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mash [69]
3 years ago
5

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical accel

eration of the rocket is given by ay=(2.70m/s3)t, where the +y-direction is upward.What is the speed of the rocket when it is 325 m above the surface of the earth?
Physics
2 answers:
Inessa [10]3 years ago
8 0

Explanation:

The vertical acceleration of the rocket is given by :

a_y=2.7t

Acceleration is given by, a=\dfrac{dv}{dt}

\dfrac{dv}{dt}=2.7t

v=\int\limits{2.7\ t.dt}  

v=\dfrac{2.7t^2}{2}..............(1)

Velocity is given by, v=\dfrac{dx}{dt}

\dfrac{dx}{dt}=\dfrac{2.7t^2}{2}

x=\int\limits{\dfrac{2.7}{2}t^2}.dt

x=\dfrac{2.7}{6}t^3

From above equation, we can find the value of t at x = 325 m

325=\dfrac{2.7}{6}t^3

t = 8.97 s

Now put t = 8.97 s in equation (1) as :

v=\dfrac{2.7(8.97)^2}{2}

v = 108.62 m/s

So, the speed of the rocket when it is 325 meters above the surface of earth is 108.62 m/s. Hence, this is the required solution.

Mazyrski [523]3 years ago
5 0
a = 2.7t 
v =  \int\limits^t_0 {2.7t} \, dt =  \frac{2.7}{2} t^2 
x =  \int\limits^t_0 {\frac{2.7}{2} t^2} \, dt =  \frac{2.7}{6} t^3

Solve for v with x = 325.
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Explanation:

The complete question is given thus;

A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).

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q = -52.5 μC

cheers i hope this helped !!!

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