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mote1985 [20]
1 year ago
15

Give three examples of objects in equilibrium found in classroom?​

Physics
1 answer:
Charra [1.4K]1 year ago
7 0

Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

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Jupiter's satellite Europa orbits Jupiter with a period of 3.55 d at an orbital radius of 6.71 108 m (assume the orbit to be cir
yawa3891 [41]

Answer:

(a)

M = 1.898 x 10^27 kg

(b)

v = 13.74 km/s

(c) E = 0.28 N/kg

Explanation:

Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s

Radius, r = 6.71 x 10^8 m

G = 6.67 x 10^-11 Nm^2/kg^2

(a) T=2\pi \sqrt{\frac{r^{3}}{GM}}

M=\frac{4\pi ^{2}r^{3}}{GT^{2}}

M=\frac{4\times3.14^{2}\times 6.71^{3}\times 10^{24}}{6.67\times 10^{-11}\times 306720^{2}}

M = 1.898 x 10^27 kg

(b) Let v be the orbital velocity

v=\frac{2\pi r}{T}

v=\frac{2\times 3.14\times 6.71\times 10^{8}}{306720}

v = 13739.5 m/s

v = 13.74 km/s

(b) The gravitational field E is given by

E = \frac{GM}{r^{2}}

E = \frac{6.67\times10^{-11}\times 1.898\times 10^{27}}{6.71^{2}\times 10^{16}}

E = 0.28 N/kg

6 0
2 years ago
1. A 700 kg satellite is in orbit 2.4 x 106 meters from the center of the Earth. What is the force of
nevsk [136]

Answer:

4.8 \cdot 10^4 N

Explanation:

The force of gravity acting on the satellite is given by:

F=\frac{GMm}{r^2}

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

m is the mass of the satellite

r is the distance of the satellite from the Earth's centre

Here we have

m = 700 kg

r=2.4\cdot 10^6 m

Substituting into the equation, we find:

F=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(700)}{(2.4\cdot 10^6)^2}=4.8 \cdot 10^4 N

<em>Note that the distance mentioned in the problem (2.4 x 10^6 meters) is not realistic, since it is less than the radius of the Earth (6.37 x 10^6 meters).</em>

3 0
3 years ago
Which chemical reaction is fastest
Nady [450]

B        BRIANLIEST PLS

Explanation:

3 0
2 years ago
100mL of 4°C water is heated to 37 °C . Assume the density of the water is 1g/mL. The specific heat of water is 4.18 J/g(°C). Wh
swat32

Answer:

13807.2  J/g°C

Explanation:

I just took the test and got it correct

6 0
3 years ago
Freezing involves the ______________ of latent heat and a change is state from the ______________ phase to the solid phase.
DaniilM [7]
<span>Freezing involves the decrease in value of latent heat by 80 Cal/gm and a change of state from the Liquid phase to the solid phase.

So, in short, Fill in the blank as follows:
1st blank = Release/decrease
2nd blank = Liquid

Hope this helps!

</span>
7 0
3 years ago
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