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3 years ago

3. What is the kinetic energy of a 1300 kg car moving at 26.3 m/s?

Physics

1 answer:

Alexxx [7]3 years ago

3 0

Answer:K.E=449598.5j

Explanation:

Kinetic energy of a moving car=1/2mv^2

Where m is the mass of the car

And V is the velocity of the car

K.E=1/2 ×1300×26.3^2

K.E=449598.5j

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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

7. Explain why observations of comets and shooting stars presented challenges to ancient astronomers using

Lelu [443]

According to Ptolemy's model, he, too, believed in a geocentric Universe and that the planets and stars were perfect spheres, though Earth itself was not.

He further thought that the movements of the planets and stars must be circular since they were perfect and, if the motions were circular, then they could go on forever.

<h3>What is comets and shooting stars?</h3>

Shooting stars are very different from comets, although the two can be related. A Comet is a ball of ice and dirt, orbiting the Sun (usually millions of miles from Earth). ... A shooting star on the other hand, is a grain of dust or rock (see where this is going) that burns up as it enters the Earth's atmosphere.

Learn more about ptolemy's model:

brainly.com/question/12639459

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3 years ago

you are doing an experiment outside on a sunny day you find the temperature of some sand is 28 degress Celsius you also find the

OLga [1]

E the temperature of a substance. Water has a very high specific heat. That means it needs to absorb a lot of energy before its temperature changes. Sand , on the other hand, have lower specific heats. This means that their temperatures change more quickly. When the summer sun shines down on them, they quickly become hot.

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3 years ago

Find the orbital speed v for a satellite in a circular orbit of radius R.Express the orbital speed in terms of G, M, and R.

AlekseyPX

<h2>Answer: $V=\sqrt{G\frac{M}{R}}$ </h2>

The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G$ is the gravity constant

$M$ the mass of the massive body around which the satellite is orbiting

$R$ the radius of the orbit (measured from the center of the planet to the satellite).

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>

In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.

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4 years ago

What is a fact about incline planes?

SVETLANKA909090 [29]

They are incline hope this helps!

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3 years ago