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zaharov [31]
3 years ago
10

A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.

00 s, how much power was developed, to the nearest watt?
Physics
2 answers:
Vika [28.1K]3 years ago
6 0
You need to calculate the work needed. This is force x distance in this case.
After you do that, divide the work by the time.
Brilliant_brown [7]3 years ago
3 0

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

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A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude
finlep [7]
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
                   = (5 * 10) Kg m/s
                   = 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
5 0
3 years ago
Which one of the following is a correct statement......
IceJOKER [234]

Answer:

Just 3

Explanation:

I believe the other two are incorrect

5 0
3 years ago
Which action might lead scientists to develop new explanations about the
VikaD [51]

Answer:

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5 0
3 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

6 0
4 years ago
For a steady two-dimensional flow, identify the boundary layer approximations.
Georgia [21]

Answer:

  • The velocity component in the flow direction is much larger than that in the normal direction ( A )
  • The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

4 0
3 years ago
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