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mario62 [17]
3 years ago
7

In a game of tug of war, Team A pulls with a force 850N, Team B pulls with force of 975N. Calculate the net force on the rope. B

e sure to include which direction.
Physics
1 answer:
galben [10]3 years ago
7 0
I’d need to know which direction team a is pulling in and which direction b is pulling in to determine which direction.

But the net force is 125N in B’s direction (since B’s force is stronger than A’s by 125N)
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When a basketball player dribbles a ball, it falls to the floor and bounces up. Is a force required to make it bounce? Why? If a
Marizza181 [45]
Energy is the one that is stored in the ball when it drops. Just before it hits the ground, the energy depends on the mass of the ball and its velocity. When the ball hits, it is compressed and the energy is stored in the compression of the air in the ball and the elasticity of the material that the ball is made from. Some is also converted to heat. The stored energy in the ball causes a force to make the ball back into a round shape and this force presses against the propels and floor the ball back up. The small amount lost as heat is the reason that the ball bounces up with less energy than when it hit.
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Which of these forms due to the force of compression?
Georgia [21]

Answer:

anticline

Explanation:

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4 0
3 years ago
Read 2 more answers
During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c
IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

5 0
3 years ago
If someone looks far enough into space, they should be able to see the beginning of the universe true of false
kodGreya [7K]
The answer is no, it would be impossible to see the beginning of the universe
8 0
3 years ago
Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

Speed  = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

  Speed  = \frac{300}{18}  = 16.67m/s

3 0
3 years ago
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