Potential energy at the bottom of the ramp = m * g * h2

Where:

Mass of the box (m) = 53 kg

Acceleration due to gravity (g) = 10 m/s^2

Height difference between the first floor and the truck bed (h2) = 0.90 m

Calculate the potential energy at the bottom of the ramp.

Equating the potential energy at the bottom of the ramp to the final kinetic energy, solve for the speed of the box at the bottom of the ramp.

(4) What is the speed of the box when it reaches the bed of the pickup truck?

Since the ramp connecting the first floor to the truck bed is frictionless, there is no external force doing work on the box. The mechanical energy of the box is conserved as it moves from the bottom of the ramp to the truck bed.

Using the same principle of conservation of energy, equate the final potential energy at the bottom of the ramp to the final kinetic energy at the truck bed.

Potential energy at the truck bed = m * g * h

Answer:

Explanation:

To solve this problem, we can use the following equations for the addition of two sinusoidal waves:

y1 = A1 sin(ωt + φ1)

y2 = A2 sin(ωt + φ2)

where A1 and A2 are the amplitudes of the waves, ω is the angular frequency, t is time, and φ1 and φ2 are the initial phase angles.

(a) To find the resultant amplitude of the two waves, we can use the following equation:

Ar = √(A1^2 + A2^2 + 2A1A2cos(φ2 - φ1))

where Ar is the resultant amplitude.

Substituting the given values, we get:

Ar = √((3.0 × 10^(-6))^2 + (4.0 × 10^(-6))^2 + 2(3.0 × 10^(-6))(4.0 × 10^(-6))cos(150° - 60°))

Ar ≈ 5.03 × 10^(-6) m

Therefore, the resultant amplitude is approximately 5.03 × 10^(-6) m.

(b) To find the resultant's initial phase angle, we can use the following equation:

tan(φr) = (A1sin(φ1) + A2sin(φ2))/(A1cos(φ1) + A2cos(φ2))

where φr is the initial phase angle of the resultant wave.

Substituting the given values, we get:

tan(φr) = (3.0 × 10^(-6)sin(60°) + 4.0 × 10^(-6)sin(150°))/(3.0 × 10^(-6)cos(60°) + 4.0 × 10^(-6)cos(150°))

φr ≈ 142.85°

Therefore, the resultant's initial phase angle is approximately 142.85°.

(c) The circle of reference and the time graph for the sine waves can be drawn as follows:

Sine Waves

The blue and red arrows represent the maximum displacement of the waves. The black arrow represents the displacement of the resultant wave. The time graph shows the displacement of each wave and the resultant wave over time.

Answer:

Object A runs into Object B. The force that object A has will transfer from Object A to Object B. But during this transfer you will loose some energy, therefore object B will not travel as fast.

Explanation:

The moment of inertia of the uniform cylindrical grinding wheel is 2,150 kgm².

What is the moment of inertia?

This refers to the angular mass or rotational inertia can be defined with respect to the rotation axis, as a property that shows the amount of torque needed for a desired angular acceleration or a property of a body due to which it resists angular acceleration. The unit is kgm².

From the question:

Mass,M =4.2kg

Radius, R=32Cm

The formula for calculating the moment of inertia for uniform cylindrical grinding wheel:

moment of inertia, I =1/2MR²

I =\(\frac{1}{2}\) * 4.2 * 32²

  =2,150.4 kgm²

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Tension in the left cable is 4395.9 N.

Tension is defined as a pulling force that acts along the length of a flexible medium like rope or cables.

Here,

The total mass of the riders is given, m = 330 kg

Let the tension in the left cable be T₁ and that in the right cable be T₂.

From the figure,

T₁ cos 15 = T₂ cos 26

T₁ = T₂ cos 26/cos 15

Also,

T₁ sin15 + T₂ sin 26 = mg

Substituting values,

(T₂ cos 26/cos 15) sin 15 + T₂ sin 26 = 330x 9.8

0.241 T₂ + 0.438 T₂ = 3234

0.679 T₂ = 3234

T₂ = 4762.8 N

Therefore, Tension in the left cable, T₁ = 0.930x 4762.8

         T₁ = 4395.9 N

Hence, The tension in the left cable is 4395.9 N

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Your question was incomplete. Attaching the image here.

Answer:

Acceleration is 2.55m/s²

Explanation:

Force(F)=mass of object(m) ×acceleration(a)

in the scenario above,a force of 15N is used to push the box and because the floor is not having a smooth surface,frictional force(8N) will oppose the 15N used to push the box thereby decreasing the acceleration. so the actual force moving the box is 7N.

Force =Normal Force - Frictional Force

F= 15N - 8N

F= 7N

therefore the 7N is the actual force causing the box to move.

To find the acceleration,insert the mass of the box and the actual force into the formula (F=m×a)

F=m×a

7N =2.75kg × a

make a the subject by dividing both sides of the equation by 2.75kg.

7/2.75 = (2.75×a)/2.75

a= 7/2.75

therefore a= 2.55m/s²,where a is the Acceleration.

1.2 g/cm³  is the density of the cheese

density= mass/volume

mass=120 g

volume=100cm³

density=120 g/=100cm³

density=1.2 g/cm³

The mass of a material substance per unit volume is known as density. The equation for density is d = M/V, where d represents density, M is mass, and V is volume. The standard unit of measurement for density is grams per cubic centimeter. For instance, water has a density of 1 grams per cubic centimeter, compared to Earth's density of 5.51 grams. The kilograms per cubic meter unit is another way to express density (in meter-kilogram-second or SI units)

Density is a simple way to discover the relationship between mass and volume. For instance, M = Vd is the formula for calculating a body's mass, while V = M/d is the formula for calculating a body's volume.

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Answer:

\(6.5\; {\rm N}\).

Explanation:

When an object travel at a speed of \(v\) in a circle of radius \(r\), the (centripetal) acceleration of that object would be \(a = (v^{2} / r)\).

In this question, the ball is travelling at \(v = 3\; {\rm m\cdot s^{-1}}\) in a circle of radius \(r = 1.8\; {\rm m}\). The (centripetal) acceleration of this ball would be:

\(\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}\).

By Newton's Laws of Motion, for an object of mass \(m\), if the acceleration of that object is \(a\), the net force on that object would be \(m\, a\). Since the acceleration of this ball is \(a = 5\; {\rm m\cdot s^{-2}}\), the net force on this ball would be:

\(\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}\).

The radial acceleration formula is

Where r = 0.870 m and v = 7.85 m/s. Let's replace these values and solve

4.44×\(10^{9}\) kg/s is the rate at which the sun mass is decreasing.

The Sun radiates energy through a process called nuclear fusion, where hydrogen atoms combine to form helium, releasing a tremendous amount of energy in the process. According to Einstein's mass-energy equivalence principle (E=mc²), this energy release corresponds to a decrease in mass.

To calculate the rate at which the Sun's mass is decreasing, we can use the formula ΔE = Δmc², where ΔE is the change in energy, Δm is the change in mass, and c is the speed of light.

Given that the Sun radiates energy at a rate of 4×10^26 W, we can substitute this value into the equation as ΔE and solve for Δm.

ΔE = 4×10^26 W

c = 3×10^8 m/s (speed of light)

Using the equation ΔE = Δmc² and rearranging it, we get Δm = ΔE / c².

Substituting the values, we have:

Δm = (4×10^26 W) / (3×10^8 m/s)²

Evaluating this expression, we find that the rate at which the Sun's mass is decreasing is approximately 4.44×10^9 kg/s.

This calculation demonstrates that the Sun's mass is gradually decreasing as it continuously radiates energy into space, primarily through the process of nuclear fusion in its core.

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Answer:

It is showing the wavelength.

Explanation: Hope it helps you:)))

have a good day

Answer:

0

Explanation:

Physically:

Beginning point means initial point so you're basically going into circles which means a displacement of zero.

Mathematically:

\(d = x _{f} - x _{i}\)

We know

\(x _{f} =x _{i}\)

So we get

\(d = x _{f} - x _{f}\)

\(d = 0\)

In A. part, the magnetic field at (0, 50) is in west direction. In B. part, the strength of the field at (0,50) is 2.06 G.

A. The current is flowing up for west as shown by the front view figure at the bottom of the gadget. Your fingers will curve to the west if you wrap your right hand around the wire with your thumb up. Put a compass at (0,50) to check the direction as well. It indicates west.

B. By using the Pythagorean Theorem to determine the strength of the combined field, the strength of the field at (0,50) is 2.06 G.

The earth's magnetic field strength= 0.50 G

The induced current magnetic field strength= 2.0

B is given by=

\(\sqrt{0.50^{2} - 2.00^{2} }\\ =\sqrt{0.25-4.00}\\ =2.06\)

Hence, we can also check by putting the probe on (0,50) and the probe reads 2.06 G.

Therefore, the strength of the field at (0,50) is 2.06 G.

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If the power output of the engines is increased by 100% to 2P0, the airplane will be able to fly at a new speed of 1.26\(v_{0}\).

As, P = F v

So, P = kv³

Then, v/v0 = \(\sqrt[3]{P/P0}\).

Therefore, after substituting the values, the speed will be 1.26\(v_{0}\).

What is power?

In physics, power is the amount of energy transferred or converted per unit of time. In the International System of Units, the unit of power is the watt, which is equivalent to 1 joule per second. In older works, force is sometimes called activity. Power is a scalar quantity.

Power is related to other quantities. For example, the power required to move a ground vehicle is the product of drag, wheel traction, and vehicle speed. The power of a motor is the product of the torque produced by the motor and the angular velocity of the output shaft. Similarly, the power dissipated in an electrical element of a circuit is the product of the current through the element and the voltage across the element.

If the power output of the engines is increased by 100% to 2P0, the airplane will be able to fly at a new speed of 1.26\(v_{0}\).

As, P = F v

So, P = kv³

Then, v/v0 = \(\sqrt[3]{P/P0}\).

Therefore, after substituting the values, the speed will be 1.26\(v_{0}\).

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The magnitude of OC is approximately 5.8 units, and the angle between the x-axis and vector OC is approximately 59.0 degrees

To evaluate the magnitude of vector OC, which is represented by vector C = (-3, -5), we can use the formula for the magnitude of a vector. The magnitude of a vector (denoted as ||v||) can be calculated using the Pythagorean theorem.

Magnitude of OC = ||C|| = √((-3)^2 + (-5)^2) = √(9 + 25) = √34 ≈ 5.8

To find the angle between the x-axis and vector OC, we can use trigonometry. The angle can be determined by finding the arctan of the y-component divided by the x-component of vector C.

Angle = arctan(-5 / -3) = arctan(5/3) ≈ 59.0 degrees

Therefore, the magnitude of OC is approximately 5.8 units, and the angle between the x-axis and vector OC is approximately 59.0 degrees (rounded to the nearest tenth).

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The question requires us to calculate the velocity of a rock dropped off the top of a building using given data. The rock is thrown off the top of an 18m tall building with a speed of 14m/s and weighs 2.5kg. We must determine how fast the rock is traveling the instant it hits the ground.

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Answer:

Water

\( \dashrightarrow \: { \sf{H _{2}O}}\)

Explanation:

By observation at the image attached, each hydrogen atom has one electron, so in total the two hydrogen atoms contribute two electrons.

Then the oxygen atom has six (6) electrons, (3 lone pairs)

So, by covalent bonding, the two electrons are shared from hydrogen to Oxygen two make oxygen stable.

Since stable oxygen atom must have 8 electrons, so the 2 electrons from hydrogen move to the oxygen-6 to make oxygen-8.

As a result, a polar compound called Water is formed

Pack B has twice the mass of pack A, and thus twice the kinetic energy, but has the same final velocity as pack A.

a. The ultimate kinetic energy of an object is given by the formula KE = 1/2 × m × v². where m is the object's mass and v is its velocity. Since both pucks are pulled at the same distance by a string of the same tension, the final kinetic energy of both pucks will be the same. However, pack B has twice the mass of pack A, so the final kinetic energy is twice that of pack A. The final velocity of the

b. Object is given by the equation v = √(2 × KE / m). Both packs have the same final kinetic energy, but pack B has twice the mass of pack A, so the final velocity is √(2 × KE / mB) = √(2 × KE / (2 × mA)) = √(KE / mA ), which is the square root of the terminal velocity of pack A.

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The question is -

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by strings with the same tension a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.

Answer: the radius of the circular path is approximately 1637.58 meters.

Explanation:

The centripetal force acting on the airplane is provided by the component of the gravitational force that acts towards the center of the circular path. This component is given by:

F_c = m * g * tan(banking angle)

Where:

F_c is the centripetal force

m is the mass of the airplane

g is the acceleration due to gravity

tan(banking angle) is the tangent of the banking angle

Now, the centripetal force is also given by the formula:

F_c = (m * v^2) / r

Where:

v is the speed of the airplane

r is the radius of the circular path

Equating the two expressions for F_c, we get:

(m * g * tan(banking angle)) = (m * v^2) / r

Canceling out the mass (m) on both sides of the equation, we have:

g * tan(banking angle) = v^2 / r

Solving for r, we get:

r = (v^2) / (g * tan(banking angle))

Substituting the given values:

v = 400 km/h = 400,000 m/h

g = 9.8 m/s^2

banking angle = 20°

Converting the speed to m/s:

v = 400,000 m/h * (1/3600) h/s = 111.11 m/s

Converting the banking angle to radians:

banking angle = 20° * (π/180) rad/° = 0.3491 rad

Now, substituting the values into the formula:

r = (111.11^2) / (9.8 * tan(0.3491))

r ≈ 1637.58 meters

Therefore, the radius of the circular path is approximately 1637.58 meters.

The total distance covered by the car is 300 miles.

The total displacement covered by the car is zero.

The average speed of the car is 17.88 m/s.

The average velocity of the car is also zero.

Distance between the points A and B, d = 150 miles

Time taken by the car to travel from A to B, t₁ = 3 hours

Time taken by the car to travel from B to A, t₂ = 5 hours

a) Given that the car travelled from A to B and then back to A.

Therefore, the total distance covered by the car is,

Distance = 2 x d

Distance = 2 x 150

Distance = 300 miles

Since the car is travelling from A to B and then returning back to the initial point A, the total displacement covered by the car is zero.

b) The speed with which the car travelled from A to B is,

v₁ = d/t₁

v₁ = 150/3

v₁ = 50 miles/hr

v₁ = 22.35 m/s

The speed with which the car travelled from B to A is,

v₂ = d/t₂

v₂ = 150/5

v₂ = 30 miles/hr

v₂ = 13.41 m/s

Therefore, the average speed of the car is,

v = (v₁ + v₂)/2

v = (22.35 + 13.41)/2

v = 17.88 m/s

As, the total displacement of the car is zero, the average velocity of the car is also zero.

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The runner who reaches their maximum speed more quickly has the larger maximum speed. In a 100-meter dash, reaching maximum speed more quickly generally indicates a higher level of acceleration, which is related to maximum speed.

The runner who reaches their maximum speed more slowly may have a longer time to build up speed, but once both runners have reached their maximum speed, they are both running at the same speed.

So, the runner who reached maximum speed more quickly will have had a higher maximum speed.

Speed is known as the rate of change of position of an object in any direction. Speed is calculated as the ratio of distance to the time in which the distance was covered.

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Answer:

Ultrasound

Explanation:

ultra = above     ultrasound = above hearing sound