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Afina-wow [57]
2 years ago
5

Calculate the force of Earth’s gravity on a spacecraft 2.00 Earth radii above the Earth’s surface (That would be 3.00 Earth radi

i from the center of Earth) if its mass is 1850 kg.
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
igomit [66]2 years ago
3 0

Answer:

2014.44 N

Explanation:

mass of spacecraft, m = 1850 kg

distance r = 3 x R

where r be the radius of earth.

g be the acceleration due to gravity on the surface of earth and g' be the acceleration due to gravity at height

\frac{g'}{g}=\left (\frac{R}{r}  \right )^{2}

\frac{g'}{g}=\left (\frac{R}{3R}  \right )^{2}

g' = g / 9

g' = 9.8 / 9 = 1.089 m/s²

Force of gravity on the space craft

F = m g' = 1850 x 1.089

F = 2014.44 N

Thus, the force of gravity on the space craft at height is 2014.44 N.

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A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?
zloy xaker [14]
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<span> NaCl - it's a salt (we can't measure the pH)
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He needs to use HF (Hydrogen fluoride) to decrease the pH.


7 0
2 years ago
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By standard convention, both the electric potential and the the electric potential energy between two charges is taken to be zer
shusha [124]

Answer: at when distance r = infinity.

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V = q/4πεr

V = electric potential (volts)

q = magnitude of electric charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

In the equation above, it can be seen that only electric potential (v) and distance (r) is a variable, and there is an inverse relationship between them (an increase in one leads to a decrease in the other)

Thus to have zero value of electric potential (v= 0) we have to have the largest value of r ( r = infinity).

Same goes for electric potential energy between two charges, the formulae is given below as

W = q1 *q2/4πεr

W= electric potential energy

q1 = magnitude of first charge.

q2 = magnitude of second charge

ε = permittivity of free space

r = distance between arbitrary point and charge.

Also, all values are constant aside from electric potential energy (w) and distance (r) which have an inverse relationship.

Thus to have zero value of electric potential energy (w =0), we have to get an infinite value of distance ( r =infinity)

6 0
2 years ago
50 J of work was performed in 20 seconds. How much power was used to do this task?
yuradex [85]
Power=work/time
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7 0
2 years ago
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At its lowest setting a centrifuge rotates with an angular speed of ω1 = 250 rad/s. When it is switched to the next higher setti
dalvyx [7]

Answer:

Part(a): The angular acceleration is 5.63~rad~s^{-2}.

Part(b): The angular displacement is 2629~rad.

Explanation:

Part(a):

If \omega_{1},~\omega_{2}~and~\alpha be the initial angular speed, final angular speed and angular acceleration  of the centrifuge respectively, then from rotational kinematic equation, we can write

\alpha = \dfrac{\omega_{2} - \omega_{1}}{t}......................................................(I)

where 't' is the time taken by the centrifuge to increase its angular speed.

Given, \omega_{i} = 250~rad~s^{-1}, \omega_{f} = 750~rad~s^{-1} and t = 9.5~s. From equation (I), the angular acceleration is given by

\alpha = \dfrac{750 - 250}{9.5}~rad~s^{-2} = 5.63~rad~s^{-2}

Part(b):

Also the angular displacement (\Delta \theta) can be written as

&&\Delta \theta = \omega_{1}~t + \dfrac{1}{2}\alpha~t^{2}\\&or,& \Delta \theta = (250 \times 9.5 + \dfrac{1}{2} \times 5.63 \times 9.5^{2})~rad = 2629~rad

8 0
2 years ago
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