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prohojiy [21]
1 year ago
6

51. A balloon starts rising from the ground, vertically upwards, uniformly at the rate of 1 m s-1. At the end of 4 seconds, a bo

dy is released from the balloon. Calculate the time taken by the released body to reach the ground. 51. A balloon starts rising from the ground , vertically upwards , uniformly at the rate of 1 m s - 1 . At the end of 4 seconds , a body is released from the balloon . Calculate the time taken by the released body to reach the ground​
Physics
1 answer:
strojnjashka [21]1 year ago
5 0

Answer:

4s

Explanation:

when a bOdy rises into the air,the time it takes to reach a particular height is the same as the time it will take the body to fall from that height to the ground.

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A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
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The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

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1 year ago
Share your thoughts about this statement by John Wesley<br> "Electricty is the soul of universe"
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<em>I think this statement is true at least in modern day times. The world runs on nothing but technology. People use more technology instead of old school textbooks and papers. Imagine living in a world without technology… that means no cars, no trains, no devices, no machines like your stove or printer, no lights and lots more. John Wesley is 100 percent correct with this statement. Electricity is indeed the powerhouse of the universe. </em>

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What has been happening to the cosmic microwave background radiation since the Big Bang?​
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Answer:

Explanation:

Cosmologists refer to a "surface of last scattering" when the CMB photons last hit matter; after that, the universe was too big. So when we map the CMB, we are looking back in time to 380,000 years after the Big Bang, just after the universe was opaque to radiation. But the CMB was first found by accident.

plz mark as brainliest

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3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
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Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

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3 years ago
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