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skad [1K]
3 years ago
15

A student uses 150 N to push a block of wood up a ramp at constant velocity of 2.6m/s. What is his power output?

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

P= 390 W

Explanation:

In physics, power is the rate of doing work or of transferring heat, i.e. the amount of energy transferred or converted per unit time.  Work  is a force F applied over a distance x. Matemathicaly it means

P = dW/dt ≈ d(F * x)/dt = xdF/dt + Fdx/dt. If force is constant dF/dt=0 so P=F dx/dt = P*v, where v is velocity, the rate of distance per unit time.

We have force and velocity. Newton is unit of Kg*m/s2, hence

P= 150 kgm/s2 * 2.6 m/s =390 Kgm2/s3 = 390 W, where W is Watts and is an unit of power

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A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.
Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = 2.5\times 10^{- 6}\ m

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

n\lambda = dsin\theta

n = 1

\lambda = dsin\theta            (1)

Also,

For very small angle, \theta:

sin\theta ≈ tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225

Using the above value in eqn (1):

\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm

3 0
3 years ago
Δtf = kf × m this formula relates the drop in freezing point (δtf) to the molal freezing point depression constant (kf) and the
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5 0
3 years ago
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The ammeter below shows the current produced by a series of solar cells that contain zinc plates being used to power a simple se
nataly862011 [7]
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By the Ohm`s Law:
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V = 15 A · 0.2 Ohms
V = 3 V
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8 0
3 years ago
How much time (in seconds does it take light to travel 1.20 billion km?
butalik [34]
Ok so light goes at a speed of <span>299,792 kilometers per second
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7 0
3 years ago
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The electric field 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's char
Sonbull [250]

Answer:

2.1×10¹⁸ C

Explanation:

Using,

E = kq/r²...................... Equation 1

Where E = Electric field, q = charge, r = distance, k = coulombs constant.

make q the subject of the equation

q = Er²/k.................. Equation 2

Given: E = 180000 N/C, r = 2.8 cm = 0.028 m

Constant: k = 9×10⁹ Nm²/C².

Substitute these values into equation 2

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q = 2.1×10¹⁸ C

Hence the object charge is 2.1×10¹⁸ C

4 0
3 years ago
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