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Neporo4naja [7]
2 years ago
6

A cat is trying to cross the street. Its velocity v as a function of time t is given in the graph below where

Physics
2 answers:
bija089 [108]2 years ago
8 0

Answer:

3.8

Explanation:

Vesna [10]2 years ago
5 0

Answer:

rightwards is the positive velocity

Rightwards is the positive elicitation:

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A 0.45-kilogram football traveling at a speed of 22 meters per second is caught by an 84-kilogram stationary receiver. if the fo
Troyanec [42]

Impulse = Change in momentum.  
The ball was moving with a momentum of 0.45 * 22 = 9.9  
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0  
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9
3 0
3 years ago
Two technicians are discussing U-joints. Technician A says that a defective U-joint could cause a loud clunk when the transmissi
Nookie1986 [14]

Answer:

Technician A and Technician B are correct.

Explanation:

6 0
3 years ago
The field around a solenoid is similar to the field around a bar magnet.<br><br> true or false?
OlgaM077 [116]
The answer would be true :)
8 0
2 years ago
A uniform disk with a mass of 5.0 kg and diameter 30 cm rotates on a frictionless fixed axis through its center and perpendicula
Marina86 [1]

Answer:\alpha =10.66 rad/s^2

Explanation:

Given

mass of disk m=5 kg

diameter of disc d=30 cm

Force applied F=4 N

Now this force will Produce  a  torque of magnitude

T=F\cdot r

T=4\dot 0.15

T=0.6 N-m

And Torque is given Product of moment of inertia and angular acceleration (\alpha )

T=I\cdot \alpha

Moment of inertia for Disc I= \frac{Mr^2}{2}

I=0.05625 kg-m^2

0.6=0.05625\cdot \alpha

\alpha =10.66 rad/s^2

3 0
3 years ago
An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by an amoun
trasher [3.6K]

Answer:

2d

Explanation:

For any instance equivalent force acting on the body is

mg-kd= m\frac{d}{dt}\frac{dx}{dt}

Where

m is the mass of the object

k is the force constant of the spring

d is the extension in the spring

and

d/dt(dx/dt)=  is the acceleration of the object

solving the above equation we get

x= Asin\omega t +d

where

\omega= \sqrt{\frac{k}{m} } = \frac{2\pi}{T}

A is the amplitude of oscillation from the mean position.

k= spring constant , T= time period

Here  we are assuming  that at t=T/4

x= 0   since, no extension in the spring

then

A=- d

Hence

x=- d sin wt + d

now, x is maximum when sin wt=- 1

Therefore,

x(maximum)=2d

7 0
3 years ago
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