Is this right? plzz anwser soon

Physics

2 answers:

-Dominant- [34]3 years ago

6 0

Answer:

yes

Explanation:

Y_Kistochka [10]3 years ago

6 0

Answer:

y e p

Explanation:

You might be interested in

Batman (95kg) is standing on top of a 50m high building looking out over the city of Gotham. Given that he uses the potential en

Oksanka [162]

Answer:

47 kJoules (kJ)

Explanation:

Potential enegy on Earth is given by the relationship:

P.E. = mgh, where m is mass, g is the acceleration due to Earth's gravity, and h is height. Since we are given metric values, we will look for an answer that is consistent with Joules, the metric measure of energy. 1 Joule is defined as 1 kg*m^2/s^2, so we wnat units of kg, m, and sec.

We are given:

m = 95kg

h = 50 meters

Earth's gravity, g is 9.8 m/s^2

Enter the data:

P.E. = mgh

P.E. = (95kg)(9.8m/s^2)(50m)

P.E. = 46550 kg*m^2/s^2 or 46550 Joules(J)

Since we only have 2 sig figs, and since 1kJ =- 1000J

We can state the potential energy is 47kJ.

Spiderman has 47kJ of potential energy for the start of any dive back to Earth. [He needed that same amount of energy to reach that height, but we don't know from where it came. A jump, helicopter, beamed up by Scotty, or tossed up by Doctor Octopus.]

3 0

1 year ago

This illustration represents the compoundA)carbon oxide.B)carbon dioxide.C)carbon monoxide.EliminateD)monocarbon oxide.

pentagon [3]

B.) Carbon Dioxide because the carbon is surrounded by oxygen

3 0

3 years ago

Read 2 more answers

Why do the passengers in high-altitude jet planes feel the sensation of weight while passengers in an orbiting space vehicle, su

kykrilka [37]

Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.

In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.

Remember that the Force of Gravity is given under the principle

$F_g = \frac{GMm}{r^2}$