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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.]

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Can someone help please and thank you:)
Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0
3 years ago
A pilgrim uses a block and tackle to lift a 225 N crate a distance of 16.5 m. She pulls 33.0 m of rope with a force of 125 N. Wh
kenny6666 [7]

Answer:

Ideal mechanical advantage of the machine is 2

Explanation:

Ideal mechanical advantage of the machine is

MA=\frac{op}{ip} \\

here we have to consider the distance

MA=\frac{33}{16.5} \\MA=2

7 0
3 years ago
Millikan is doing his oil drop experiment. He has a droplet with radius 1.6 µm suspended motionless in a uniform electric field
swat32

Answer:

The charge on the droplet is 3.106\times10^{-16}\ C.

Yes, quantization of charge is obeyed within experimental error.

Explanation:

Given that,

Radius = 1.6μm

Electric field = 46 N/C

Density of oil = 0.085 g/cm³

We need to calculate the charge on the droplet

Using formula of force

F= qE

mg=qE

V\times\rho\times g=qE

q=\dfrac{V\times\rho}{E}

q=\dfrac{\dfrac{4}{3}\pi\times r^3\times\rho\times g}{E}

Put the value into the formula

q=\dfrac{\dfrac{4}{3}\times\pi\times(1.6\times10^{-6})^3\times85\times9.8}{46}

q=3.106\times10^{-16}\ C

We need to calculate the quantization of charge

Using formula of quantization

n = \dfrac{q}{e}

Put the value into the formula

n=\dfrac{3.106\times10^{-16}}{1.6\times10^{-19}}

n=1941.25

Yes, quantization of charge is obeyed within experimental error.

Hence, The charge on the droplet is 3.106\times10^{-16}\ C.

Yes, quantization of charge is obeyed within experimental error.

7 0
3 years ago
Read 2 more answers
A person is drinking a glass of soda with ice.
prisoha [69]

The relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.

The relative kinetic energy of gas molecules increases with increase in the mean distance between the gas molecules.

Also, relative kinetic energy of gas molecules increases with in the temperature of the gas molecules and decreases with a decrease in the temperature of the of the gas molecules;

ΔK.E ∝ T

The ice in the soda lowers the temperature of the gas molecules, thereby reducing their average speed which in turn reduces the average kinetic energy of the gas molecules in the soda.

Above the soda in the glass, the concentration of the gas molecules is less and their mean distance is greatest when compared to inside the soda. This results to an increase in the speed of the gas molecules which increases their average kinetic energy.

Thus,  the relative kinetic energy of molecules in the soda is least energy and above the soda in the glass is greatest energy.

Learn more about temperature and kinetic energy here: brainly.com/question/305606

3 0
3 years ago
NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
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