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3 years ago

What are the cahnges that a force can bring out on a body ? Give examples

2 answers:

5 0

Hi Pupil Here is your answer ::

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1 The shape of the Body

Example : The shape of the ball lying on a floor can be changed by pressing it.

2 Direction of the Body

Example : The direction of motion of moving ball can be changed by hitting it with a bat.

3 The speed of the Body

Example : A ball at rest can be set in motion if force is applied only

4. Size of the Body

Example : The length of a spring tied and on one end can be increased by pulling it.

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Hope this helps .......

Serggg [28]3 years ago

5 0

See above and enjoy

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If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁ , F₂ = m * a₂

fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

3 0

3 years ago

-x times -x what is the answer

Brilliant_brown [7]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

3 a A motorcyclist starts from rest and reaches

andrew-mc [135]

The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v - u/t = a

a = 6 - 0/3

= 2 m/s^2

Again;

v - u/t = a

a = 0 - 6/5