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Rainbow [258]
3 years ago
15

A block starting from rest slides down the length of an 18 plank with an acceleration of 4.0 meters per second. How long does th

e block take to reach the bottom?
Physics
1 answer:
Snowcat [4.5K]3 years ago
4 0

Answer:

\boxed{\text{\sf \Large 3.0 s}}

Explanation:

Use distance formula

\displaystyle d=ut+\frac{1}{2} at^2

u= \text{\sf  initial velocity}\\d= \text{\sf  distance}\\a= \text{\sf  acceleration}\\t= \text{\sf  time taken}

\displaystyle 18=0 \times t+\frac{1}{2} \times 4 \times t^2

t=3

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8 0
3 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
A detrital rock is named
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Sedimentary rocks also known as clastic sedimentary rock
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3 years ago
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Describe the importance of conservative forces to conservation of energy.
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6 0
3 years ago
A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electr
Katen [24]

Answer:

ФE = 9.403W

Explanation:

In order to calculate the magnitude of the electric flux trough the sheet, you use the following formula:

\Phi_E=\vec{A}\cdot \vec{E}=AEcos\theta       (1)

A: area of the rectangular sheet = (0.400m)(0.600m) = 0.24m^2

E: magnitude of the electric field = 95.0N/C

θ: angle between the direction of the electric field and the normal to the surface of the sheet

You replace the values of the parameters in the equation (1):

\Phi_E=(0.24m^2)(95.0N/m)cos(20\°)=9.304W

The magnitude of the electric flux is trough the sheet is 9.403W

5 0
2 years ago
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