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daser333
11 days 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 [1K]11 days 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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2 days ago
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If the coefficient of static friction between a table and a uniform massive rope is μs, what fraction of the rope can hang over
inna [987]
Here's the procedure explained: Assume F represents the portion of the rope that is extending over the table. In this scenario, the frictional force that holds the rope on the table can be calculated using the formula: Ff = u*(1-f)*m*g. Additionally, it is important to determine the gravitational force that attempts to pull the rope off the table, Fg, calculated through: Fg = f*m*g. You then need to set these two equations equal to each other and resolve for f: f*m*g = u*(1-f)*m*g leads to f = u*(1-f) = u - uf. Simplifying gives f + uf = u, which results in f = u/(1+u) representing the fraction of the rope. This will lead you to the final answer.
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1 day ago
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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
Yuliya22 [1153]

Answer:

The typical weight of a human heart is approximately 0.93 lbs.

Explanation:

Based on this,

the heart's weight constitutes about 0.5% of total body mass.

Total human weight = 185 lbs

Let the entire body weight be represented as w and the heart's weight as w_{h}.

We aim to determine the heart's weight for a human

Using the provided information

w_{h}=0.5\times w

Where, h = heart weight

w = human weight

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

The final weight of a human heart is 0.93 lbs.

8 0
10 days ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [942]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

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7 days ago
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The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring
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Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

E=PE_{2}-PE_{1}

E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2

Substitute the values into the formula

E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2

E=0.0072\ J

E=7.2\ mJ

Therefore, a total of 7.2 mJ is needed to extend the pen.

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11 days ago
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