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liberstina [14]
4 years ago
7

What is the coefficient of kinetic friction between the ramp and the couch?

Physics
1 answer:
Cloud [144]4 years ago
4 0

Answer:

maybe c. because if you do 70

Explanation:

-50 you get 20 than you add that to the 25 and you get 45 and the nearest closest number is 47

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In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
4 years ago
What is the chemical name for lime
Sergio039 [100]
Calcium Oxide
CaO I think
5 0
3 years ago
Which of the following statements are evidence that gases do not always behave ideally? Check all that apply.?
yaroslaw [1]

Your answer will be A

Hope this helps you out a little


8 0
3 years ago
Read 2 more answers
Record breaking low temperatures are associated with which air mass?
Papessa [141]

Answer:

Arctic Air masses

Explanation:

The Arctic Air masses are the world's most cold air masses. Specifically they are called Continental Arctic. Extremely cold and with very little moisture(extremely dry). These originate from north arctic region of 24 hour darkness that make the wind extremely cold and dry. The temperature in these regions are well below -50°C.

8 0
3 years ago
A boy walks a distance of 100 meters to the right at a steady speed of 1.00 m/s. Then he stops for 30 seconds. Then returns back
Mamont248 [21]

Explanation:

speed=\frac{Distance}{time}

In the figure attached:

A boy travels AB distance of 100 m, with speed of 1.00 m/s

AB = 100 m

1.00 m/s=\frac{100 m}{t_1}

t_1=100 s

After reaching at point B he stops there at for 30 seconds.

t_2= 30 s

After , 30 seconds he he comes back to his initial position that is A with steady speed of 1.00 m/s.

Distance covered from B to A= 100 m

Time taken by him during coming back=t_3

1.00=\frac{100 m}{t_3}

t_3=100 s

After returning to to point A he turns left and moves towards point C with speed of 1.5 m/s for 2 minutes.

Distance of AC = ?

t_4=2 min= 120 s

1.5 m/s=\frac{AC}{120 s}

AC = 180 m

The total time of his round trip is:T

T=t_1+t_2+t_3+t_4=100 s+30 s(stop)+ 100 s(returning)+120 s=350 s

The total distance: D

D = AB + BA (returning) + AC=100 m + 100 m + 180 m = 380 m[/tex]

The total displacement of boy:

Displacement is the shortest distance between the initial point and final point.

He first walked to point B from A and then came back to A . And after that walking to point C from A.So, the final displacement (d) is from A to C.

d = AC = 180 m

The total displacement of boy is 180 m.

The average speed of the boy is given by:

=\frac{AB+BA+AC}{t_1+t_2+t_3+t_4}=\frac{D}{T}

\frac{D}{T}=\frac{380 m}{350 s}=1.085 m/s

The average velocity of the boy is given by:

=\frac{AC}{t_1+t_2+t_3+t_4}=\frac{d}{T}

\frac{d}{T}=\frac{180 m}{350 s}=0.5142 m/s

5 0
4 years ago
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