How far would a horse that rides at 5.5 m/s travel in 6.3 minutes?

A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter f

V125BC [204]

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

<u>So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment. </u>

So, option B is the answer

4 0

3 years ago

The slow, steady downhill flow of loose, weathered Earth materials is called A. flow. B. slide. C. creep. D. slump.

Art [367]

I beileve its b
Hope this helps☺

7 0

3 years ago

Read 2 more answers

A body-centered cubic lattice has a lattice constant of 4.83 Ă. A plane cutting the lattice has intercepts of 9.66 Å, 19.32 Å, a

anastassius [24]

Answer:

Miller Indices are [2, 4, 3]

Solution:

As per the question:

Lattice Constant, C = $4.83 \AA$

Intercepts along the three axes:

$\bar{x} = 9.66 \AA$

$\bar{x} = 19.32 \AA$

$\bar{x} = 14.49 \AA$

Now,

Miller Indices gives the vector representation of the atomic plane orientation in the lattice and are found by taking the reciprocal of the intercepts.

Now, for the Miller Indices along the three axes:

a = $\frac{1}{9.66}$

b = $\frac{1}{19.32}$

c = $\frac{1}{14.49}$

To find the Miller indices, we divide a, b and c by reciprocal of lattice constant 'C' respectively:

a' = $\frac{\frac{1}{9.66}}{\frac{1}{4.83}} = \frac{1}{2}$

b' = $\frac{\frac{1}{19.32}}{\frac{1}{4.83}} = \frac{1}{4}$

c' = $\frac{\frac{1}{14.49}}{\frac{1}{4.83}} = \frac{1}{3}$

7 0

3 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

3 years ago

How much energy is required to ionize hydrogen when it is in the n = 5 state?

PtichkaEL [24]

Answer:

0.544 eV

Explanation: