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wolverine [178]

3 years ago

11

A 5 kg block is sliding down a plane inclined at 30^0 with a constant velocity of 4 m/s. To determine the coefficient of frictio

n between the block and the plane, what pieces of information do you need?

2 answers:

Leya [2.2K]3 years ago

3 0

Answer:

The necessary information is if the forces acting on the block are in equilibrium

The coefficient of friction is 0.577

Explanation:

Where the forces acting on the object are in equilibrium, we have;

At constant velocity, the net force acting on the particle = 0

However, the frictional force is then given as

F = mg sinθ

Where:

m = Mass of the block

g = Acceleration due to gravity and

θ = Angle of inclination of the slope

F = 5×9.81×sin 30 = 24.525 N

Therefore, the coefficient of friction is given as

24.525 N = μ×m×g × cos θ = μ × 5 × 9.81 × cos 30 = μ × 42.479

μ × 42.479 N= 24.525 N

∴ μ = 24.525 N ÷ 42.479 N = 0.577

bearhunter [10]3 years ago

3 0

Answer:

We only need the angle of inclination.

The coefficient of friction is 0.58

Explanation:

Check attachment for diagram

For a book placed on an inclined plane, the force acting on the object causing it to slide downwards is the moving Force (Fm).

Fm = Wsin(theta) = mgsin(theta)

The force opposing the moving force is the frictional force Ff

Ff = nR where

n is the coefficient of friction

R is the normal reaction = mgcos(theta)

For the body to have a constant velocity (not accelerate), then Fm =Ff and W = R

n = Ff/R

n = Fm/R

n = Wsintheta/Wcostheta

n = sintheta/costheta

n = tan(theta)

Given theta = 30°

n = tan 30°

n = 0.58

the coefficient of friction between the block and the plane is 0.58

The calculation shows that the only information we need to get the coefficient of friction is the angle is inclination

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Question 11 of 11 | Page 11 of 11

KiRa [710]

Answer:

Decreases the time period of revolution

Explanation:

The time period of Cygnus X-1 orbiting a massive star is 5.6 days.

The orbital velocity of a planet is given by the formula,

v = √[GM/(R + h)]

In the case of rotational motion, v = (R +h)ω

ω = √[GM/(R + h)] /(R +h)

Where 'ω' is the angular velocity of the planet

The time period of rotational motion is,

T = 2π/ω

By substitution,

<em>T = 2π(R +h)√[(R + h)/GM] </em>

Hence, from the above equation, if the mass of the star is greater, the gravitational force between them is greater. This would reduce the time period of revolution of the planet.

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A student conducted an experiment to measure the acceleration due to gravity 'g' of a simple pendulum.

valina [46]

I think i used calulater and it gives me 47.5

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3 years ago

The diagram shows the flow of water through a hydroelectric power station. The falling water

KonstantinChe [14]

Answer:

KE = $\frac{1}{2}$ m $v^{2}$

Explanation:

In the generation of energy from hydroelectric power station, the motion of water, and the turbines are paramount. The falling flowing water turns the blades of the turbine, which in-turn causes the movement of a coil within a strong magnetic field.

The motion of the coil which cuts the strong magnetic field induces current. Thus, the system generates electrical energy.

The equation that links kinetic energy (KE), mass (m) and speed (v) can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

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In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.

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