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Gala2k [10]
2 years ago
6

A student is holding a 1 kg bag. What is the normal force on the bag from the ground?

Physics
1 answer:
baherus [9]2 years ago
3 0

Answer:

gravitional pull

Explanation:

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

\omega= Angular velocity

Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

\Omega = \frac{2\pi rad}{T}

T = \frac{2\pi rad}{\Omega}

T = \frac{2\pi}{1.17}

T = 5.4 s

Therefore the precession period is 5.4s

7 0
2 years ago
According to Newton's<br> Ist law, what will an<br> object in motion tend<br> to do?
SCORPION-xisa [38]

Answer:

According to <em>Newton's first law of motion:</em>

<u>An object in motion tends to remain in motion unless an external force acts upon it.</u>

<u>It stays in motion with the same speed and goes in the same direction.</u>

<u></u>

<em>Hope this helped </em>

<em>:)</em>

4 0
3 years ago
Read 2 more answers
Based on the mass and speed data, which object will have the greatest amount of kinetic energy
Alexus [3.1K]
The objects will have to be listed
5 0
2 years ago
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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0
3 years ago
How long will it take to go 150,000 m traveling at 50 km/hr
Vanyuwa [196]
150 km / 50 km/h = 3h

5 0
3 years ago
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