A ____________ is the measure of how far the particles in the medium move away from their normal rest position.

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and t

andrew11 [14]

Answer:

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f = $\frac{r}{2}$

f = $\frac{0.40}{2} = 0.20 m$

Now,

m = - $\frac{v}{u}$

- 2 = - $\frac{v}{u}$

$\frac{v}{u}$ = 2 (2)

Now, by lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

v = $\frac{uf}{u - f}$ (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u = $\frac{uf}{u - f}$

2 = $\frac{f}{u - f}$

2(u - 0.20) = 0.20

u = 0.30 m

6 0

3 years ago

A 250KG ROCKET HAS 1,953,125 JOULES OF ENERGY. HOW FAST IS IT<br>MOVING?

UNO [17]

Answer:

cnbv gncvjhmv

Explanation:

6 0

4 years ago

A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag

Kobotan [32]

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5 0

3 years ago

a 0.199 kg snowball moving west makes an inelastic collision with a 2.89 kg box moving 0.523 m/s west. afterward,they move west

kogti [31]

Answer:

The initial velocity of the snowball was 22.21 m/s

Explanation:

Since the collision is inelastic, only momentum is conserved. And since the snowball and the box move together after the collision, they have the same final velocity.

Let $m_1$ be the mass of the ball, and $v_1$ be its initial velocity; let $m_2$ be the mass of the box, and $v_2$ be its velocity; let $v_f$ be the final velocity after the collision, then according to the law of conservation of momentum:

$m_1v_1+m_2v_2=v_f(m_1+m_2)$ .

From this we solve for $v_1$ , the initial velocity of the snowball:

$\boxed{v_1=\frac{v_f(m_1+m_2)-m_2v_2}{m_1}}$

now we plug in the numerical values $m_1=0.199\:kg$ , $m_2=2.89\:kg$ , $v_2=0.523\:m/s$ , and $v_f=1.92\:m/s$ to get:

$v_1=\frac{1.92*(0.199+2.89)-2.89*0.523}{0.199}$

$\boxed{v_1=22.21\:m/s}$

The initial velocity of the snowball is 22.21 m/s.

<em>P.S: we did not take vectors into account because everything is moving in one direction—towards the west.</em>

4 0

3 years ago

How can you use the position-time graphs for two in-line skaters to determine if and when one in-line skater will pass the other

Jet001 [13]

Explanation:

Position-time graphs measure/express the position of a skater over time relative to the start or finish of the race (depends on how it is used). Note: are the skaters in line vertically or horizontally? Like is one directly behind the other or are they next to each other?

If the two skaters are in line horizontally with each other, then their position will be the same relative to the start or finish of the race. This means if one passes the other one, the position would be different for all times after they pass. On the graph, it would look like one single line at the start (as position is same) which splits into 2 (representing the new difference in position due to 1 passing the other.

If the two skaters are in line vertically, their lines on the graph will appear parallel to each other (assuming they are going same speed) because the position is changing at the same rate, one is just reaching the same point after the other. If the skater behind overtakes the one in front. The lines on the graph will cross and continue either in parallel but with the other line on top to represent the moment where their position is the same right before they pass and after, where the second skater is now in front.

Hope this helped!

8 0

3 years ago