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Dmitriy789 [7]
3 years ago
5

A net force of 15 N is applied to a cart with a mass of 2.1 kg. a. What is the acceleration of the cart? b. How long will it tak

e the cart to travel 2.8 m, starting from rest?
Physics
1 answer:
Gennadij [26K]3 years ago
6 0
The first question's answer is :
If, F=ma
Then, 15N= 2.1kg (a)
15/2.1=a
7.14=a
Therefore, acceleration = 7.14m/s^2

sorry I am not sure about the second question :-(
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What happens when an electron emits a photon?
Vera_Pavlovna [14]

Answer:

An electron emits a photon and falls from a higher energy level to a lower energy level.

Explanation:

A quantized form of electromagnetic radiation is called a photon.

Since the electromagnetic radiation has dual characteristics. The photons are the energy wave packets of the radiation.

When an electron is exposed to such radiation, it absorbs some of the energies of photons and goes to the excited state.

The excited state is only a temporary state. So, the electron releases some energy that comes to a lower energy level.

7 0
4 years ago
Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth
scoundrel [369]

Answer:

a) 24 Hs. b) 0.224 m/s² c) 448 N

Explanation:

a) As satellites in a geosynchronous orbits, stay directly over a point fixed on the Equator while the Earth rotates, the only way that this can be possible, if the period of the satellite (time to complete a full orbit) is equal to the time that the Earth uses to complete a spin itself, which is exactly one day.

b)

The value of g, is just the acceleration due to the gravitational attraction between the satellite and the Earth.

According the Universal Law of Gravitation, this force can be written in this way:

Fg = ms . a = G me. ms / (re+rs)² ⇒a=g= G me / (re + rs)²

Replacing by the values of G, me, re, and rs, we get:

g = 6.67. 10⁻¹¹ . 5.97.10²⁴ / (6.37 10⁶ + 3.58.10⁷)² m/s²

g= 0.224 m/s²

c) If we call "weight" to the magnitude of the gravitational force on the satellite (as we do with masses on Earth), we can find this value, just solving the equation for Fg, as follows:

Fg = G me . ms / (re + rs)²

Replacing by the values, we find:

Fg = 448 N

5 0
3 years ago
Hi! Whoever can answer this question the best I will give the brainliest answer!!
oksano4ka [1.4K]
A battery is a device that stores chemical energy and converts it to electrical energy. Two terminals made of different chemicals (typically metals), the anode and the cathode; and the electrolyte, which separates these terminals.
Hope I get it and hope this helps
6 0
2 years ago
A constant force of 120 N pushes a 55 kg wagon across an 8 m level surface. If the wagon was initially at rest, what is the fina
Illusion [34]

Answer:

The kinetic energy of the wagon is 967.0 J

Explanation:

Given that,

Force = 120 N

Mass = 55 kg

Height = 8 m

We need to calculate the kinetic energy of the wagon

Using newtons law

F = ma

\dfrac{120}{55}=a

a =2.2\ m/s^2

Using equation of motion

v^2 =u^2+2as

Where,

v = final velocity

u = initial velocity

s = height

Put the value in the equation

v^2=0+2\times2.2\times8

v=5.93\ m/s

Now, The kinetic energy is

K.E=\dfrac{1}{2}mv^2

K.E=\dfrac{1}{2}\times55\times(5.93)^2

K.E=967.0\ J

Hence, The kinetic energy of the wagon is 967.0 J

3 0
3 years ago
camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin
son4ous [18]

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined  for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

v=\frac{\Delta x}{\Delta t}

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

v=\frac{37.1}{5.4}

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

7 0
3 years ago
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