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slamgirl [31]
3 years ago
7

How does an inclined plane change distance and how does it change direction

Physics
1 answer:
ElenaW [278]3 years ago
7 0
 Objects are known to go down because of a unbalanced force
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A student removes a 10.5 kg stereo amplifier from a shelf that is 1.82 m high. The amplifier is lowered at a constant speed to a
Nonamiya [84]

Answer:

(a) the work done by the student is 110.1 J

(b) The gravitational force that acts on the amplifier is 102.9 N

Explanation:

Given;

mass of the amplifier, m = 10.5 kg

initial position of the amplifier, x₀ = 1.82 m

final position of the amplifier, x₁ =0.75 m

The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m

(b) The gravitational force that acts on the amplifier;

F = mg

F = 10.5 x 9.8

F = 102.9 N

(a) the work done by the student is calculated as;

W = FΔx

W = 102.9 x 1.07

W = 110.1 J

7 0
3 years ago
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =
Sloan [31]

Answer:

The current is  I  = 8.9 *10^{-5} \  A

Explanation:

From the question we are told that

     The  radius is r =  3.17 \  mm  =  3.17 *10^{-3} \ m

      The current density is  J =  c\cdot r^2  =  9.00*10^{6}  \ A/m^4 \cdot r^2

      The distance we are considering is  r =  0.5 R  =  0.001585

Generally current density is mathematically represented as

          J  =  \frac{I}{A }

Where A is the cross-sectional area represented as

         A  =  \pi r^2

=>      J  =  \frac{I}{\pi r^2  }

=>    I  =  J  *  (\pi r^2 )

Now the change in current per unit length is mathematically evaluated as

        dI  =  2 J  *  \pi r  dr

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

         I  = 2\pi  \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr

         I  = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr

        I  = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ]  | \left    0.001585} \atop 0}} \right.

        I  = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]

substituting values

        I  = 2 *  3.142  *  9.00 *10^6 *   [ \frac{0.001585^4}{4} ]

        I  = 8.9 *10^{-5} \  A

5 0
4 years ago
Why is it that we can see blood cells only with a microscope? A) The cells form clumps. B) The cells keep moving about. C) The c
Snezhnost [94]
D the cells are very small
7 0
3 years ago
Read 2 more answers
What is one use for infrared waves? A radar B Medical imaging C Thermal imaging cameras
Vilka [71]

Answer:

Im answering for free points sry

Explanation:

...

7 0
3 years ago
An object is dropped from a​ tower, 10241024 ft above the ground. the​ object's height above ground x seconds after the fall is
irga5000 [103]
Your question is asked in a confusing way. Can you share more information about the equation?
4 0
4 years ago
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