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3 years ago

What is wind energy?

Physics

1 answer:

eimsori [14]3 years ago

6 0

<span>d. the conversion of kinetic energy created by the movement of gasses into a useful form of energy</span>

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Zoe is setting up a track for a toy car. The track has a ramp that is 32° above horizontal. If Zoe wants the car to travel as a

jolli1 [7]

Answer:

Explanation:

Not enough information.

IF we ASSUME she wants the car to be at LAUNCH LEVEL after 1 second of flight.

THEN

The highest point will have zero vertical velocity and will have taken ½ second to get there. This means that the initial vertical velocity was

v = gt

vy₀ = 9.8(0.5)

vy₀ = 4.9 m/s

vsinθ = vy₀

v = vy₀/sinθ

v = 4.9/sin32

v = 9.2466...

v = 9.2 m/s

8 0

2 years ago

A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is

grin007 [14]

I’ve got 120 n, hope it helps!

7 0

2 years ago

What gravitational force does the moon produce

saul85 [17]

Answer:

$1.94\cdot 10^{20} N$

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

The gravitational force is always attractive.

In this problem, we have:

$m_1 = 5.98\cdot 10^{24}kg$ is the mass of the Earth

$m_2 = 7.34\cdot 10^{22} kg$ is the mass of the Moon

$r=3.88\cdot 10^8 m$ is the separation between the Earth and the Moon

Therefore, the gravitational force between them is

$F=(6.67\cdot 10^{-11})\frac{(5.98\cdot 10^{24})(7.34\cdot 10^{22})}{(3.88\cdot 10^8)^2}=1.94\cdot 10^{20} N$

6 0

3 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

3 years ago

Which of the following is most like an indicator of a chemical change

qaws [65]

Answer:

methyl orange, methyl red,phenoptalin, merhy red

Explanation:

all this following are indicators use to check the end point of a reaction

5 0

3 years ago