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3 years ago

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WARRIOR [948]3 years ago

3 0

Oop i have no clue but i need to answer questions so that i can ask some :)

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*PLEASE HELP WILL GIVE BRAINLIEST*

Lorico [155]

More mass will in turn create more force in a collision.

The number of marbles in a collision will affect the outcome because they could bounce off each other and what not.

4 0

3 years ago

Explain how to identify a starting position<br> on a line.

olga nikolaevna [1]

Answer:

We can easily find out the beginning point of the line by using dot representation. When it comes to position vector, it expresses the exact position of certain object from the starting point of the coordinate system. The vector is a straight line that has a certain end which is fixed to its body

4 0

3 years ago

Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of

zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$

Radius of the barrel is, $R=2.90\ m$

Number of revolutions are, $n =10$

Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

$T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s$

Now, angular velocity is related to time period as:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s$

Now, centripetal force acting on the child is given as:

$F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N$

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0

3 years ago

A ball is dropped from the top of a tower 40 m high. What is its velocity when it has covered 20 m? What would be its velocity w

stellarik [79]

Given :

Height from which ball is dropped , h = 40 m .

Acceleration due to gravity , g= 10 m/s² .

Initial velocity , u = 0 m/s .

To Find :

Velocity when ball covered 20 m and velocity when it hit the ground .

Solution :

Now , height when ball covered 20 m distance is , 40 - 20 = 20 m .

By equation of motion :

$v^2=u^2+2gh\\\\v=\sqrt{2\times 10\times 20}\ m/s\\\\v=20\ m/s$

Now , distance covered when body reaches ground is , 40 m .

Putting value h = 40 m in above equation , we get :

$v=\sqrt{2\times 10\times 40}\ m/s\\\\v=20\sqrt{2}\ m/s$

Hence , this is the required solution .

7 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.20×10^6 N, one at an angle 14.0∘ west of north,

laila [671]

Answer:

1.45544 J

Explanation:

See attachment

5 0

3 years ago