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Paraphin [41]
2 years ago
7

Why is the same side of the Moon always visible from Earth?

Physics
1 answer:
son4ous [18]2 years ago
8 0

Answer:

Explanation:

Answer

The true fact is that C is what happens in outer space.  Both rotations take 27.3 days.  

A: The exact opposite is true. It does rotate about it's axis.

B: Again this is just plain false. Given the way we observe it, the moon must be rotating around the earth.

D. they don't. 27.3 hours and 24 hours are not the same.

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Calculate the kinetic energy of a car of mass 800kg moving at 100 kmph.
valentina_108 [34]

Answer:

KE = 308642.02469136 J

4 0
2 years ago
A graduated cylinder.measures 15.3 mL. Convert this measurement to DaL
ololo11 [35]

Answer:

0.000153DaL  

Explanation:

We have been given:

         15.3mL to convert to DaL

DaL is a unit of volume which indicates a decaliter.

 This implies that;

             1 Da L  = 1 x 10²L

So:

               1 mL  = 1 x 10⁻³L

       So 15.3mL will give 15.3 x 10⁻³L

So;

           1 x 10²L   =  1 DaL  

      15.3 x 10⁻³L  will give \frac{15.3 x 10^{-3} }{1 x 10^{2} }   = 15.3 x 10⁻⁵DaL

Therefore, this is 0.000153DaL  

           

5 0
3 years ago
Can we divide two vectors?
Likurg_2 [28]
Yes i think so im pretty sure
8 0
3 years ago
A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal
Natalija [7]

Answer : The specific heat of unknown sample is, 8748.78J/kg^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-[q_2+q_3]

m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]

where,

c_1 = specific heat of unknown sample = ?

c_2 = specific heat of water = 4186J/kg^oC

c_3 = specific heat of copper = 390J/kg^oC

m_1 = mass of unknown sample = 72.0 g  = 0.072 kg

m_2 = mass of water = 203 g  = 0.203 kg

m_2 = mass of copper = 187 g  = 0.187 kg

T_f = final temperature of calorimeter = 39.4^oC

T_1 = initial temperature of unknown sample = 80.0^oC

T_2 = initial temperature of water and copper = 11.0^oC

Now put all the given values in the above formula, we get

0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]

c_1=8748.78J/kg^oC

Therefore, the specific heat of unknown sample is, 8748.78J/kg^oC

7 0
3 years ago
If a stone with an original velocity was of 0 Is falling from a ledge and takes 8 seconds to hit the ground what is the final ve
Llana [10]
A=(v-u)/t
10=v/8
v=80(unit)
3 0
3 years ago
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