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1 year ago

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the picture shows six panels (a, b, c, d, e, and f), each displaying two snapshots of a box moving on a rough flor. forces may b

e applied to the boxes but are not shown. rank the change in kinetic energy for the panels.

1 answer:

Agata [3.3K]1 year ago

8 0

K = 1/2mv^2 of kinetic energy. The change in the object's kinetic energy is equal to the net work performed on it.

<h3>What causes the kinetic energy to change?</h3>

Equations. Mass and the square of the velocity are directly related to translational kinetic energy. The difference between the end and starting kinetic energies is known as change in kinetic energy.

<h3>In solar panels, is there kinetic energy?</h3>

employing semiconductor-cell-based panels. technique that uses solar thermal systems to store solar energy. This heat is used directly or transformed into concentrated solar power, or the sum of the potential energy and kinetic energy of an object or system, and electricity.

Learn more about kinetic energy here:

brainly.com/question/26472013

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A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe

Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

$I_R=\frac{1}{12}ML^2$

where

$M=3.30\cdot 10^{-2} kg$ is its mass

L = 0.450 m is its length

Substituting,

$I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2$

The moment of inertia of the two rings at the beginning is

$I_r = 2mr^2$

where

m = 0.200 kg is the mass of each ring

$r=5.20\cdot 10^{-2} m$ is their distance from the center of the rod

Substituting,

$I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2$

So the total moment of inertia at the beginning is

$I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2$

The initial angular velocity of the system is

$\omega_1 = 35.0 rev/min$

The angular momentum must be conserved, so we can write:

$L=I_1 \omega_1 = I_2 \omega_2$ (1)

where $I_2$ is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

$r=\frac{0.450 m}{2}=0.225 m$

so the moment of inertia of the rings is

$I_r=2(0.200)(0.225)^2=0.0203 kg m^2$

and the total moment of inertia is

$I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2$

Substituting into (1), we find the final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min$

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

$I_2 = I_R = 5.57\cdot 10^{-4}kg m^2$

So using again equation of conservation of the angular momentum:

$L=I_1 \omega_1 = I_2 \omega_2$

We find the new final angular speed:

$\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min$

7 0

3 years ago

A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?

Advocard [28]

Answer:

$v=697.2km/h$

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

$V=\frac{d}{t}$

The distance is clearly 1743 km and the time is:

$t=2h+30min*\frac{1h}{60min} =2.5h$

Thus, the velocity turns out:

$v=\frac{1743km}{2.5h}\\ \\v=697.2km/h$

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

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3 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

lawyer [7]

I really think it could be a 23 kg

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Easy question. Answer should include one line reasoning

KATRIN_1 [288]

C.figure 3 is the answer had the same and got is right

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mezya [45]

Average atomic mass of carbon is 12.01 amu

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