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faltersainse [42]
3 years ago
14

A rifle of mass 2 kg is horizontally suspended by a pair of strings so that recoil can be measured. The rifle fires a bullet of

mass 1/100 kg at a speed of 200 m/s. The recoil velocity of the rifle is about ___________.
Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

Vg = 1m/s

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A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal. What is the range of the ball?
just olya [345]

Answer:

the range or the ball is 48.81 m

Explanation:

given;

Nicole throws a ball at 25 m/s at an angle of 60 degrees abound the horizontal.

find:

What is the range of the ball?

solution:

let Ф = 25°

Vo = 25 m/s

<u>consider x-motion using time of fight: x = Vox * t</u>

where x = R = range

t =<u> 2 Voy </u>

      g

R =<u> Vo² sin (2Ф)</u>

           g

plugin values into the formula:

R = <u>(25)² sin (2*25) </u>

               9.81

R = 48.81 m

therefore, the range or the ball is 48.81 m

4 0
4 years ago
Read 2 more answers
Acceleration is defined as the rate of change for which of the following? Question 4 options: time position velocity displacemen
vovangra [49]

Acceleration can be defined as the rate of change in the velocity of an object. Option C is correct.

<h3>What is Acceleration?</h3>
  • It is defined as the rate of change in velocity.
  • It can also be defined as the rate of change in position in a particular direction.

a =  \dfrac {v-u}t\ \ \ \ \rmor}\\\\a = \dfrac {\Delta v }t

Where,

a - acceleration

\Delta v - change in velocity

t - time

Therefore, acceleration can be defined as the rate of change in the velocity of an object.

Learn more about Velocity:

brainly.com/question/2239252

4 0
2 years ago
A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest
enyata [817]

Answer:

v = 2,425 m / s

Explanation:

A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.

highest point

           Em₀ = U = m g y

lowest point

          Em_{f} = K = ½ m v²

         Em₀ = Em_{f}

        mg y = ½ m v²

        v = √ 2gy

let's calculate

        v = √ (2 9.8 0.3)

        v = 2,425 m / s

3 0
3 years ago
You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 130-turn coil that has
dimaraw [331]

Answer:

B =  0.129 T

Explanation:

Given,

frequency, f = 60 Hz

maximum  emf = 5200 V

Number of turns, N = 130

Area per turn = 0.82 m²

We know,

ω = 2 π f

ω = 2 π x 60 = 376.99 rad/s

now, Magnetic field calculation

B =\dfrac{\epsilon_{max}}{NA\omega}

B =\dfrac{5200}{130\times 0.82\times 376.99}

B =  0.129 T

Hence, the magnetic field is equal to B =  0.129 T

3 0
3 years ago
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