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kolezko [41]
3 years ago
6

The electric motor of a model train accelerates the train from rest to 0.620 m/s in 21ms. The total mass of the train is 825 g.

find the average power delivered to the train during its acceleration.
Physics
1 answer:
Rudiy273 years ago
7 0

Explanation:

Average power = change in energy / change in time

P = ΔE / Δt

P = (½ mv²) / t

P = (½ (0.825 kg) (0.620 m/s)²) / (0.021 s)

P = 7.55 Watts

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A 20-kg barrel is rolled up a 20-m ramp to the back of a truck whose floor is 5.0 m above the ground. What work is done in loadi
oee [108]

The angle of inclination is calculated using sin function,

sin θ = 5 m / 20 m = 0.25

θ = 14.4775° 

 

<span>The net force exerted is then calculated:
F net = m g sin θ = 20 * 9.8 * 0.25 </span>

F net = 49N 

<span>Work is product of net force and distance:
W = F net * d = 49 * 20 </span>

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4 0
3 years ago
What subatomic particle acts like a mini-magnet?
Semenov [28]
 The subatomic particles that acts like a mini-magnet is electron. Electrons are negatively charged sub atomic particles in an atom. The electron spin is a property of an electron that makes it behave like it's spinning; a spinning electron produces a magnetic field that makes it behave like a tiny magnet in an atom. 
5 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
What is the correct answer?
inysia [295]
The answer to your question is metaphase
3 0
3 years ago
Triangle ABC is similar to triangle A'B'C'. Which sequence of similar transformations could map △ABC onto △A'B'C'?
Olin [163]

Answer:

b.

Explanation:

Dilation and Translation

6 0
3 years ago
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