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vazorg [7]
4 years ago
12

If you know the distance an object has traveled in a certain amount of time you can determine

Physics
2 answers:
lidiya [134]4 years ago
6 0
You can determine the velocity it is moving at. 

Nutka1998 [239]4 years ago
5 0
With that information, you can determine the object's speed.
Just divide the distance covered by the time to cover the distance.

If you also know the direction the object moved, then you can
determine its velocity.  If you don't, then you can't.
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Describe the changes in the velocity of a ball thrown straight up into the air. then describe the changes in the ball's accelera
expeople1 [14]

There is a displacement. Just because the ball is thrown up, and not crossways, doesn't mean its location is not moving. Remember, positive displacement is together a displacement in the direction east, right, and up.  The velocity is the distance over time. To compute that, you must look how high the ball moved before falling back down. Acceleration is expected to be constant at 9.80m/s^2. That is the force of gravity. But remember that you are disregarding air friction when you are computing the acceleration.

7 0
3 years ago
A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th
kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

P_1, V_1, T_1 & P_2, V_2, T_2

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container  therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

-Q=W+U_1----1

Q=-W+U_2-----2

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

U_1 & U_2 change in internal Energy of gas

Thus from 1 & 2 we can say that

U_1+U_2=0

n_1c_v(T-T_1)+n_2c_v(T-T_2)=0

T(n_1+n_2)=n_1T_1+n_2T_2

T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}

where n_1=\frac{P_1V_1}{RT_1}

n_2=\frac{P_2V_2}{RT_2}

T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}

T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}

and P=\frac{P_1V_1+P_2V_2}{V_1+V_2}

6 0
3 years ago
I need to find 1).a,b,c
Aleksandr [31]
Let's cut through the weeds and the trash
and get down to the real situation:

                  A stone is tossed straight up at  5.89 m/s .
                  Ignore air resistance.


Gravity slows down the speed of any rising object by  9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

                     (5.89 m/s / 9.8 m/s²)  =  0.6 seconds.

At that timer, he has run out of upward gas.  He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is  (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of  2.945 m/s,
he topped out at

                    (2.945 m/s) (0.6 s) =  1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down)  =  1.2 seconds until he hits rubber again.



 
5 0
4 years ago
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