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choli [55]
3 years ago
8

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

h with enough speed to make it to the moon. Some information that will help during this problem:mearth = 5.9742 x 1024 kgrearth = 6.3781 x 106 mmmoon = 7.36 x 1022 kgrmoon = 1.7374 x 106 mdearth to moon = 3.844 x 108 m (center to center)G = 6.67428 x 10-11 N-m2/kg2On your first attempt you leave the surface of the earth at v = 5534 m/s. How far from the center of the earth will you get?
Physics
1 answer:
SCORPION-xisa [38]3 years ago
7 0

Answer:

The answer is 5102,91 m.

Explanation:

To find<u> escape velocity</u>:

\frac{1}{2} mV^{2} =G\frac{Mm}{r^{2} }. Therefore:

V=\sqrt{\frac{2GM}{r} }

Using the necessary informations:

5534=\sqrt{\frac{2(6,67428.10^{-11)(5,9742.10^{24}) } }{r}}

as we solve the equation we find r=5102,91m

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3 years ago
Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.
ivolga24 [154]

Answer: 2.068*10^{-14}m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602*10^{-16}

h= planck constant = 6.626*10^{-34}

c= speed of light =2.998* 10^{8}

v= potential difference= 6*10^{4}

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= \frac{hc}{qv}

λ = 6.626*10^{-34} * 2.998* 10^{8} / 1.602*10^{-16} * 6*10^{4}

λ = \frac{19.878*10^{-26} }{9.612*10^{-12} }

by doing the necessary calculations, we have that

λ = 2.068*10^{-14}m

8 0
3 years ago
Use the values provided to calculate the initial voltage
EleoNora [17]

Answer: V1= 360 V

Explanation:

7 0
3 years ago
Read 2 more answers
How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?
ExtremeBDS [4]

Answer:

The value is T =  122036.2 \  N

Explanation:

From the question we are told that

     The mass of the car is  m  = 6526 \  kg

      The acceleration  is  a=  8.9 \  m/s^2

Generally the net force applied on the rope is mathematically represented as

          F_{net}   =  T -  W

Here W is the weight of the car which is evaluated as

         W =  m * g

=>      W =  6526  * 9.8

=>       W =  63954.8 \  N

Generally the net force can also be mathematically represented as

       F =  m * a

So

        m * a  =  T  -  63954.8

=>     6526  *  8.9 =  T  -  63954.8

=>      T =  122036.2 \  N

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3 years ago
If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.
Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

If we use the formula our point of investigation is Height of image,

H2= D2/D1× H1

H2 = 2D2/D1 × H1

H2 = 2H1

6 0
3 years ago
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