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agasfer [191]
3 years ago
15

Power output of an engine is with work (w=60,000) during 60 s Calculate the power of this work?*

Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

<em>The power output of the engine is 1,000 Watt</em>

Explanation:

<u>Mechanical Power </u>

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

The engine does a work of W=60,000 J during t=60 s. Calculate the power:

\displaystyle P=\frac {60,000}{60}

P = 1,000 Watt

The power output of the engine is 1,000 Watt

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What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

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3 years ago
A vector points -43.0 units
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Answer:

Explanation:

To find the direction of this vector we need o find the angle that has a tangent of the y-component over the x-component:

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3 years ago
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

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