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Marysya12 [62]
3 years ago
14

A box is pulled to the right with a force of 65 N at an angle of 58 degrees to the horizontal. The surface is frictionless. The

free body diagram is shown. What is the net force in the x-direction?
Physics
2 answers:
Citrus2011 [14]3 years ago
5 0
The free-body diagram is missing, but I assume the only forces acting on the box are the force F pushing the box, the weight of the object and the normal reaction of the surface.

Since the weight and the normal reaction acts in the vertical (y) direction, the only force acting on the box in the horizontal (x) direction is the horizontal component of the force F, which is given by
F_x = F \cos 58^{\circ} = (65 N)(\cos 58^{\circ} )=34.4 N
And so this is the net force in the x-direction.
Alex787 [66]3 years ago
3 0

Answer:

34 N

Explanation:

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Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains
tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

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Q = 80,850 in^3/min. = v*A = v*50.24in^2

v =  (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0
3 years ago
How much work is needed to lift a 3kg create a vertical displacement of 22m
Ulleksa [173]
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6 0
3 years ago
You have a tungsten sphere (emissivity ε = 0.35) of radius 25 cm at a temperature of 25°C. If the sphere is enclosed in a room w
egoroff_w [7]

Answer:

Explanation:

Stefan's formula for emission of radiation is

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E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²

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8 0
2 years ago
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marissa [1.9K]

Answer:

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I=\dfrac{Q}{t}

t = \dfrac{Q}{I}

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6 0
2 years ago
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Hope it helps
7 0
3 years ago
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