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2 years ago

6

A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

1 answer:

Fiesta28 [93]2 years ago

5 0

The first option 8.4 J

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As keisha runs the generator, which best describes what should happen to the needle that measures electric current? it will move

SOVA2 [1]

The diagram shows a simple electric generator. The needle that measures electric current will move back and forth between a largely positive and a large negative value.

What is an electric generator?

An electric generator is physically equivalent to an electric motor. but it converts mechanical energy into electrical energy.
The electrical field generated is dependent on the inclination of the wire with respect to magnetic field lines, and this inclination changes over time,

because of that she will experience a varying electrical field, and thus a varying electric current will be zero.

The maximum positive value will occur when the wire is perpendicular to the magnetic field lines after one-fourth of rotation, and then zero.

Hence option C is correct.

The diagram shows a simple electric generator. The needle that measures electric current will move back and forth between a largely positive and a large negative value.

Learn more about electric generator here:

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2 years ago

An arrow leaves a bow at 60 m/s at an angle of 30 degrees to the horizon.

Orlov [11]

Answer: 11

Explanation:

4 0

3 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

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3 years ago

Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it

MakcuM [25]

Answer:

Explanation:

Let the vertical height by which it descends be h . Let it acquire velocity of v .

1/2 mv² = mgh

v² = 2gh

As it leaves the surface of sphere , reaction force of surface R = 0 , so

centripetal force = mg cosθ where θ is the angular displacement from the vertex .

mv² / r = mg cosθ

(m/r )x 2gh = mg cosθ

2h / r = cosθ

cosθ = (r-h) / r

2h / r = r-h / r

2h = r-h

3h = r

h = r / 3

5 0

3 years ago

What is the DISPLACEMENT of the motorbike rider in the picture?

lutik1710 [3]

120m north east hope this helps

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3 years ago