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liraira [26]
3 years ago
8

HELP ITS FOR SCIENCE

Physics
2 answers:
yuradex [85]3 years ago
7 0
For number 2:
Speed is related to kinetic energy because kinetic energy is the energy of motion.

Hope this helped!
Rashid [163]3 years ago
7 0

Answer:

A wrecking ball has the capability to destroy a building due to the greater mass and kinetic energy than the yo-yo.

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3 years ago
What standard of measurement do most countries use and discuss why such a standard is beneficial
frosja888 [35]

I mean i guess its the metric system?

6 0
3 years ago
The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10−12w/m2. Find the
Monica [59]

Answer:

The distance d from the car at which the sound from the stereo can still be discerned = 97720.5 m

Explanation:

Sound intensity heard at distance is related to the distance with the relation = (power of sound at the source)/(surface area of the wall of an imaginary sphere at the distance in question)

I = P/4πd²

Assuming the car has 2 speakers,

P = 0.06 W × 2 = 0.12 W

d = ?

For the intensity of the least discernible sound,

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

8 0
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
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