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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

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