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2 years ago

5

2 How does Descartes' "quality of motion" differ from the modern momentum?

1 answer:

Rzqust [24]2 years ago

6 0

Answer: Descartes was more of speed which defers from modern day velocity.

Explanation:

Descartes law if conservation referred or defined “motion” rather than “momentum” as what is obtainable in today's world as ”speed” the rate at which something moves rather than “velocity” which is a product of speed and direction. So in conclusion Descartes was more of speed which defers from modern day velocity.

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Which of these equations will you used to find the final velocity if the initial

Svetach [21]

Answer:

2.77 would be the answer for this

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2 years ago

What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied

bija089 [108]

Answer:

400N/m

Explanation:

K = F/x

= 4/0.01

= 400 N/m

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2 years ago

If the engine receives 6.45 kJ of heat energy from the reservoir at 520 K in each cycle, how many joules per cycle does it rejec

Oxana [17]

Answer:

3.72 kJ

Explanation:

QH = 6.45 kJ

TH = 520 K

Tc = 300 K

Qc = ?

By use of Carnot's theorem

Qc / QH = Tc / TH

Qc / 6.45 = 300 / 520

Qc = 3.72 kJ

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2 years ago

A runaway railroad car, with mass 30x10^4 kg, coasts across a level track at 2.0 m/s when it collides with a spring loaded bumpe

Natalija [7]

Answer:

0.775 m

Explanation:

As the car collides with the bumper, all the kinetic energy of the car (K) is converted into elastic potential energy of the bumper (U):

$U=K\\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

where we have

$k=2\cdot 10^6 N/m$ is the spring constant of the bumper

x is the maximum compression of the bumper

$m=30\cdot 10^4 kg$ is the mass of the car

$v=2.0 m/s$ is the speed of the car

Solving for x, we find the maximum compression of the spring:

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(30\cdot 10^4 kg)(2.0 m/s)^2}{2\cdot 10^6 N/m}}=0.775 m$

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2 years ago

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a_sh-v [17]

Fa sho it's 30j edge said so

6 0

2 years ago

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