Physics, 30.06.2019 07:20 Jasten

Aspaceship hovering over the surface of venus drops an object from a height of 17 m. how much longer does it take to reach the surface than if dropped from the same height on earth? neglect air resistance in both cases.

Answers: 3

Show answers

Answers

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

$t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s$

Answer from: shilohtito

Explanation:

It is given that,

A spaceship hovering over the surface of Venus drops an object from a height of 17 m, h = 17 m

Acceleration due to gravity on the surface of Venus, $g'=8.87\ m/s^2$

The second equation of motion is given by :

$s=ut+\dfrac{1}{2}gt^2$ , u = 0

$s=\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2s}{g'}}$

$t=\sqrt{\dfrac{2\times 17}{8.87}}$

t' = 1.95 s..............(1)

Acceleration due to gravity on the surface of Earth, $g=9.8\ m/s^2$

Again using second equation of motion to find the time taken on the surface of earth.

$s=ut+\dfrac{1}{2}gt^2$ , u = 0

$s=0+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2\times 17}{9.8}}$

t = 1.86.................(2)

Change in time, $\Delta t=t'-t=1.95-1.86=0.089\ s$

So, on the surface of earth it will taken 0.089 seconds more. Hence, this is the required solution.

Answer from: Quest

Commer= truck drivertravel clerk=bachelors actuary= techinal program

Answer from: Quest

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

Another question on Physics

Physics, 21.06.2019 20:20

Astronomers discover an exoplanet (a planet of a star other than the sun) that has an orbital period of 3.87 earth years in its circular orbit around its sun, which is a star with a measured mass of 3.59 x 1030 kg. find the radius of the exoplanet's orbit.

Answers: 3

Answer

Physics, 22.06.2019 13:00

Which of the following correctly describes what happens when an atomic bomb explodes? small pieces of fissionable material are joined and form a body with a mass greater than the critical mass, the relative number of neutrons escaping decreases, and a chain reaction and explosion result. large pieces of fissionable matter are brought together quickly and form a body with a mass smaller than the critical mass, the relative number of escaping neutrons increases, and a chain reaction and explosion result.

Answers: 2

Answer

Physics, 22.06.2019 18:30

Against his financial advisor's advice, frank has decided to invest his money in some risky stocks because he once made quite a bit of money in the stock market. his decision illustrates a. the representativeness heuristic b. overconfidence c. the availability heuristic d. confirmation bias select the best answer from the choices provided

Answers: 3

Answer

Physics, 23.06.2019 00:30

Arope has one end tied to a vertical support. you hold the other end so that the rope is horizontal. if you move the end of the rope back and forth with a frequency of 4 hz, the transverse wave you produce has a wavelength of 0.5 m. what is the speed of the wave in the rope?

Answers: 1

Answer

You know the right answer?

Aspaceship hovering over the surface of venus drops an object from a height of 17 m. how much longer...

Questions