A vector is 9.55 m long and

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

3 years ago

The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I

jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0

2 years ago

.In a bizarre but harmless accident, Superman fell from the top of the Eiffel Tower. How fast was Superman traveling when he hit

ELEN [110]

Answer:

(7.8) x (9.8 m/s) = 76.44 m/s

during the time he spent falling.

Since his falling speed was zero when he 'stepped' off of the top,

he hit the ground at 76.44 m/s.

That's about 170 miles per hour.

I'll bet he left one serious crater!

I hope this helps too! :D

Explanation:

4 0

3 years ago

An experiment discovered that a particular material was a poor conductor of electricity. This material would most likely be used

LenaWriter [7]

Someone’s mindset has a powerful influence on his/her perspective because a mindset __________.
A.involves the way he/she is used to thinking about things B.can easily be changedC.is not affected by his/her valuesD.cannot be influenced by what happened to someone in the past

8 0

3 years ago

Read 2 more answers

A spring hangs at rest from a support. If you suspend a 0.46 kg mass from the spring it elongates 0.79m.

Fudgin [204]

a. The restoring force in the spring has magnitude

F[spring] = k (0.79 m)

which counters the weight of the mass,

F[weight] = (0.46 kg) g = 4.508 N

so that by Newton's second law,

F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m

b. Using the same equation as before, we now have

F[weight] = (0.75 kg) g = 7.35 N

so that

(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m

4 0

2 years ago