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Sindrei [870]
3 years ago
5

If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s than the recoil speed of the cannon will be *

Physics
1 answer:
Allushta [10]3 years ago
4 0

Answer:

D. 10 m/s

Explanation:

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Which subatomic particle has the smallest mass?
spin [16.1K]

Answer:

quark

Explanation:

neutrón= 1.675 x 10-27 kg

proton=1,673 x 10-27 kg

electron=9,11 x 10-31 kg

quark=7,13 - 14,26 · 10−30 kg

3 0
2 years ago
A certain moving electron has a kinetic energy of 0.991 × 10−19 J. Calculate the speed necessary for the electron to have this e
alisha [4.7K]

Answer: The speed necessary for the electron to have this energy is 466462 m/s

Explanation:

Kinetic energy is the energy posessed by an object by virtue of its motion.

K.E=\frac{1mv^2}{2}

K.E= kinetic energy = 0.991\times 10^{-19}J

m= mass of an electron = 9.109\times 10^{-31}kg

v= velocity of object = ?

Putting in the values in the equation:

0.991\times 10^{-19}J=\frac{1\times 9.109\times 10^{-31}kg\times v^2}{2}

v=466462m/s

The speed necessary for the electron to have this energy is 466462 m/s

4 0
3 years ago
Read 2 more answers
Suppose that the height of the incline is h = 14.7 m. Find the speed at the bottom for each of the following objects. 1.solid sp
tensa zangetsu [6.8K]

Answer:

1. 14.4 m/s  2. 13.2 m/s 3. 12.0 m/s 4. 13.9 m/s

Explanation:  

Assuming no friction present, the different objects roll without slipping, so there is a constant relationship between linear and angular velocity, as follows:

ω= v/r

If no friction exists, the change in total kinetic energy must be equal in magnitude to the change in the gravitational potential energy:

∆K = -∆U

 ½ *m*v² + ½* I* ω²  = m*g*h

Simplifying and replacing the value of the angular velocity:

½ * v² + ½ I *(v/r)² = g*h (1)

In order to answer the question, we just need to replace h by the value given, and I (moment of inertia) for the value for each different object, as follows:

  •  Solid Sphere I = 2/5* m *r²

                Replacing in (1):

                ½ * v² + ½ (2/5 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(10/7*9.8 m/s2*14.7 m)  = 14. 4 m/s

  • Spherical shell I=2/3*m*r²

                Replacing in (1):

                ½ * v² + ½ (2/3 *m*r²) *(v/r)² = g*h

                Replacing by the value given for h, and solving for v:

                v = √(6/5*9.8 m/s2*14.7 m)  = 13.2 m/s

  • Hoop   I= m*r²

                Replacing in (1):

                ½ * v² + ½ (m*r²) *(v/r)² = g*h

               Replacing by the value given for h, and solving for v:

               v = √(9.8 m/s2*14.7 m)  = 12.0 m/s

  • Cylinder I = 1/2 * m* r²

                 Replacing in (1):

                ½ * v² + ½ (1/2 *m*r²) *(v/r)²= g*h

                 Replacing by the value given for h, and solving for v:

                v = 2*√(1/3*9.8 m/s2*14.7 m)  = 13.9 m/s

5 0
3 years ago
What are some examples of gravitational forces that keep an object in orbit? Pls give at least 2-3 example if you can.​
Lelu [443]

Answer:

Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.

1. satellite in orbit around the earth             (motion of earth is negligible)

2. moon in orbit around the earth     (center of motion several thousand miles        

                                                             from center of earth)

3. earth in orbit around sun    (center of rotation close to center of sun)

4. binary stars         (if masses of stars are equal center of rotation is in middle)

3 0
3 years ago
Which of the following statements is true of a gas?
xenn [34]

Answer:

it takes the shape and size of the container that it is in

Explanation:

4 0
3 years ago
Read 2 more answers
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