1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Snowcat [4.5K]
3 years ago
8

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

3.8-kg shell horizontally with an initial velocity of 503 m/s. Suppose the cannon is then unbolted from the earth, and no external force hinders its recoil. What would be the velocity of a shell fired from this loose cannon
Physics
1 answer:
Nata [24]3 years ago
3 0

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

v_1 = \frac{m_2v_2}{m_1}

Apply the principle of conservation kinetic energy

K = \frac{1}{2}m_1v_1^2 +  \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\

v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14  \ m/s

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

You might be interested in
This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag
Dmitry [639]

Answer:

Part a)

F = 135.7 N

Part b)

F = 62.5 N

Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg + F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg + \mu(Fsin\theta)

F(cos\theta - \mu sin\theta) = mg

F = \frac{mg}{cos\theta - \mu sin\theta}

F = \frac{55}{cos50 - 0.310(sin50)}

F = 135.7 N

Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

Fcos\theta = mg - F_f

Fsin\theta = F_n

so we have

Fcos\theta = mg - \mu(Fsin\theta)

F(cos\theta + \mu sin\theta) = mg

F = \frac{mg}{cos\theta + \mu sin\theta}

F = \frac{55}{cos50 + 0.310(sin50)}

F = 62.5 N

6 0
3 years ago
A 70kg man runs at a pace of 4 m/s and a 50g meteor travels at 2 km/s. Which has the most kinetic energy?.
kirill115 [55]

Answer:

the 70kg man

Explanation:

because he has more weight and is moving faster

4 0
3 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
2 years ago
Energy is defined as
vodka [1.7K]
I believe that your answer is going to be C. The ability to do work
3 0
3 years ago
Read 2 more answers
Other questions:
  • Which of the following is a ferromagnetic material
    13·2 answers
  • In order to get the bike to move and stay moving, what friction must be overcome
    9·1 answer
  • a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
    9·1 answer
  • What is the energy in an atoms nucleus
    8·1 answer
  • to travel at a constant velocity, a car exerts force that is 551N to balance air resistance. how much work does the car do on th
    9·1 answer
  • How much charge passes through a wire in 4.0 s if the current is 3.0 A?
    6·1 answer
  • A ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40
    6·1 answer
  • An induced charge is when charged particles can be attracted to neutral objects.<br> True or false?
    5·1 answer
  • How are volcanoes formed at subduction zones
    6·2 answers
  • A 100 N force is applied to a box at an angle of 60° to the
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!