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WARRIOR [948]
3 years ago
6

1 point

Physics
1 answer:
stiks02 [169]3 years ago
3 0

Answer:

<h2>28,000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 800 × 35

We have the final answer as

<h3>28,000 N</h3>

Hope this helps you

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A motorcycle of mass 100 kilograms travels around a flat, circular track of radius 10 meters with a constant speed of 20 meters
JulijaS [17]

Answer:

100/10 = 10 , 10 × 10 = 100÷20 = 5

I'm pretty sure its wrong

8 0
3 years ago
Read 2 more answers
A merry-go-round of radius 2 m is rotating at one revolution every 5 s. A
galben [10]

Answer:

a) The angular speed of the child is approximately 1.257 rad/s

b) The angular speed of the teenager is approximately 1.257 rad/s

c) The tangential speed of the child is approximately 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager is approximately 2.513 m/s

Explanation:

The revolutions per minute, r.p.m. of the merry-go-round = 1 revolution/(5 s)

The radius of the merry-go-round = 2 m

The location of the child = 1 m from the axis

The location of the teenager = 2 m from the axis

1 revolution = 2·π radians

Therefore, we have;

The angular speed, ω = (Angle turned)/(Time elapsed) = (2·π radians)/(5 s)

∴ The angular speed of the merry-go-round, ω = 2·π/5 radians/second

a) The angular speed of the child = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

b) The angular speed of the teenager = The angular speed of the merry-go-round = 2·π/5 radians/second ≈ 1.257 rad/s

c) The tangential speed, v = r × The angular speed, ω

Where;

r = The radius of rotation of the object

For the child, r = 1 m

The tangential speed of the child = 1 m × 2·π/5 radians/second = 2·π/5 m/s ≈ 1.257 m/s

d) For the child, r = 2 m

The tangential speed of the teenager = 2 m × 2·π/5 radians/second = 4·π/5 m/s ≈ 2.513 m/s

8 0
2 years ago
The alternative to nuclear fission reactors in a nuclear fusion reactor. Explain why it is much more difficult to get a fusion r
Sliva [168]

Answer: The major challenges are as

1) understanding of the plasma: Plasma is a soup like mixture of subatomic particles of different atoms nuclei and electrons that are shattered apart by the temperature at which plasma is formed. further research is needed to understand the behavior of plasma so that it can be put to a proper use.

2) Confinement of plasma: Once we get the plasma we need to hold it so that we can obtain heat from it to drive a steam turbine but the sheer temperature of plasma is in millions of Celsius thus currently making it impossible to confine conventionally. Scientists use a loop of electric and magnetic fields to keep it in circulatory like manner so that it can be studied.

3) finally to obtain electricity from the plasma it should be stable to produce electricity. But currently to obtain pressure, temperature so that we have a sustained supply is highly difficult in technical and economical aspects.

Inertial confinement: In order to get the nuclei of atoms close enough for fusion this type of method used compression of the nuclei into highly small volumes.This is accomplished by use of lasers which are directed towards the fuel pellets that implode and travel towards other nuclei making fusion possible. It's main advantage is that it requires lesser time to initiate fusion but the disadvantage being that a large power is used to fire the lasers and the lasers should all hit the small target.

Magnetic Confinement: In this method we use a magnetic and electric fields in a properly designed space to keep the plasma in motion. In motion the nuclei of the atoms come close enough to initiate fusion.It's advantage being less power is required to start the process as compared to inertial confinement and the disadvantage being that plasma confinement is currently not properly understood.

5 0
3 years ago
A train travels 8.81 m/s in a -51.0° direction.
Amiraneli [1.4K]

The displacement of the train after 2.23 seconds is 25.4 m.

<h3>Resultant velocity of the train</h3>

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

  • θ is the angle between the velocity = (90 - 51) + 37 = 76⁰

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

<h3>Displacement of the train</h3>

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

Learn more about displacement here: brainly.com/question/2109763

#SPJ1

8 0
1 year ago
Is Mars a gas planet or a rocky planet?
OLEGan [10]
Mars is a rocky planet
8 0
2 years ago
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