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3 years ago

6

Amy is holding one end of a jump rope. Her friend Beth is holding the other end of the rope. Amy swiftly moves the handle of the

jump rope downward, creating a large wave in the rope.
Will Beth feel a tug on the rope at her end?

2 answers:

Anuta_ua [19.1K]3 years ago

6 0

Yes Beth will feel a tug on the rope at her end! :D

Umnica [9.8K]3 years ago

3 0

Yes. If a downward force is created, it will be felt at the end of the rope as well.

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A ball is pushed into a spring-loaded launcher with a force of 20 N, which compresses the spring 0.08 m.

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Answer:

0.8J

Explanation:

Given parameters:

Force = 20N

Compression = 0.08m

Unknown:

Spring constant = ?

Elastic potential energy = ?

Solution:

To solve this problem, we use the expression below:

F = k e

F is the force

k is the spring constant

e is the compression

20 = k x 0.08

k = 250N/m

Elastic potential energy;

EPE = $\frac{1}{2}$ k e² = $\frac{1}{2}$ x 250 x 0.08²

Elastic potential energy = 0.8J

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

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= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

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2 years ago

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Answer:

Thats a personal question

Explanation:

It asks about you

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Answer:

Explanation:

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Starting at (0,0) an object travels 36 meters north and then it covers 20 meters east. What is

Svetradugi [14.3K]

Answer:

Explanation:

Using the pythagoras theorem, the displacement is expressed as;

d² = x²+y²

y = 36m (north)

x = 20m east

Substitute;

d² = 36²+20²

d² = 1296+400

d² = 1696

d = √1696

d = 41.18m

For the direction;

theta = tan^-1(y/x)

theta = tan^-1(36/20)

theta = tan^-1(1.8)

theta = 60.95°

Hence the magnitude is 41.18m and the direction is 60.95°

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3 years ago