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3 years ago

6

A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that

of the woman kw?

1 answer:

miss Akunina [59]3 years ago

4 0

Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662

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How is the direction of light changed when it travels from an optically denser medium to an optically rarer medium????? please a

ANTONII [103]

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where:

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (angle between the incoming ray and the normal to the interface)

$\theta_2$ is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

In this problem, light travels from an optically denser medium to an optically rarer medium, so

$n_1 > n_2$

Therefore, the term $\frac{n_1}{n_2}$ is greater than 1, so

$sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1$

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

4 0

3 years ago

A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

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3 years ago

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

olga nikolaevna [1]

Answer:112.82 m/s

Explanation:

Given

range of arrow=68 m

$Angle=3^{\circ}$

as the arrow travels it acquire a vertical velocity $v_y$

$v_y=u+at$

$v_y=0+9.81\times t$ -------1

Range is given by

R=ut

where u=initial velocity

$68=u\times t$

$t=\frac{68}{u}$

substitute the value of t in eqn 1

$v_y=9.81\times \frac{68}{u}$

$v_y\times u=9.81\times 68=667.08$ --------2

and $tan(3)=\frac{v_y}{u}$

$v_y=utan(3)=0.0524u$

substitute it in 2

$0.0524 u^2=667.08$

$u^2=12,728.644$

u=112.82 m/s

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Firlakuza [10]

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Gemiola [76]

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3 years ago