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Aliun [14]
2 years ago
15

beginning in 1996, a glacier lost an average of 3.7 meters of thickness each year. find the total change in its thickness by the

end of 2012.
Physics
1 answer:
Harman [31]2 years ago
6 0
51.8 meters lost (excluding 1996)
years 1996-2012(including 1996)-17 yrs*3.7 =62.9 meters
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Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up
kirza4 [7]

Answer:

I believe the answer is b

Explanation:

5 0
3 years ago
A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant
Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

6 0
3 years ago
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iragen [17]
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8 0
2 years ago
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When the biker is at the top of the ramp, he has a speed of 10 m/s at
oksian1 [2.3K]

Answer:

25000

Explanation:

100 x 25 x 10 = 25000

8 0
3 years ago
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
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