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Kryger [21]
3 years ago
8

A small, 3 kg weight is moved from 5 m from the ground to 8 m. What is the change in potential energy?

Physics
1 answer:
Ulleksa [173]3 years ago
4 0

Answer: 88.2 J

Explanation: PE, OR Potential Energy, equals to mass gravity and height.

mass - 3kg

height - 3 because (8-5)

gravity is 9.8kg

so 3 times 3 times 9.8 = 88.2J

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You walk six blocks east along 12th Street, then three blocks west, then one block east, then six blocks east, then seven blocks
Oksana_A [137]
Displacement depends upon the path taken as it is a vector.

From your problem above we would have a total displacement of;
Defining +x direction as east and -x direction as west

6east-3west+1east+6east-7west

6-3+1+6-7=3 blocks east or + x-direction

So even though they walked a total of 17 blocks it ends up only being 3 blocks total in +xdirection that was travelled by displacement.

Any questions please ask.
5 0
3 years ago
Tory has a mass of 40kg. She sleds down a hill that has a slope of 25 degrees. what is the component of her weight that is along
Fudgin [204]

1.7 x 10^2 N

or 166 N

First you find the vertical component of the weight, which is 9.8*40, (g*m), which is 392 N. You then find the angle between that and the slope, which is 90-25, which is 65. You then multiply the vertical weight by cos(65), to find the component of that that is parallel to the slope. You get 165.666 N

3 0
3 years ago
Why cant people with aids foght out the infection?
sveta [45]
They can fight the infection but not the disease
3 0
2 years ago
Read 2 more answers
Earning Goal: To be able to calculate work done by a constant force directed at different angles relative to displacement
lana [24]

Answer:

the work done by the 30N force is 4156.92 J.

For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:

W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J

8 0
2 years ago
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
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