1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Westkost [7]
1 year ago
11

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

ergy to swoop down to save his city, how much energy does he have by going up so high?
Physics
1 answer:
Oksanka [162]1 year ago
3 0

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.]

You might be interested in
During strengthening heat treatment, the _______ step traps the material in an unstable crystalline structure. a)-Quenching, b)-
ZanzabumX [31]

Answer: A) Quenching

Hope this helps

4 0
3 years ago
Consider a block on a spring oscillating on a frictionless surface. The amplitude of the oscillation is 11 cm, and the speed of
IRISSAK [1]

Answer:

The angular frequency of the block is ω = 5.64 rad/s

Explanation:

The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.

Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.

The angular frequency of the oscillation ω is

ω = v/r

ω = 62 cm/s ÷ 11 cm

ω = 5.64 rad/s

So, the angular frequency of the block is ω = 5.64 rad/s

6 0
4 years ago
A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.4 m/
Bad White [126]

Answer:

1.06 secs

Explanation:

Initial speed of sled, u = 8.4 m/s

Final speed of sled, v = 5.8 m/s

Coefficient of kinetic friction, μ = 0.25

Using the impulse momentum theory, we know that the impulse applied to the sled is equal to change in momentum of the sled:

FΔt = mv - mu

where m = mass of the object

Δt = time interval

F = force applied

The force applied on the sled is the frictional force, which is given as:

F = -μmg

where g = acceleration due to gravity

Therefore:

-μmgΔt =  mv - mu

-μmgΔt = m(v - u)

-μgΔt = v - u

Making Δt subject of formula:

Δt = (v - u) / -μg

Δt = (5.8 - 8.4) / (-0.25 * 9.8)

Δt = -2.6/ -2.45

Δt = 1.06 secs

It took the sled 1.06 secs to travel from A to B.

7 0
3 years ago
Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage
agasfer [191]

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

6 0
2 years ago
A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ ௠at a d
Angelina_Jolie [31]

A very small source of light that radiates uniformly in all directions produces an electric field with an amplitude of ܧ ௠at a distance R from the source. What is the amplitude of the magnetic field at a point 2R from the source?

If the distance from the source is doubled. The amplitude of the magnetic field is smaller 4 times.

4 0
3 years ago
Other questions:
  • The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
    11·2 answers
  • uppose an airline determines it can charge different prices to people traveling for business or emergencies and leisure traveler
    10·1 answer
  • 2. Place the pairs in order from weakest to strongest gravitational force. (Brainpop Gravity Challenge)
    5·1 answer
  • PLS HELP which ones would be made of cells? and which ones show cell walls?
    11·1 answer
  • Question one please! This is 8th grade science btw.
    12·1 answer
  • A rightward force of 4.0 N is exerted upon an object for a distance of 3.0 meters.
    14·1 answer
  • How many orbits/shells/energy levels does Rubidium have?
    12·1 answer
  • If the wavelength of a wave is quadrupied then what happens to the frequency of the wqve(Assume speed is constant) ​
    11·1 answer
  • Converting 67 m•s¹ to km•h¹​
    5·1 answer
  • 5. Your friend claims that the Moon’s repeated orbit around Earth causes the cycle of the Moon’s phases.
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!