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3 years ago

Please help me with my quiz

Physics

1 answer:

Oduvanchick [21]3 years ago

8 0

Answer:

matter: A

data: E

variable: C

Controlled: D

Physical: B

Explanation:

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While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38° with the vertical. while driving

mamaluj [8]

The car's reference frame should be your go to. Just have the car be stationary (modifying the velocity of the rain rather than the car's). From there you should be able to get the rain's velocity by a simple equation set.

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4 years ago

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2. A test model for an experimental gasoline engine does 45 J of work in one cycle

loris [4]

Answer: im just trying to get some points so i can answer my question.

sorry

Explanation:

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3 years ago

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F

muminat

Answer:

$v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

Explanation:

Given that the force of the particle is,

$F_{x}=F_{0}e^{-kt}$

Now it can be further written as

$m\frac{dv}{dt}= F_{0}e^{-kt}\\\frac{dv}{dt}=\frac{ F_{0}}{m} e^{-kt}\\dv=\frac{ F_{0}}{m}e^{-kt}dt\\ v=\frac{ F_{0}}{-km}e^{-kt}+C$

Now the initial conditions are v=1 at t=0.

So,

$1=\frac{ F_{0}}{-km}e^{0}+C\\C=1+\frac{ F_{0}}{km}$

Now the velocity will become.

$v_{x} (t)=\frac{ F_{0}}{-km}e^{-kt}+1+\frac{ F_{0}}{km}\\v_{x} (t)=1+\frac{ F_{0}}{km}(1-e^{-kt})$

And,

$\frac{dx}{dt} =1+\frac{ F_{0}}{km}(1-e^{-kt})\\dx=(1+\frac{ F_{0}}{km}(1-e^{-kt}))dt\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+C\\$

And, another initial condition is x=0 at t=0

$0=0+\frac{ F_{0}0}{km}+\frac{ F_{0}t}{k^{2} m}e^{0}+C\\C=-\frac{ F_{0}t}{k^{2} m}$

Now,

$x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}e^{-kt}+-\frac{ F_{0}t}{k^{2} m}\\x=t+\frac{ F_{0}t}{km}+\frac{ F_{0}t}{k^{2} m}(e^{-kt}-1)$

5 0

3 years ago

For a wave, the _____ the amplitude, the _____ energy the wave carries.

sattari [20]

Answer:

For a wave, the HIGHER the amplitude, the MORE energy the wave carries.

8 0

3 years ago

Two identical small charged spheres are a certain distance apart, and each initially experiences an electrostatic force of magni

Rom4ik [11]

Answer:

New force, $F'=\dfrac{1}{4}F$

Explanation:

The electrostatic force between two spheres is given by :

$F=\dfrac{kq_1q_2}{r^2}$

According to given condition, each of the spheres has lost half its initial charge, new force is given by :

$F'=\dfrac{kq_1/2q_2/2}{r^2}$

$F'=\dfrac{1}{4}\times \dfrac{kq_1q_2}{r^2}$

$F'=\dfrac{1}{4}F$

So, the new force becomes (1/4)th of the initial force. Hence, the correct option is (d).

4 0

4 years ago