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mr_godi [17]
3 years ago
6

A 12.0-g bullet is fired horizontally into a 104-g wooden block that is initially at rest on a frictionless horizontal surface a

nd connected to a spring having spring constant 154 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 83.0 cm, what was the speed of the bullet at impact with the block?
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

 v₀ = 292.3 m / s

Explanation:

Let's analyze the situation, on the one hand we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy, as the data they give us are Let's start with this second part.

We write the mechanical energy when the shock has passed the bodies

    Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

    Em_{f} = K_{e} = ½ k x²

    Em₀ = Em_{f}

    ½ (m + M) v² = ½ k x²

Let's calculate the system speed

    v = √ [k x² / (m + M)]

    v = √[154 0.83² / (0.012 +0.104) ]

    v = 30.24 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

    p₀ = m v₀

After the crash

   p_{f} = (m + M) v

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

   p₀ =  p_{f}

   m v₀ = (m + M) v

   v₀ = v (m + M) / m

let's calculate

   v₀ = 30.24 (0.012 +0.104) /0.012

   v₀ = 292.3 m / s

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GenaCL600 [577]

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

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e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  \mu mg  =  \frac{mv^2}{r}

    where \mu is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

6 0
3 years ago
Suppose the block is released from rest with the spring compressed 5.00 cm. The mass of the block is 1.70 kg and the force const
IRISSAK [1]

First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:

\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}

Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:

\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}

Therefore the speed is 1.026 m/s.

5 0
1 year ago
Determine the kinetic energy of a 55kg woman running with the velocity of 5.87m/s
Greeley [361]
The formula for kinetic energy = ½m·v<span>2

1/2 * 55 kg x 5,87 m/s ^2 = 27.5 x </span>34.4569 = <span>947.56475 Joule </span>≈ 948 J
4 0
3 years ago
How fast would you have to travel on the surface of earth at the equator to keep up with the sun (that is, so that the sun would
professor190 [17]

Answer: 1037 miles per hour

Explanation: In order to see the sun in the same position in the sky, you would have to travel against the speed of rotation of the earth, because this is what causes the sun to appear in a constantly changing position.

Because of this, we will have to calculate the speed of rotation of the earth. To get started, we must know the circumference of the earth. Assuming the circumference formula for a sphere,

Circumference=2\pi R

Where R is the radius of the earth, we find that the perimeter of the earth is approximately 24881 miles. The equation to calculate speed is given by

v=\frac{Distance}{Time}

Because the earth completes one rotation in 24 hours, we have to find the speed of rotation as the perimeter of the earth divided by 24 hours.

The obtained result is 1037 miles per hour.

You would have to travel at 1037 miles per hour in the direction opposite to the direction the rotation is ocurring in.

5 0
3 years ago
A 0.12 kg body undergoes simple harmonic motion of amplitude 8.5 cm and period 0.20 s. (a) What is themagnitude of the maximum f
Neporo4naja [7]

Answer:

a)F=698.83 N

b)K=8221.56 N/m

Explanation:

Given that

mass ,m = 0.12 kg

Amplitude ,A= 8.5 cm

time period ,T = 0.2 s

We know that

T=\dfrac{2\pi}{\omega}

{\omega}=\dfrac{2\pi }{0.2}\ rad/s

{\omega}=31.41\ rad/s

We know that

{\omega}^2=m\ K

K=Spring constant

K=\dfrac{\omega^2}{m}

K=\dfrac{31.41^2}{0.12}\ N/m

K=8221.56 N/m

The maximum force F

F= K A

F= 8221.56 x 0.085 N

F=698.83 N

a)F=698.83 N

b)K=8221.56 N/m

3 0
3 years ago
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