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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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A satellite in outer space is moving at a constant velocity of 20.5 m/s in the +y direction when one of its on board thruster tu
Katyanochek1 [597]

Answer:

a)  V_f = 25.514 m/s

b)  Q =53.46 degrees CCW from + x-axis

Explanation:

Given:

- Initial speed V_i = 20.5 j m/s

- Acceleration a = 0.31 i m/s^2

- Time duration for acceleration t = 49.0 s

Find:

(a) What is the magnitude of the satellite's velocity when the thruster turns off?

(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.

Solution:

- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:

                                   V_f = V_i + a*t

                                   V_f = 20.5 j + 0.31 i *49

                                   V_f = 20.5 j + 15.19 i

- The magnitude of the velocity vector is given by:

                                   V_f = sqrt ( 20.5^2 + 15.19^2)

                                   V_f = sqrt(650.9861)

                                  V_f = 25.514 m/s

- The direction of the velocity vector can be computed by using x and y components of velocity found above:

                                 tan(Q) = (V_y / V_x)

                                 Q = arctan (20.5 / 15.19)

                                 Q =53.46 degrees

- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.

4 0
3 years ago
Find the magnitude of the sum<br> of these two vectors:<br><br> 101 m<br> 60.0 °<br> 85.0 m
attashe74 [19]

Answer: 161.3

I have a acellus too and got this question correct, so I hope this helps y’all out

8 0
2 years ago
. If she
I am Lyosha [343]

Answer:

9 meters

Explanation:

Given:

Mass of Avi is, m=40\ kg

Spring constant is, k=176,400\ N/m

Compression in the spring is, x=20\ cm=0.20\ m

Let the maximum height reached be 'h' m.

Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.

So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.

Decrease in elastic potential energy is given as:

EPE=\frac{1}{2}kx^2\\EPE=\frac{1}{2}\times 176400\times (0.20)^2\\EPE=88200\times 0.04=3528\ J

Now, increase in gravitational potential energy is given as:

GPE=mgh=40\times 9.8\times h=392h

Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,

392h=3528\\\\h=\frac{3528}{392}\\\\h=9\ m

Therefore, Avi will reach a maximum height of 9 meters.

6 0
3 years ago
An amoeba has 1.00 x 1016 protons and a net charge of 0.300 pC. Assuming there are 1.88 x 106 fewer electrons than protons, If y
Xelga [282]

Answer:

The fraction of the protons would have no electrons =1.88\times 10^{-10}

Explanation:

We are given that

Amoeba has total number of protons=1.00\times 10^{16}

Net charge, Q=0.300pC

Electrons are fewer than protons=1.88\times 10^6

We have to find the fraction of protons would have no electrons.

The fraction of the protons would have no electrons

=\frac{Fewer\;electrons}{Total\;protons}

The fraction of the protons would have no electrons

=\frac{1.88\times 10^{6}}{1.00\times 10^{16}}

=1.88\times 10^{-10}

Hence, the fraction of the protons would have no electrons =1.88\times 10^{-10}

6 0
2 years ago
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kykrilka [37]
I think the answer is B
4 0
3 years ago
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