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Fantom [35]
3 years ago
15

Write your opinion about achievement made by during rana rule​

Physics
1 answer:
Sedaia [141]3 years ago
4 0

Answer:

Your opinion about achievement made by during rana rule

Explanation:

April Fools !

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If an automobile engine delivers a power of 50.0 hp, how much time will it take for the engine to do 6.40 x 10^4 j of work? (1 h
aleksandrvk [35]

Given data

Power (P) = 50 hp,

                = 50 × 746,  we know that 1 hp = 746 W.

                = 37300 Watts      (Watt = J/s)  

Work  = 6.40 ×10⁴ J

Power is defined as rate of doing work and the unit of power is<em> Watt.</em>

Mathematically,

         Power = (Work / time)   Watts

                     = 6.40 ×10⁴ / time

          37300 W = 6.40 ×10⁴ J /time      (Where time in seconds)

         => time = Work/Power

                      = 6.40 ×10⁴/37300

                      = <em>1.74 seconds  </em>

<em>  </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>

<em> </em>


7 0
3 years ago
Describe why drawing a line of best fit is useful.
Amiraneli [1.4K]

Answer:

Cause life

Explanation:

Cause life

8 0
3 years ago
What are two factors<br> affecting friction? How do<br> they affect friction?<br> ch
AlexFokin [52]
The roughness of the surface, the mass of the object, and the area of contact.
5 0
3 years ago
In
son4ous [18]
(Direction) for the fact that it will continue having the momentum at the constant speed in which the engines turned off.
6 0
3 years ago
A 75 kg baseball player runs at a velocity of 6 m/s before sliding to a stop at second base. a. What is the kinetic energy of th
lana [24]

Answer:

a. \displaystyle k_o=1350\ J

b. \displaystyle k_1=0\ J

c. \Delta k=-1350\ J

d. W=-1350\ J

e. F=-675\ N

Explanation:

<u>Work and Kinetic Energy </u>

When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

\Delta k=k_1-k_0

Knowing that

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle \Delta k=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The work done by the force who caused the change of velocity (acceleration) is

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

If we know the distance x traveled by the object, the work can also be calculated by

W=F.x

Being F the force responsible for the change of velocity

The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops

a. Before the slide, his initial kinetic energy is

\displaystyle k_o=\frac{mv_0^2}{2}

\displaystyle k_o=\frac{(75)6^2}{2}

\boxed{\displaystyle k_o=1350\ J}

b. Once he reaches the base, the player is at rest, thus his final kinetic energy is

\displaystyle k_1=\frac{(75)0^2}{2}

\boxed{\displaystyle k_1=0\ J}

c. The change of kinetic energy is

\Delta k=k_1-k_0=0\ J-1350\ J

\boxed{\Delta k=-1350\ J}

d. The work done by friction to stop the player is

W=\Delta k=k_1-k_0

\boxed{W=-1350\ J}

e. We compute the force of friction by using

W=F.x

and solving for x

\displaystyle F=\frac{W}{x}

\displaystyle F=\frac{-1350\ J}{2\ m}

\boxed{F=-675\ N}

The negative sign indicates the force is against movement

6 0
2 years ago
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