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3 years ago

A wind turbine has a total input power of 2 500 kW.

Physics

1 answer:

Sonbull [250]3 years ago

6 0

Answer:

Output power = 500 KW

Explanation:

Given the following data;

Efficiency = 20%

Input power = 2500 KW

To find the output power;

$Efficiency = \frac {Out-put \; power}{In-put \; power} * 100$

Substituting into the equation, we have;

LET Output power = OP

$20 = \frac {OP}{2500} * 100$

Cross-multiplying, we have;

$20 * 2500 = OP * 100$

$50000 = OP * 100$

$OP = \frac {50000}{100}$

Output power = 500 KW

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According to the principle of conservation of momentum, A. the amount of momentum of all the objects in the universe is constant

anygoal [31]

Answer:

D. in a closed system, an object's momentum before a collision will equal its momentum after the collision.

Explanation:

As we know by newton's II law we have

$F = \frac{\Delta P}{\Delta t}$

so we will have

F = net force on the system

$\Delta P$ = change in momentum

so we can say if the momentum is conserved or the initial momentum of the system will be equal to the final momentum of the system

then we will say that

$\Delta P = 0$

so net force on the system must be ZERO

so in such type of questions the system must be isolated and there is no external force on it

so correct answer will be

D. in a closed system, an object's momentum before a collision will equal its momentum after the collision.

6 0

3 years ago

A projectile is launched from ground level at angle u and speed v0 into a headwind that causes a constant horizontal acceleratio

ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is $v_0$

Horizontal acceleration is $a_x=a$

At maximum height velocity is zero therefore

$v_f=v_i-gt$

$0=v_0\sin u-gt$

$t=\frac{v_0\sin u}{g}$

Total time of flight $T=2t=\frac{2v_0\sin u}{g}$

During this time horizontal range is

$R=v_o\cos u\cdot 2t-\frac{a(2t)^2}{2}$

$R=\frac{2v_0^2\sin u\cos u}{g}-\frac{2av_0^2\sin ^u}{g^2}$

For maximum range $\frac{\mathrm{d} R}{\mathrm{d} u}=0$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2\cos 2u}{g}-\frac{4av_0^2\sin u\cos u}{g^2}$

$\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0$

$\tan 2u=\frac{g}{a}$

$u=\frac{1}{2}tan ^{-1}\frac{g}{a}$

(b)If a =10% g

$a=0.1g$

thus $u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}$

$u=42.14^{\circ}$

7 0

3 years ago

Water displacement is used to find what? a. volume b. density

Alex

Answer:

Volume

Explanation:

,,,,,,,,,,,,,,,,,,,

4 0

2 years ago

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h

noname [10]

The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s

7 0

3 years ago

A particle is moving along the x-axis so that its position at any time t is greater than and equal to 0 is given by x(t)=2te^-t?

erastovalidia [21]

For speed you can differentiate the equation, for acceleration you can again differentiate the equation .
at t=0 the particle is slowing down , when you get equation for velocity put t=0 then only -1 is left

6 0

3 years ago