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marin [14]
3 years ago
12

Two golf balls with the same mass were hit at the same time. Golf ball A went a shorter distance than golf ball B. What could be

a reason that this happened?
Golf ball B experienced more friction.
Golf ball A experienced less friction.
Golf ball B was hit with a weaker force.
Golf ball A was hit with a weaker force.
Physics
2 answers:
Firdavs [7]3 years ago
7 0

Answer:

Golf ball A was hit with a weaker force

Explanation:

Golf ball B went a larger distance than golf ball A, so we could think friction affected golf ball A rather than golf ball B. That is why the first two choices are wrong.

The strongest forced applied made golf ball B go further.

GrogVix [38]3 years ago
5 0
Golf ball a was hit with a weaker force
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A cart was pulled for a distance of 1 kilometer and the amount of work accomplished equaled 40,000 joules. With what force was t
goblinko [34]

Answer:

40 N

Explanation:

3 0
3 years ago
if the mass is 10 gram and the volume is 7 cubic centimetre,find the density in kilogram per cubic metre
wel

Answer:

1.42

Explanation:

devide 10 by 7

7 0
3 years ago
A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.​
GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

7 0
2 years ago
A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the
zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

a=2.5 m/s^2

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

d=0+\frac{1}{2}(2.5)(35)^2=1531 m

8 0
3 years ago
The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r
r-ruslan [8.4K]

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

          v = x / t

          x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

           x = 2d

          2d = v t

          d = v t/2

     

let's calculate

          d = 1500 0.5 / 2

          d = 375 m

3 0
3 years ago
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