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2 years ago

5

A dog walking to the right with a speed of 1.5 m/s sees a cat and speeds up with a constant rightward acceleration of magnitude

12 m/s^2. What is the velocity of the dog after speeding up for 3.0m

1 answer:

joja [24]2 years ago

8 0

Answer: 8.62

Explanation: It’s correct answer:)

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Speed and Motion you went from the starting line to the finish line at different rates. If you repeated the activity while carry

Phantasy [73]

Answer:

It will cause kinetic energy to increase.

Explanation:

Given that Speed and Motion you went from the starting line to the finish line at different rates.

If you repeated the activity while carrying weights but keeping your times the same, the weight carried will add up to the mass of the body.

And since Kinetic energy K.E = 1/2mv^2

Increase in the mass of the body will definitely make the kinetic energy of the body to increase.

Since the time is the same, that means the speed V is the same.

Weight W = mg

m = W/g

The new kinetic energy will be:

K.E = 1/2(M + m)v^2

This means that there will be increase in kinetic energy.

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2 years ago

A balloon filled with helium has a small hole on the right side of the balloon. How would Pascal explain why the entire balloon

Morgarella [4.7K]

Answer:

delete this answer please.

4 0

2 years ago

Any transparent material that allows light of only certain color to pass through is called a filter . true or false

Dahasolnce [82]

it is TRUE hope this helps

7 0

2 years ago

Problema en la cual aplicaste velocidades, impulso, conservación del movimiento y de la energía.

Kazeer [188]

Huh huh what? ¿Can’t you translate?

6 0

2 years ago

Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil

MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

$emf_{max}=NBA\omega$

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

$A=\pi r^2 =\pi(0.23m)^2=0.16m^2$

$emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V$

Finally, the peak current is given by:

$I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA$

b)

we have that

$I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A$

$P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W$

hope this helps!!

6 0

2 years ago