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Yakvenalex [24]
2 years ago
9

Students set up a controlled experiment (please help I am so tired and unable to function lol)

Physics
1 answer:
Veronika [31]2 years ago
4 0
The temp...............
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A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg ball moving to the left at 5 m/s. What is the velocity
Vinil7 [7]

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Oh wow so this is why you stole those points from me

Explanation:

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2 years ago
URGENTTT PLEASE HELP
mel-nik [20]
This doesn’t make any sense
3 0
2 years ago
Which has a higher frequency - visible or infrared radiation?
bearhunter [10]
Visible has a higher frequency
4 0
3 years ago
Consider the following objects and their locations. Object a 2-kg object 5 cm above the floor Object b 2-kg object 120 cm above
Nataly [62]

Answer:

The gravitational acceleration is same for all objects.

a = b = c = d

Explanation:

Acceleration due to gravity or gravitational acceleration is the force exerted by Earth on unit mass of an object.

Acceleration due to gravity doesn't depend on the height of the object when the height is object is near to the surface of the Earth. Only when the height is comparable to the radius of the Earth, the value of gravitational acceleration changes.

But for the objects here, the gravitational acceleration is independent of the mass or height of the objects and has a constant value of 9.8 m/s².

Therefore, the gravitational acceleration of all the objects is same.

If 'a', 'b', 'c', and 'd' represent gravitational accelerations of objects 'a', 'b', 'c', and 'd' respectively, then a = b = c = d.

5 0
3 years ago
A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What
kramer

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let p_{1} = momentum of the 1st ball

p_{2} = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}

or

(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}

Since we are dealing with identical balls, all the m terms cancel out so we are left with

(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} +  (v_{2})_{f}\cos \theta_{2}

Putting in the numbers, we get

1.33 = (0.750) \cos(33.30)  + (v_{2})_{f} \cos \theta_{2}

=  > (v_{2})_{f} \cos \theta_{2} = 0.703

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}

or

=  > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30)  = 0.412

Taking the ratio of the sine equation to the cosine equation, we get

\frac{ \sin \theta _{2}}{ \cos \theta_{2} }  =  \tan \theta_{2}  =  \frac{0.412}{0.703}  = 0.586

or

\theta_{2}  =  { \tan}^{ - 1} (0.586) = 30.4

Solving now for (v_{2})_{f},

(v_{2})_{f}  =  \frac{0.412}{ \sin(30.4) }  = 0.815 \:  \frac{m}{s}

3 0
3 years ago
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