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3 years ago

The strong nuclear force felt by a single proton in a large nucleus

1 answer:

Crank3 years ago

4 0

Your question: The strong nuclear force felt by a single proton in a large nucleus _______________________.

Answer: is about the same as that felt by a single proton in a small nucleus.

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13 points and brainlyest if possible. Thanks.

nikdorinn [45]

Most likely it would be C not completely sure

3 0

3 years ago

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If a topographic map included a 6,000 ft. mountain next to an area of low hills, which would best describe the contour lines on

VMariaS [17]

The answer to the given question above would be option B. If a topographic map included a 6,000 ft. mountain next to an area of low hills, the statement that best describe the contour lines on the map is this: <span>The contour lines around the mountain would be very close together. Hope this helps.</span>

8 0

3 years ago

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A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?

amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

3 0

3 years ago

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Dimensional Analysis : 3 days to seconds

LuckyWell [14K]

Hi,

To convert 3 days to seconds write this.

1h = 3600s
24h = 3600 · 24 = 86400
3 days = 3 · 86400 = 259200sec

Hope this helps.
r3t40

8 0

3 years ago

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n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pend

Kitty [74]

Answer:

68cm

Explanation:

You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get

$p_f=p_i\\mv_1+Mv_2=(m+M)v$

m: mass of the bullet

M: mass of the pendulum

v1: velocity of the bullet = 410m/s

v2: velocity of the pendulum =0m/s

v: velocity of both bullet ad pendulum joint

By replacing you can find v:

$(0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}$

this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:

$E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2$

g: 9.8/s^2

h: height

By doing h the subject of the equation and replacing you obtain:

$(0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m$

hence, the heigth is 68cm

4 0

3 years ago