What are the differences between refraction and diffraction?

A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff

igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

Acceleration, a= 32 ft/s^2

initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

substituting the appropriate values,

(88)^2 - 0 = 2*32* h

h=(88)^2 / 64

h= 121 ft

hence

the height (in feet) of the cliff is 121 ft

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3 0

11 months ago

A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri

Alla [95]

Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

$q=1\times10^{-8}C$

Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

7 0

2 years ago

Skater begins to spend with arms held out at shoulder height. The skater wants to match the speed of the spin to the beat of the

Aleksandr [31]

Answer:

the moment of inertia with the arms extended is Io and when the arms are lowered the moment

I₀/I > 1 ⇒ w > w₀

Explanation:

The angular momentum is conserved if the external torques in the system are zero, this is achieved because the friction with the ice is very small,

L₀ = L_f

I₀ w₀ = I w

w = $\frac{I_o}{I}$ w₀

where we see that the angular velocity changes according to the relation of the angular moments, if we approximate the body as a cylinder with two point charges, weight of the arms

I₀ = I_cylinder + 2 m r²

where r is the distance from the center of mass of the arms to the axis of rotation, the moment of inertia of the cylinder does not change, therefore changing the distance of the arms changes the moment of inertia.

If we say that the moment of inertia with the arms extended is Io and when the arms are lowered the moment will be

I <I₀

I₀/I > 1 ⇒ w > w₀

therefore the angular velocity (rotations) must increase

in this way the skater can adjust his spin speed to the musician.

7 0

2 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

2 years ago

Read 2 more answers

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

8 0

2 years ago