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Alla [95]
3 years ago
13

A load of bricks is being lifted by a crane at a steady velocity of 5.0m/s when one bricks falls off 6.0 m above the ground

Physics
1 answer:
Ganezh [65]3 years ago
3 0
LOOK OUT BELOW  !   !   !
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What was the hikers average velocity during part c of the hike?
Brrunno [24]

Answer:

6 km/h west

Explanation:

During part c of the hike, the hiker moved 6 km west, and the time was 9.45 am to 10.45 am.

So, we have:

- displacement: d = 6 km west

- time taken: from 9.45 am to 10.45 am = 1 hour

Therefore, the average velocity is given by the ratio between displacement and time taken:

v=\frac{d}{t}=\frac{6 km}{1 h}=6 km/h

and the direction is the same as the displacement (west)

6 0
3 years ago
Read 2 more answers
An electric pole casts a shadow of 24 m long . If the tip of the shadow is 25m far from the top of the pole , how high is the po
Vikentia [17]

Answer:

7 m

Explanation:

a^ + b^=c^

a^ + 24^=25^

a^+576=625

a^=625-576

a^=49

a=7

7 0
3 years ago
Billy jumps straight up with a positive velocity of 100 m/s. When Billy
ZanzabumX [31]

Answer:......................................................................................

                   .Explanation:

7 0
2 years ago
A long cylindrical insulating shell has an inner radius of a = 1.41 m and an outer radius of b = 1.67 m. The shell has a constan
Natasha2012 [34]

Answer:

a. E = 122.4 N/C

b. E = 58.2 N/C

c. E = 0

Explanation:

The electric field at an arbitrary point away from the axis of the cylinder can found by applying Gauss’ Law, which states that an electric flux through a closed surface is equal to the total charge enclosed by this surface divided by electric permittivity.

In order to apply this law, we have to draw an imaginary cylindrical surface of arbitrary height ‘h’ and radius ‘r’, which is equal to the point where the E-field is asked.

A. For the outside of the cylinder, we will draw our imaginary surface with r = 1.97.

E2\pi rh = \frac{\lambda V}{\epsilon_0} = \frac{\lambda \pi (b^2 - a^2)h}{\epsilon_0}\\E2\pi (1.97)h = \frac{(5.3\times 10^{-9})\pi(1.67^2 - 1.41^2)h}{\epsilon_0}\\E = 122.4~N/C

B. This time our imaginary surface should be inside the cylinder, therefore the enclosed charge will be less than that of part A.

E2\pi rh = \frac{\lambda V_{enc}}{\epsilon_0} = \frac{\lambda \pi (r^2 - a^2}h{\epsilon_0}\\E2\pi (1.51)h = \frac{5.3\times 10^{-9})\pi(1.51^2 - 1.41^2)h}{\epsilon_0}\\E = 58.2~N/C

C. In this case our imaginary surface will be inside the cylinder, where there is no charge at all. Therefore, the enclosed charge will be zero and the electric field will be zero.

6 0
3 years ago
How is friction reduced between an air hockey puck and the table?
irakobra [83]

Cause it's Corona Time

3 0
3 years ago
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