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daser333
3 months ago
15

The muzzle velocity of a gun is the velocity of the bullet when it leaves the barrel. The muzzle velocity of one rifle with a sh

ort barrel is greater than the muzzle velocity of another rifle that has a longer barrel. In which rifle is the acceleration of the bullet larger
Physics
1 answer:
Keith_Richards [3.2K]3 months ago
7 0

Answer: small barrel gun

Explanation:

It is noted that short barrel guns have a higher muzzle velocity for bullets compared to longer barrel guns.

Acceleration is determined by the change in velocity with respect to time.

a=\dfrac{\Delta v}{\Delta t}

For short barrel guns, the bullet reaches its muzzle velocity more quickly, leading to greater acceleration than that of bullets from long barrel guns.

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Molybdenum (Mo) has a BCC crystal structure, an atomic radius of 0.1363 nm, and an atomic weight of 95.94 g/mol. Compute and com
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Answer:

The book is titled Solid State or Condensed Matter

Explanation:

8 0
2 months ago
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Draw the vector C⃗ =1.5A⃗ −3B⃗ . The length and orientation of the vector will be graded. The location of the vector is not impo
Yuliya22 [3333]
I created the illustration found in the accompanying file.

There are two images included.

The upper one illustrates the impacts of:

- scaling vector A by a factor of 1.5, depicted in red with a dashed line.

- scaling vector B by -3, shown in purple with a dashed line.

The lower image displays the resultant vector: C = 1.5A - 3B.

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Next, an arrow is drawn from the tail of 1.5A to the position of -3B after this shift.

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3 months ago
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two kittens are on opposite sides of a field, 250 m apart. kitten the runs at a constant speed of 25 m/s due east on a collision
ValentinkaMS [3465]

Set the initial location of kitten A on the left side of the field (designated as point A) at the origin, running east which is the positive direction. Kitten B starts at position {x_B}_0=250\,\mathrm m, while kitten A’s beginning spot is {x_A}_0=0\,\mathrm m.

Kitten A moves with a velocity of v_A=25\,\dfrac{\mathrm m}{\mathrm s}, and kitten B with v_B=-12\,\dfrac{\mathrm m}{\mathrm s}. Their positions over time are described by

x_A=\left(25\,\dfrac{\mathrm m}{\mathrm s}\right)t

x_B=250\,\mathrm m+\left(-12\,\dfrac{\mathrm m}{\mathrm s}\right)t

The collision occurs when the positions are the same, i.e. when x_A=x_B. Solving this gives

\left(25\,\dfrac{\mathrm m}{\mathrm s}\right)t=250\,\mathrm m+\left(-12\,\dfrac{\mathrm m}{\mathrm s}\right)t

\implies\left(37\,\dfrac{\mathrm m}{\mathrm s}\right)t=250\,\mathrm m

\implies t=\dfrac{250\,\mathrm m}{37\,\frac{\mathrm m}{\mathrm s}}=6.76\,\mathrm s

Which results in approximately 6.8 seconds, considering significant figures.

3 0
3 months ago
At the end of the school day, at exactly 2:30 pm, a group of students run out of the school building and reach the edge of the s
Yuliya22 [3333]

Explanation:

The initial time is t₁ = 2:30 pm

The final time is t₂ = 2:30:45

We must analyze the students' motion concerning time. The final time exceeds the initial time by 45 seconds.

Change in time,

\Delta t=t_2-t_1\\\\\Delta t=2:30:45-2.30\\\\\Delta t=45\ s

Therefore, this is the solution we need.

3 0
2 months ago
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