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klasskru [66]
2 years ago
15

Two students are playing paddle ball with a 5 kg spongy ball. If the ball is thrown at the batter with a speed of 5 m/s and boun

ces off with a speed of 10 m/s. What is the magnitude of the impulse imparted on the ball by the bat?
Physics
1 answer:
fenix001 [56]2 years ago
3 0

Answer:

75 kgm/s

Explanation:

Impulse: This can be defined as the product of mass and change in velocity. The S.I unit is kgm/s.

From the question,

I = m(v-u)................... Equation 1

Where I = impulse, m = mass, v = final velocity, u = initial velocity.

Let the direction of the initial velocity be the positive direction.

Given: m = 5 kg, v = -10 m/s (bounce off), u = 5 m/s.

Substitute into equation 1

I = 5(-10-5)

I = 5(-15)

I = -75 kgm/s.

The negative sign tells that the impulse act in the same direction as the final velocity of the ball

Hence,

I = 75 kgm/s

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A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?
evablogger [386]
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
4 0
2 years ago
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2. A pebble is dropped down a well and hits the water 1.5 s later. Using the equations for motion with constant acceleration, de
kow [346]

Answer:

S = 11.025 m

Explanation:

Given,

The time taken by the pebble to hit the water surface is, t = 1.5 s

Acceleration due to gravity, g = 9.8 m/s²

Using the II equations of motion

                          S = ut + 1/2 gt²

Here u is the initial velocity of the pebble. Since it is free-fall, the initial velocity

                                u = 0

Therefore, the equation becomes

                            S = 1/2 gt²

Substituting the given values in the above equation

                              S = 0.5 x 9.8 x 1.5²

                                 = 11.025 m

Hence, the distance from the edge of the well to the water's surface is, S = 11.025 m

3 0
2 years ago
A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
Makovka662 [10]

Answer:

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

Explanation:

Given that

v(t) = 5 t i + t² j - 2 t³ k

We know that acceleration a is given as

a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k

a=5 i+2t j - 6\ t^2k

Therefore the acceleration function a will be

a=5 i+2t j - 6\ t^2k

The acceleration at t = 2 s

a= 5 i + 2 x 2 j - 6 x 2² k  m/s²

a=5 i + 4 j -24 k m/s²

The magnitude of the acceleration will be

a=\sqrt{5^2+4^2+24^2}\ m/s^2

a= 24.83 m/s²

The direction of the acceleration a is given as

a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

a)a=5 i+2t j - 6\ t^2k

b)a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2

5 0
2 years ago
Richardson pulls a toy 3.0 m across the floor by a string, applying a force of0.50 N. During the first meter, the string is para
Anastasy [175]

Answer:

Total Work done =0.65 joule

Explanation:

Work done is given Mathematically as

W=F *d

Where w=work done in joules

F=applied force

d= distance moved

The work done to move the toy accros the first meter is

W1=0.5*1

W1=0.5joule

The work done to move the toy across the next 2m at an angle of 30° is

.W2=0.5*2cos30

W2=0.5*2*0.154

W2=0.154joule

Hence total work done is

W1+W2=0.5+0.154

Total Work done =0.65 joule

7 0
2 years ago
A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a
Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0
2 years ago
Read 2 more answers
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