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andrew-mc [135]

2 years ago

6

If you increase the force acting on an object, what will happen to the object's motion

1 answer:

Mama L [17]2 years ago

6 0

Force is directly proportional to rate of change of velocity so it increasing, velocity (motion of the object) will also increase.

Hope this helps!

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Kipish [7]

Answer:

Velocity

Explanation:

"The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s."

^^This explanation is from physicsclassroom.com

3 0

2 years ago

A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

7 0

2 years ago

A bowling ball is 21.6 cm in diameter. What is the angular speed of these ball whenit is moving at 3.0 m/s?

sergejj [24]

Answer:

Angular speed = 27.78 rad/s (Approx)

Explanation:

Given:

Diameter = 21.6 cm

Speed = 3 m/s

Find:

Angular speed

Computation:

Radius = 21.6 / 2 = 10.8 cm = 0.108 m

Angular speed = v / r

Angular speed = 3 / 0.108

Angular speed = 27.78 rad/s (Approx)

3 0

2 years ago

What is a planet’s period of rotation?

Annette [7]

The answer is the letter "C" ( I have honors science I am good at this type of stuff )

Hope I helped :) ( ask me for help when u need it :)

5 0

3 years ago

Read 2 more answers

A block of mass 22 kg is sliding along the ice at constant speed 5.0 m/s just ahead of it is q block of mass 29 kg sliding in th

mars1129 [50]

Answer:

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

Mass of block

Mass of another block

Initial speed of the block

Initial speed of the another block

Initial speed of the another block

For conservation of momentum, we have

Substitute all the values and solving for final speed of the 22kg block is

new speed of the 22-kg block is 1.57 m/s

Couldnt write the answer so check picture

8 0

2 years ago