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Romashka [77]
3 years ago
8

A 2800 kg speedboat starting from rest attains a speed of 16 m/s in 8.0 s as a combination of 1200 N of air resistance and water

drag act on the boat. How much power is being expended by speedboat at the maximum speed?

Physics
1 answer:
denis23 [38]3 years ago
6 0

Answer:

For me

Explanation:

power=workdone/time

power= 128×1200/8

power=19200

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What is the process called when the moon begins to fade from a full moon to the new moon?
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
castortr0y [4]

The vertical component of force exerted by the hi.nge on the beam will be,142.10N.

To find the answer, we need to know more about the tension.

<h3>How to find the vertical component of the force exerted by the hi.nge on the beam?</h3>
  • Let's draw the free body diagram of the system.
  • To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.

                      F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\

  • To find the answer, we have to find the tension,

                     Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N

  • Thus, the vertical component of the force exerted by the hi.nge on the beam will be,

                F_V=(29*9.8)-(169.43*sin57)=142.10N

Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.

Learn more about the tension here:

brainly.com/question/28106868

#SPJ1

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Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa
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<h2>Answer: B. Gravitational potential energy </h2>

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That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which  <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy U_{p} will be:

U_{p}=mgh  

Where m is the mass of the object, g the acceleration due gravity and h the height of the object.

As we can see, the value of U_{p} is directly proportional to the height.

6 0
3 years ago
Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis.
amm1812

Answer:

38 cm from q1(right)

Explanation:

Given, q1 = 3q2 , r = 60cm = 0.6 m

Let that point be situated at a distance of 'x' m from q1.

Electric field must be same from both sides to be in equilibrium(where EF is 0).

=> k q1/x² = k q2/(0.6 - x)²

=> q1(0.6 - x)² = q2(x)²

=> 3q2(0.6 - x)² = q2(x)²

=> 3(0.6 - x)² = x²

=> √3(0.6 - x) = ± x

=> 0.6√3 = x(1 + √3)

=> 1.03/2.73 = x

≈ 0.38 m = 38 cm = x

8 0
3 years ago
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