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3 years ago

A 50 kg boy runs and jumps with a forward velocity of 1.5 m/s into a 125 kg stationary boat.

Physics

2 answers:

Dafna11 [192]3 years ago

8 0

Answer:

0.43 m/s, forward

Explanation:

i just took the test

Vesnalui [34]3 years ago

4 0

Answer:

A. 0.43 m/s, forward

Explanation:

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Goryan [66]

Answer:

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7 0

2 years ago

Read 2 more answers

A woman does 260 J of work to move a table 1.9 m across the floor. What is the magnitude of the force applied by the woman to th

balandron [24]

Answer:136.8N

Explanation:

Work=260joules

Distance=1.9m force=?

Force=work/distance

Force=260/1.9

Force=136.8N

4 0

3 years ago

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach

Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

3 0

2 years ago

Read 2 more answers

A cyclotron designed to accelerate protons has a magnetic field of magnitude 0.15 T over a region of radius 7.4 m. The charge on

klio [65]

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, $q=1.60218\times 10^{-19}\ C$

Mass of a proton, $m=1.67262 \times 10^{-27}\ kg$

The cyclotron frequency is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}$

f = 2286785.40 Hz

$\omega=14368296.44\ rad/s$

$\omega=1.43\times 10^7 rad/s$

Hence, this is the required solution.

8 0

2 years ago

A beam of light strikes one face of a window pane with an angle of incidence of 30.0°. The index of refraction of the glass is 1

natka813 [3]

Answer:

(a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

Explanation:

Given that,

Angle of incidence = 30.0°

Index of reflection of glass = 1.52

(a). We need to calculate the angle of refraction for the ray inside the glass

Using snell's law

$\dfrac{\sin i}{\sin r}=\mu$

$\sin r=\dfrac{\sin i}{\mu}$

Put the value into the formula

$\sin r=\dfrac{\sin 30}{1.52}$

$r=\sin^{-1}(0.329)$

$r=19.26^{\circ}$

(b). We know that,

The incident ray and emerging ray is equal then the ray will be parallel.

We need to prove that the rays in air on either side of the glass are parallel to each other

Using formula for emerging ray

$\dfrac{\sin e}{\sin r}=\mu$

$\sin e=\sin r\times \mu$

Put the value into the formula

$\sin e=0.3289\times 1.52$

$e=\sin^{-1}(0.499)$

$e=29.9\approx 30^{\circ}$

So, $\sin i=\sin e$

This is proved.

Hence, (a). The angle of refraction is 19.26°.

(b). That is proved that the rays in air on either side of the glass are parallel to each other

6 0

3 years ago