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4 years ago

13

If the particles of an object have more kinetic energy than the particles of an object, what must be true?

1 answer:

son4ous [18]4 years ago

7 0

If the objects comprise a gas, then the first object contains more
thermal energy (heat) than the second object.

If the objects are solid, then you can't draw any conclusion unless
both objects have the same total mass. If that's the case, then the
first object must be moving faster than the second one.

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-What do you think happened to make the Moon look the way it does?

goldenfox [79]

Answer: The physics of evolution had made the moon like it is today....Please watch this video from you tube about the evolution of the moon.

Explanation:

https://youtu.be/UIKmSQqp8wY

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3 years ago

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

3 years ago

A 300.0-kg speedboat is moving across a lake at 35.0 m/s.

Varvara68 [4.7K]

The answer is 300kg times the 35 m/s10,500 kg•m/s

5 0

3 years ago

Read 2 more answers

An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?

MatroZZZ [7]

Answer:

Explanation:

Given

Inclination $\theta =32.9^{\circ}$

Distance of landing point $R=7.78\ m$

Considering athlete to be an Projectile

range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

where u=launch velocity

$7.78=\frac{u^2\sin (2\times 32.9)}{9.8}$

$u^2=83.58$

$u=\sqrt{83.58}$

$u=9.142\ m/s$

(b)If u is increased by 8% then

new velocity is $u'=9.87\ m/s$

$R'=\frac{u'^2\sin 2\theta }{g}$

$R'=9.07\ m$

3 0

3 years ago

Read 2 more answers

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