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Vsevolod [243]
3 years ago
5

Question: Point B is 25 km due east of point A. Starting from point A, a camel walks 20 km in

Physics
1 answer:
Korolek [52]3 years ago
8 0

<em>Resultant angle; θ = 25.59°  </em>

This question is dealing with bearings and distance.

We are told that from point A, the camel walks 20 km at 15° in the south of east direction.

Thus, d_s,e = 20 km

Resolving along the horizontal east direction gives; d_e = 20 cos 15

d_e = 19.32 km

Also, resolving along the vertical south direction gives; d_s = 20 sin 15

d_s = 5.18 km

Net vertical distance; d_vert = 8km - 5.18km = 2.72 km

Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km

Now, the resultant angle is given by;

tan θ = d_vert/d_hor

tan θ = 2.72/5.68

tan θ = 0.4789

θ = tan^(-1) 0.4789

θ = 25.59°

Read more at; brainly.com/question/22518031

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Answer:

Answers at the bottom!!

Explanation:

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Gamma rays and x-rays consist of high-energy waves that can travel great distances at the speed of light and generally have a great ability to penetrate other materials. For that reason, gamma rays (such as from cobalt-60) are often used in medical applications to treat cancer and sterilize medical instruments.

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Hope this helps!!

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What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

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3 years ago
A cannonball is fired horizontally at the same time a ball being dropped from the same height How do the times it takes them to
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Answer:

The cannonball and the ball will both take the same amount of time before they hit the ground.

Explanation:

For a ball fired horizontally from a given height, there is only a vertical acceleration on it towards the ground. This acceleration is equal to the acceleration due to gravity (g = 9.81 m/s^2). A ball dropped from a height will also only experience the same vertical acceleration downwards which is also equal to g = 9.81 m/s^2.

Therefore both the cannonball and the ball will take the same amount of time to hit the ground if they are released/fired from the same height.

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