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just olya [345]

3 years ago

7

A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle

of 30 degrees above the horizontal.Find the work done on the box.

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

W = 34.64 ft-lbs

Explanation:

given,

Horizontal force = 4 lb

distance of push, d = 10 ft

angle of ramp, θ = 30°

Work done on the box = ?

We know,

W = F.d cos θ

W = 4 x 10 x cos 30°

W = 40 x 0.8660

W = 34.64 ft-lbs

Hence, work done on the box is equal to W = 34.64 ft-lbs

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alekssr [168]

Answer:

P = 3800 W

Explanation:

Power is equal to energy divide by time P = E/t

plug in 4560000 J for energy

convert 20 minutes to seconds

20 * 60 = 1200 seconds

plug in 1200 seconds for time

P = 4560000/1200

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What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these

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Answer:

E. all of these

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Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

aev [14]

Answer:

Explanation:

Given

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third Particle which charge +q must be placed left of $2.5\mu C$ because it will repel the q charge while $-3.5\mu C$ will attract it

suppose it is placed at a distance of x m

$F_{1q}=\frac{kq(2.5)}{x^2}$

$F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}$

$F_{1q}+F_{2q}=0$

$\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0$

$\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}$

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3 years ago

What does the mass defect represent?

Lina20 [59]

Answer:

C.) The amount of mass used up in holding a nucleus together.

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Regards!

5 0

3 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

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$\Delta x = (\mu - 1) t$

here we know that

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now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

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$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

3 0

3 years ago