A metre rod on wedge is example of unstable equilibrium.why

What force is required to accelerate to 10 kg object to 5.9 m/s/s?

g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

Mass = 10 kg

Acceleration = 5.9 m/s^2

Force = Mass * Acceleration

Force = 10 kg * 5.9 m/s^2

Force = 59 kg m /s^2 = 59 N

3 0

3 years ago

Gravity pulls downward on a rock with a force of 800 N. If you pull upward on the rock with a force of 400 N, what is the total

wel

The answer is
a. 400 N downward

3 0

3 years ago

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A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

$\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev$

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago

A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

ac = 3.92 m/s²

5 0

3 years ago

Julio blows air across his hot bowl of soup. The tiny ripples he creates are similar to _____.

Allushta [10]

The tiny ripples on the soup are not only similar to wind-generated
waves ... they ARE wind-generated waves. This is a big part of the
reason why they bear such an uncanny resemblance.

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3 years ago

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