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PIT_PIT [208]
3 years ago
7

un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l

os 13 segundos
Physics
1 answer:
Dmitriy789 [7]3 years ago
8 0

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

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What is the current in ampere through a 100<br> watt and 120 volt device?​
Gwar [14]

Answer:

0.833 amps

Explanation:

mark me brainliest pls

5 0
3 years ago
The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[
AysviL [449]

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

P = \frac{E}{t}

where;

P is power in watts

E is energy in Joules

t is time in seconds

P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W

The average current that this cell phone draws when turned on:

P = IV

I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

5 0
3 years ago
You push the ball with aforce of 22.8N which induces a -2.3N frictional force. What is the net force while you push?
inna [77]
The net force is 22.8-2.3 or 20.5 N
5 0
3 years ago
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
For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas
nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

F=\dfrac{mv^2}{R}, m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

F=\dfrac{(16)^2m}{60}=4.26m\ N

Case 2.

R = 15 m v = 8 m/s

F=\dfrac{(8)^2m}{15}=4.26m\ N

Case 3.

R = 30 m v = 4 m/s

F=\dfrac{(4)^2m}{30}=0.54m\ N

Case 4.

R = 45 m v = 4 m/s

F=\dfrac{(4)^2m}{45}=0.36m\ N

Case 5.

R = 30 m v = 16 m/s

F=\dfrac{(16)^2m}{30}=8.54m\ N

Case 6.

R = 15 m v =12 m/s

F=\dfrac{(12)^2m}{15}=9.6m\ N

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0
3 years ago
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