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pychu [463]
3 years ago
5

Two ice boats (one of mass m, one of mass 2m) hold a race on a frictionless, horizontal, frozen lake. Bath ice boats start at re

st, and the wind exerts the same constant force on both iceboats. After moving a same distance s, which iceboat crosses the finish line with a quicker speed v
Physics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

They both cross the finish line with the same kinetic energy

Explanation:

Same force, same displacement, so same KE at the location of the finish line. They don’t cross the line at the same time, but that was not the question!

∆KA= (mA/2)vF,A2

∆KA=∆KBso vF,B/vF,A= (MA/MB)1/2∆KB= (mB/2)vF,B2

Lighter boat goes faster so reaches finish line 1st

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pshichka [43]

Answer:

A. transverse waves

vote me brainliest?? 5/5

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8 0
2 years ago
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Which instrument changes kinetic energy into electrical energy?
Pachacha [2.7K]

Answer:

Dynamo

Explanation:

Dynamo started to rotate which is known as kinetic energy.When dynamo is in running it produces electricity.dynamo specially used for generating electricity.

5 0
3 years ago
ITS TIMED PLEASE HELP !!!!!
dusya [7]

Answer: direct linear

Explanation:

3 0
2 years ago
What is the kinetic energy of a 1,350kg car traveling at a speed of 12m/s?
Ainat [17]
Kinetic energy= 1/2 m v^2

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kinetic energy = 97200Joules
7 0
2 years ago
A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency \omega =2\pi f=2\times 3.14\times 120=753.6rad/sec

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt

From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

4 0
3 years ago
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