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3 years ago

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12. Rainbows occur at __________. A. 40–42 degrees above the horizon B. at or below 25 degrees above the horizon C. the horizon

D. 60–75 degrees above the horizon.

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

the right answer is option C

Explanation:

at the horizon

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A wind generator, such as the one shown , uses the power of wind to generate electricity What do you think is happening inside t

sesenic [268]

Answer:

A wind turbine captures the wind, which then produces a renewable energy source. The wind makes the rotor spin; as the rotor spins, the movement of the blades drives a generator that creates energy. The motion of the blades turning is kinetic energy. It is this power that we convert into electricity.

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3 years ago

Read 2 more answers

A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

3 0

3 years ago

A projectile is launched at an angle of 60° from the horizontal and at a velocity of

gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

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3 years ago

Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t

Mrrafil [7]

Answer:

a) Then net force acting on the wagon 30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g = 10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force Fn = P = 98 [N] positive ( surface is frictionless)

∑Fy = 0 P - Fn = 0 P = Fn = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N] = 10 Kg * a [N] = Kg * m /s²

-30 [kg*m/s²] = 10 * Kg * a

-30/10 = - 3 m/s² = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon 30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

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3 years ago

Jimmy is standing 300 feet away from a rocket that is being shot up into the air away from him at an angle of elevation of 70 de

ira [324]

Answer: Vertical height of rocket is 824.24ft

Explanation: see attachment.

Opp/ Adj = Tan 70°

h/ 300 = Tan 70°

h= 300tan 70°

h = 824.24ft

4 0

3 years ago