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OverLord2011 [107]
3 years ago
11

What is a watt equivalent to in terms of kilograms, meters and seco

Physics
1 answer:
Ierofanga [76]3 years ago
5 0

1 watt = 1 joule per second

1 joule = 1 newton-meter

1 newton = 1 kg-meter per second²

Put it all together and you have . . .

1 watt = 1 <em>kg-meter² / second³


</em>
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Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot
almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0 (Ec. 1)

\Sigma F_{y} = N - W = 0 (Ec. 2)

Where:

P - Pushing force, measured in newtons.

T - Tension, measured in newtons.

\mu_{k} - Coefficient of kinetic friction, dimensionless.

N - Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P+T -\mu_{k}\cdot W = 0

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

\mu_{k} =\frac{P+T}{m\cdot g}

If we know that P = 400\,N, T = 290\,N, m = 300\,kg and g = 9.807\,\frac{m}{s^{2}}, then:

\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 0.235

The kinetic coefficient of friction of the crate is 0.235.

5 0
3 years ago
You’re an electrical engineer designing an alternator (the generator that charges a car’s battery). Mechanical engineers specify
makvit [3.9K]

Answer:

I will specify a value of 0.009T for the alternator’s magnetic field

Explanation:

E_peak = 14 V

d = 10cm = 0.1m, so r = 0.1/2 =0.05m

N = 250 turns

f = 1200rpm = (1200rp/m x 1m/60sec) = 20 revolutions per second

At peak performance, peak voltage is given by the equation;

E_peak = NABω

Let's make the magnetic field B the subject;

B = E_peak/(NAω)

Now we know that ω = 2πf

Thus, ω = 2π x 20 revs/s = 125.664 revs/s.

Let's convert it to the standard unit which is rad/s.

1 rev/s = 6.283 rad/s

Thus, 125.664 revs/s = 125.664 x 6.283 = 789.55 rad/s

Area (A) = πr² = π x 0.05² = 0.007854 m²

Thus, plugging in the relevant values to get;

B = 14/[(250 x 0.007854 x 789.55)] = 0.009T

8 0
3 years ago
What is another name for the range of motion in your joint
Neporo4naja [7]

Answer:Gliding

Explanation:

Because it’s an ball and socket joint

3 0
3 years ago
How does light travel??????????
adell [148]
Light travels through waves. 
6 0
4 years ago
A police officer uses a radar gun that emits electromagnetic waves with a frequency of 10.525 GHz (1.0525 1010 Hz). Assume the p
Paraphin [41]

Answer:

Explanation:

For Doppler effect in radar the formula is as follows

Δf = f₀ x 2v /c

Δf is change in frequency in reflected wave , f₀ is original frequency , v is velocity of source and c is velocity of right .

Δf  = 2460 Hz , f₀ = 1.0525 x 10¹⁰ Hz , v = ? c is velocity of light .

2460 =  1.0525 x 10¹⁰ x 2 v / 3 x 10⁸

2460 = 105.25 x 2 v / 3

v = 2460 x 3 / (105.25 x 2 )

v = 35 .06 m /s .

(b)

Since the observed frequency is less , the source is moving away from the observer .

option (2) is correct .

4 0
3 years ago
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