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2 years ago

IF SOME ONE SENDS U A LINK DO NOT PUSH ON IT CUZ THEY WILL HACK UR STUFF .................. BE CAREFUL PLZ

Physics

1 answer:

Karolina [17]2 years ago

7 0

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A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass

natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet = 3.56×10²⁶ Kg.

5 0

2 years ago

Martin needs to push a box up a ramp.

creativ13 [48]

Uh I think it is inertia but I could be wrong

8 0

2 years ago

While driving north at 25 m/s during a rainstorm you notice that the rain makes an angle of 38° with the vertical. While driving

Natali [406]

Answer: $21.33^{\circ}$

Explanation:

Given

velocity of driver $v_1$ =25 m/s w.r.t ground towards north

driver observes that rain is making an angle of $38^{\circ}$ with vertical

While returning $v_2$ =25 m/s w.r.t. ground towards south

suppose $u_1$ =velocity of rain drop relative car while car is going towards north

$u_2$ =velocity of rain drop relative car while car is going towards south

z=vector sum of $u_1 & v_1$

Now from graph

$\tan 38=\frac{v_1+v_2}{u_2}$

$u_2=\frac{25+25}{\tan 38}=64 m/s$

$z=\vec{u_2}+\vec{v_2}$

therefore magnitude of z is given by

$|z|=\sqrt{u_2^2+v_2^2}$

$|z|=\sqrt{64^2+25^2}$

$|z|=68.70 m/s$

$\tan A=\frac{v_2}{u_2}$

$\tan A=\frac{25}{64}=0.3906$

$A=21.33^{\circ}$

Thus rain drops make an angle of $21.33^{\circ}$ w.r.t to ground

6 0

2 years ago

Identify the charge(s) with increasing electric potential energy.

kap26 [50]

Cznxjsjssbsishsisbsisbsisbsbeoeb

4 0

2 years ago

A person pushes two boxes with a horizontal force F of magnitude of 100 N.

Monica [59]

The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".

According to the question,

Mass of box A,

$m_A = 9\ kg$

Mass of box B,

$m_B = 2 \ kg$

Horizontal force,

$F_{app} = 100 \ N$

From the Newton's law,

→ $F_{app} = (\frac{F_{app}}{m_A+m_B} )a$

or,

→ $a = \frac{F_{app}}{(m_A+m_B)}$

Bu substituting the values, we get

→ $= \frac{100}{9+2}$

→ $= \frac{100}{11}$

→ $9.10 \ m/s^2$

We can see that between the two boxes, the action-reaction pair exist.

then,

→ $F_{action-reaction} = m_b \ a$

→ $=2\times 9.10$

→ $= 18.2 \ N$ (magnitude)

Thus the above solution is appropriate.

Learn more about the magnitude here:

brainly.com/question/13545862

7 0

2 years ago