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victus00 [196]
3 years ago
8

which has a higher acceleration:a 10kg object acted upon with a net force of 20N or an 18kg object acted on by a net force of 20

N
Physics
2 answers:
MA_775_DIABLO [31]3 years ago
7 0
<span>Answer: The acceleration of 10 kg object is greater than that of 18 kg object.

Explanation:
According to Newton's Second law:
F = ma --- (A)

Let's find the acceleration for both 10 kg and 18 kg objects!
The net force on both of these masses = F = 20N

(1) Acceleration of 10 kg object
Mass = m = 10 kg
Plug in the values in equation (A):
20 = 10 * a
Acceleration = a = 2 m/s^2

(2) Acceleration of 18 kg object
Mass = m = 18 kg
Plug in the values in equation (A):

20 = 18 * a
Acceleration = a = 1.11 m/s^2


2 > 1.11; therefore, 10 kg object has the higher acceleration compared to the acceleration of the 18 kg object.</span>
Juliette [100K]3 years ago
6 0

Explanation:

The expression of the applied force in terms of mass and acceleration by using Newton's second law is as follows;

F= ma

Here, m is the mass of the object and a is the acceleration of the object.

Calculate the acceleration of a 10 kg object acted upon with net force of 20 N in the above expression.

Put m= 10 kg and F= 20 N in the above expression.

20= (10)a

a= 2 m/s^2

Calculate the acceleration of a 18 kg object acted upon with net force of 20 N in the above expression.

Put m= 18 kg and F= 20 N.

20= (18)a

a= 1.11 m/s^2

Therefore, the acceleration of a 10 kg object acted upon with net force of 20 N is greater than the acceleration of a 18 kg object acted upon with net force of 20 N.

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Agata [3.3K]
We have the following equation for height:
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 Where,
 a: acceleration
 vo: initial speed
 h0: initial height.
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 For t = 0 we have:
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 h (0) = h0
 h0 = 0 (reference system equal to zero when the ball is hit).
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 vo = (1/2) * (9.8) * (5.8)
 vo = 28.42
 Substituting values we have:
 h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
 h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
 Rewriting:
 h (t) = -4.9 * t ^ 2 + 28.42 * t
 The maximum height occurs when:
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 t = 28.42 / 9.8
 t = 2.9 seconds.
 Answer:
 
The ball was at maximum elevation when:
 
t = 2.9 seconds.
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Answer:

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