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charle [14.2K]
3 years ago
13

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

turkey.?
What is the velocity of the ball after the kick?

What force did the turkey feel?

Why doesn’t the turkey go flying backward since the ball flies forward? Is momentum conserved?
Physics
1 answer:
Harman [31]3 years ago
5 0

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

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At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach
Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

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2 years ago
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B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT
Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

Then, y = 90 - 30

y = 60°

Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

Learn more about displacement at: brainly.com/question/321442

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