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3 years ago

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 Watt.

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

= 1.74 seconds

Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work 

 

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Describe why drawing a line of best fit is useful.

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Answer:

Cause life

Explanation:

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What are two factors affecting friction? How do they affect friction? ch

AlexFokin [52]

The roughness of the surface, the mass of the object, and the area of contact.

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son4ous [18]

(Direction) for the fact that it will continue having the momentum at the constant speed in which the engines turned off.

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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:

Work and Kinetic Energy

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

Write your opinion about achievement made by during rana rule​

Write your opinion about achievement made by during rana rule