Object dropped from height of 25 meters, what velocity will it hit the ground?

n astronaut who weighs 800 N on the surface of the earth lifts off from planet Zuton in a space ship. The free-fall acceleration

ANTONII [103]

Answer: 0.29 kN

Explanation:

We have the following data:

$W_{E}=800 N$ is the weight of the astronaut on Earth

$g_{E}=9.8 m/s^{2}$ is the free fall acceleration due gravity on Earth (directed downwards)

$g_{Z}=3 m/s^{2}$ is the free fall acceleration due gravity on Zuton (directed downwards)

$a=0.5 m/s^{2}$ is the acceleration of the spaceship at litoff (directed upwards)

We have to find the <u>magnitude of the force</u> $F$ the space ship exerts on the astronaut.

Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:

$W_{E}=mg_{E}$ (1)

Where $m$ is the mass of the atronaut.

Isolating $m$ :

$m=\frac{W_{E}}{g_{E}}$ (2)

$m=\frac{800 N}{9.8 m/s^{2}}$ (3)

$m=81.63 kg$ (4)

Now that we know the mass of the astronaut, we can find its weight on Zuton:

$W_{Z}=mg_{Z}$ (5)

$W_{Z}=(81.63 kg)(3 m/s^{2})$ (6)

$W_{Z}=244.89 N$ (7)

Then, we can calculate the force the space ship exerts on the astronaut by the following equation:

$F-W_{Z}=m.a$ (8)

Isolating $F$ :

$F=m.a+W_{Z}$ (9)

$F=(81.63 kg)(0.5 m/s^{2})+244.89 N$ (10)

$F=285.7 N \frac{1 kN}{1000 N}=0.285 kN$ (11)

Finally:

$F=0.285 kN \approx 0.29 kN$

5 0

4 years ago

A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring balance C and D. Balance C is

vagabundo [1.1K]

A uniform metal rod with of length 80cm and a mass of 3.2kg is supported horizontally by two vertical spring balances C and D. Balance C is 20cm from one end while D is 30cm from the other end would show the reading of 1.06 Kg and 2.13 kg respectively

<h3>What is gravity?</h3>

It can be defined as the force by which a body attracts another body towards its center as the result of the gravitational pull of one body and another, The gravity varies according to the mass and size of the body for example the force of gravity on the moon is the 1/6th times of the force of gravity on the earth.

As given in the problem, A uniform metal rod of the length of 80cm and mass of 3.2kg is supported horizontally by two vertical springs balance C and D. Balance C is 20cm from one end while D is 30cm from the other end

The weight of the rod acting downward is from the center of the rod at 40 cm

Let us suppose the reading on the spring balance C and D are P and Q respectively

By using the equilibrium for the vertical force

Fv=0

P + C = 3.2

By using the equilibrium for the moment around the left corner

20×P+ 50×Q= 40 ×3.2

By solving for both P and Q from the above two equations we would get

P =1.06 and Q = 2.13

Thus, the reading on the spring balance C and D would be 1.06 Kg and 2.13 kg respectively

Learn more about gravity from here

brainly.com/question/4014727

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5 0

1 year ago

Witch statement accurately describes radiation

sammy [17]

Answer

Explanation:

radiaton is used from bombs but radiation and to cure cancer it is a toxic gas that if you smell it you can die

7 0

4 years ago

What is happening inside muscles when they become sore?

Verizon [17]

Answer:

When muscles are required to work harder than they're used to or in a different way, it's believed to cause microscopic damage to the muscle fibres, resulting in muscle soreness or stiffness. DOMS is often mistakenly believed to be caused by a build up of lactic acid, but lactic acid isn't involved in this process.

Explanation:

4 0

3 years ago

A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a

Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

$x=\sqrt {\frac {2mgh}{k}}$

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

$x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m$

5 0

3 years ago