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svp [43]
3 years ago
10

A tennis ball is dropped from 1.13 m above the

Physics
1 answer:
Alex73 [517]3 years ago
7 0

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

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A ball is thrown vertically upward with a speed of 12.0 m/s. (<br> a. How high does it rise?
sdas [7]
It rises till all of its Kinetic energy is converted into potential energy.

so, mgh=(1/2)m(v^2)

so, h=(v^2)/2g = 12*12/(2*9.81)=7.34 m
5 0
3 years ago
Which of the following are examples of projectile motion?
denis-greek [22]

Answer:

A or C    I would pick C

Explanation:

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3 years ago
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A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
3 years ago
Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
3 years ago
A car speeds up from 18.54 m/s to<br> 29.52 m/s in 13.84 s.<br> The acceleration of the car is:
valkas [14]

Answer:

.7934m/s^{2}

Explanation:

Acceleration = change in velocity / change in time

A = 10.98m/s / 13.84s

A = .7934m/s^{2}

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3 years ago
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