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3 years ago

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Consider two wind turbines, each located in sites with the same wind speed, and each with the same efficiency. Turbine 1 has a r

adius of 25 meters. Turbine 2 has a radius of 50 meters. Turbine 2 generates ___ times as much power as turbine 1. (Fill in the blank.)

1 answer:

Alex787 [66]3 years ago

7 0

Answer:

shut up nerd

Exption:

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A system releases 255 cal of heat to the surroundings while delivering 428 cal of work. what is the change in internal energy of

vichka [17]

Answer:-683 cal

Explanation:

Given

Heat released by system Q=-255 cal

as heat released is taken as negative and vice-versa

Work done by system W=428 cal

From First law of thermodynamics

$Q=\Delta +W , where \Delta U$ =change in internal Energy

$\Delta U=-255-428=-683$

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4 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

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4 years ago

What type of issue does the effects manufacturer need

SOVA2 [1]

Answer:

A) Availability

Explanation:

Right on Edge 2021

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3 years ago

(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r

Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface $U_1=\frac{GM_em}{R_e}$

Gravitational Potential Energy at height h is $U_2=\frac{GM_em}{R_e+h}$

Energy required to lift the satellite $E_1=U_1-U_2$

$E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}$

Now Energy required to orbit around the earth

$E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}$

$\Delta E=E_1-E_2$

$\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}$

$E_1=E_2$ (given)

$\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0$

$\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0$

$h=\frac{R_e}{2}$

$h=3.19\times 10^6\ m$

(b)For greater height $E_1$ is greater than $E_2$

thus energy to lift the satellite is more than orbiting around earth

4 0

3 years ago

What is the relation between acceleration due to gravity and radius of the earth?

emmasim [6.3K]

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

Please Mark me brainliest

8 0

3 years ago

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