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bagirrra123 [75]
4 years ago
14

A space shuttle can accelerate at a rate of 2.5m/s squared. Given that, and the fact that the engines of the space shuttle can p

roduce a force of about 1,600,000N, determine the mass(in kilograms) of the space shuttle
Physics
2 answers:
Afina-wow [57]4 years ago
8 0

Answer:

640000 kg

Explanation:

Formula is

Force = mass * acceleration

Mass = force / acceleration

Mass = 1600000 / 2.5

Mass = 640000 kg

ira [324]4 years ago
4 0
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Tides occur in oceans but not in lakes why?​
lawyer [7]

Answer:

Explanation:

Tides occur mainly in oceans because that is basically one huge body of water that is free to move all over the earth. Lakes and rivers do not cover enough area to have their water be moved significantly by gravity, or in other words, to have tides.

7 0
3 years ago
Read 2 more answers
you are sitting behind the bus driver on a moving bus in relation to a person standing on the sidewalk you are what
grigory [225]

You are sitting behind the bus driver on a moving bus in relation to a person standing on the sidewalk you are what

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relative to sidewalk you are moving with the driver

4 0
4 years ago
The masses of the earth and moon are 5.98 x 1024 and 7.35 x 1022 kg, respectively. Identical amounts of charge are placed on eac
dybincka [34]

Answer:

The magnitude of charge on each is 5.707\times 10^{13} C

Solution:

As per the question:

Mass of Earth, M_{E} = 5.98\times 10^{24} kg

Mass of Moon, M_{M} = 7.35\times 10^{22} kg

Now,

The gravitational force of attraction between the earth and the moon, if 'd' be the separation distance between them is:

F_{G} = \frac{GM_{E}M_{M}}{d^{2}}        (1)

Now,

If an identical charge 'Q' be placed on each, then the Electro static repulsive force is given by:

F_{E} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}           (2)

Now, when the net gravitational force is zero, the both the gravitational force and electro static force mut be equal:

Equating eqn (1) and (2):

\frac{GM_{E}M_{M}}{d^{2}} = \frac{1}{4\pi\epsilon_{o}}\frac{Q^{2}}{d^{2}}

(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22}) = (9\times 10^{9}){Q^{2}}

\sqrt{\farc{(6.67\times 10^{- 11})\times (5.98\times 10^{24})\times (7.35\times 10^{22})}{9\times 10^{9}}} = Q

Q = \pm 5.707\times 10^{13} C

7 0
3 years ago
Attempting to impress the skeptical patrol officer with your physics knowledge, you claim that you were traveling so fast that t
horsena [70]

Answer:

v_r = 1.268 × 10⁸ mi/hr

Explanation:

We are given;

wavelength of the red light; λr = 693 nm = 693 × 10^(-9) m

wavelength of the yellow light; λy = 582 nm = 582 × 10^(-9) m

Frequency is given by the formula;

f = v/λ

Where v is speed of light = 3 x 10^(8) m

Frequency of red light; f_o = [3 x 10^(8)]/(693 × 10^(-9)) = 4.33 x 10¹⁴ Hz

Similarly, Frequency of yellow light;

f = [3 x 10⁸]/(582 × 10^(-9)) = 5.15 x 10¹⁴ Hz

To find the speed of the car, we will use the formula;

f = f_o[(c + v_r)/c)]

Where c is speed of light and v_r is speed of car.

Making v_r the subject;

cf/f_o = c + v_r

v_r = c(f/f_o - 1)

So, plugging in the relevant values, we have;

v_r = 3 × 10⁸[((5.15 x 10¹⁴)/(4.33 x 10¹⁴)) - 1]

v_r = 3 × 10⁸(0.189)

v_r = 5.67 x 10⁷ m/s

Converting to mi/hr, 1 m/s = 2.23694 mile/hr

So, v_r = 5.67 × 10⁷ × 2.23694

v_r = 1.268 × 10⁸ mi/hr

5 0
4 years ago
A boy throws a ball straight up into the air. It reaches the highest point of its flight after 4 seconds.
Zanzabum

Answer:Take into account the the Earths gravity is 9.8 meters a second

Explanation:

Gravity pulls down on the ball at g=-9.81 m/s^2. Up is positive, down is negative.The ball started at a certain initial velocity of Vi m/s. Time it took is t=4s. Final velocity is Vf=0 m/s, because at the highest point the ball stops moving.

Vf=(g*t)+Vi

Rearrange for Vi.

Vi=Vf-(g*t)

Vi=0-(-9.81*4)=39.24 m/s (upward)

Think about it this way for the non-mathematical approach. The ball stops at the top. The initial velocity gets reduced by 9.81 m/s every second, and reaches 0 m/s at the top. It took 4 seconds, so 9.81*4 is equal to the initial upward velocity.

8 0
3 years ago
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