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3 years ago

When going from slower medium to a faster medium, light is bent _____.

Physics

2 answers:

Damm [24]3 years ago

8 0

Answer:

When going from slower medium to a faster medium, light is bent _downwards_.

Explanation:

hope this helps you!

Ad libitum [116K]3 years ago

8 0

Answer:

When going from slower medium to a faster medium, light is bent away from the normal line

Explanation:

"when light travels faster, it bends away from the normal, since it is entering a less dense medium. when light travels slower, it bends towards the normal, since it is entering a denser medium (Socratic)."

Hope this helps! Happy holidays.

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As you slide a heavy box across the floor, friction applies a force of -100 N

nirvana33 [79]

Answer:

A -500J

Explanation:

because W=Fs

100 × 5 = 500

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3 years ago

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How long would it take for a rock falling 97.2 m/s to reach the ground from 100 meters

mote1985 [20]

Answer;

velocity(v) = 97.2 m/s ,

distance (S) = 100 m

determine time(t) = ?

We know that,

distance (S) = velocity(v) × time(t)

So, time (t) = distance ÷ velocity

= 100 ÷ 97.2

<em>I hope this will help you.</em>

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4 years ago

Car A has twice the mass of Car B; both travel at the same speed. Compared to Car B. Car A has:

krek1111 [17]

Answer:

Car A has twice the energy of car B.

Explanation:

The kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2$

Where

m is the mass

v is the speed of the object

Car A has twice the mass of Car B. Both travel at the same speed i.e.

$m_A=2m_B$

So,

$\dfrac{K_A}{K_B}=\dfrac{(1/2)m_Av^2}{(1/2)m_Bv^2}\\\\=\dfrac{2m_B}{m_B}\\\\\dfrac{K_A}{K_B}=2\\\\K_A=2\times K_B$

So, the kinetic energy of car A is twice of the kinetic energy of car B.

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3 years ago

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Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

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3 years ago

Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h

Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

$\Delta U = mg \Delta h$

where

m is the man's mass

g is the gravitational acceleration

$\Delta h$ is the change in height of the man

In this problem, we have:

$\Delta U=4620 J$ is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

$\Delta h=8.4 m$ is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

$m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg$

6 0

3 years ago

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