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My name is Ann [436]
3 years ago
5

Providing that it does not collide with another object, what is the most likely reason that an object moving through space would

change velocity?
Physics
1 answer:
Ronch [10]3 years ago
4 0

<em>GRAVITY !</em>

The object doesn't have to collide or even TOUCH something else in order to feel a gravitational force toward it.

If there's a gravitational force between the space object and ANYTHING else, (and there always is !), then that force will change the object's speed or direction of motion.

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Which source of energy would you prefer for cooking and why?
timurjin [86]

Answer:

gas

Explanation:

because It is fast and easy

3 0
3 years ago
If two particles have equal kinetic energies, are their momenta necessarily equal? explain.
Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

K.E=\frac{1}{2}mv^{2}

Similarly the momentum is given by m\times v

For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective kinetic energies is given by

K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}

K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}

Similarly For 2 particles with masses m_{1},m_{2}and moving with velocities v_{1},v_{2} respectively the respective momenta are given by

p_{1}=m_{1}\times v_{1}

p_{2}=m_{2}\times v_{2}

Now since it is given that the two kinetic energies are equal thus we have

\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0
3 years ago
-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck
Hoochie [10]

Answer:

I would shout fore help if I was being raped or try to make him or her stop

6 0
3 years ago
A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and
s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

acceleration due to gravity (g) = 9.8 m/s^{2}

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r =  m x g x μ

v^2 / r =  g x μ  .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r =  g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0
3 years ago
Before Freddy lands on the skateboard it has a certain momentum. After landing, the skateboards momentum
Nadya [2.5K]

Answer:

remains the same

Explanation:

Momentum refers to the quantity of motion of a body. When any body of mass moves, it possess momentum. Numerically,

Momentum =  mass x velocity

i.e. momentum is the product of the mass x velocity

Momentum of a body is always conserved.

In the context, the skateboard has certain momentum before Freddy lands on it. After Freddy lands, the momentum of skateboard remains the same, there is no change in the momentum.

This is because, here the momentum is conserved. After Freddy lands on the skateboard, the total mass on the skateboard increases and so the velocity decreases making the momentum same before the landing.

3 0
2 years ago
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