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skelet666 [1.2K]
2 years ago
15

Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?

Physics
1 answer:
andreev551 [17]2 years ago
5 0

Answer:

merry go round and Ferris wheel have a constant acceleration due to the change in direction at every point.

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The smallest division value of electronic balance​
lara [203]

Answer:

0.1g to 0.0000001g hope it helps uu

4 0
2 years ago
Suppose a car of mass m is moving at a constant speed v of
SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

\arcsin(v^2/(gR))

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

w_r = mg\sin(\theta)

The cosine component is equivalent to the normal force, which we will not be using since friction is unnecessary.

Newton’s Second Law states that

F_{net} = ma = mg\sin(\theta)\\\sin(\theta) = a/g

Also, the car is making a circular motion:

a = \frac{v^2}{R}

Combining the equations:

\sin(\theta) = \frac{a}{g} = \frac{v^2/R}{g} = \frac{v^2}{gR}

Finally the angle is

\arcsin(v^2/(gR))

4 0
3 years ago
Which best describes the relationship between the direction of energy and wave motion in a transverse wave?
sammy [17]
I think the correct answer from the choices listed above is the second option. The relationship between the direction of energy and wave motion in a transverse wave would be the <span>energy direction is perpendicular to the motion of the wave. Hope this answers the question. Have a nice day.</span>
6 0
3 years ago
Read 2 more answers
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will
ExtremeBDS [4]

Answer:

α = 2,857 10⁻⁵ ºC⁻¹

Explanation:

The thermal expansion of materials is described by the expression

          ΔL = α Lo ΔT

          α = \frac{\Delta L}{L_o \ \Delta T}

in the case of the bar the expansion is

         ΔL = L_f - L₀

         ΔL=   1.002 -1

         ΔL = 0.002 m

the temperature variation is

         ΔT = 100 - 30

         ΔT = 70º C

we calculate

         α = 0.002 / 1 70

         α = 2,857 10⁻⁵ ºC⁻¹

5 0
3 years ago
Questions:<br>(a) Why should the amplitude of oscillation be small on a pendulum experiment​
Snowcat [4.5K]
The formula for the pendulum experiment is based on the assumption that the amplitude is small so that the angle is approximately equal to the Sine of the angle.
5 0
3 years ago
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