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2 years ago

How does a generator produce electrical energy?

Physics

2 answers:

Grace [21]2 years ago

8 0

Option b is the correct answer

Daniel [21]2 years ago

3 0

The correct answer is B. it converts mechanical energy to electrical energy.

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nata0808 [166]

Answer:

Its sure 2nd gare beaciuse

Explanation:

of tobi was obito was tobi his som came and start doing enginner but he faild so i gave him answer he repiled oh baby baby oh

5 0

3 years ago

A 2500-N net force acting on a 880-kg car accelerates it at a rate of ______ m/s/s

Bond [772]

Answer:

a = 2.84 m/s²

Explanation:

Given that,

Net force, F = 2500 N

Mass of the car, m = 880 kg

We need to find the acceleration of the car. Net force is given by :

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{2500\ N}{880\ m/s^2}\\\\a=2.84\ m/s^2$

So, the acceleration of the car is 2.84 m/s².

8 0

3 years ago

What is the horizontal component of a ball thrown at a 27 degree angle at 16 m/s?

Airida [17]

Answer:

14.25 m/s

Explanation:

In this problem, we need to find the horizontal component of a ball thrown at a 27 degree angle at 16 m/s.

It can be given by :

$v_x=v\cos\theta\\\\=16\times \cos(27)\\\\=14.25\ m/s$

So, the horizontal component of the ball is 14.25 m/s.

6 0

3 years ago

How can working together help scientist achieve their goals?

abruzzese [7]

When scientists work together it helps them achieve their goals when they share different ideas with eachother

7 0

3 years ago

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago