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2 years ago

9

What is the kinetie energy of a 3-kilogram ball that is rolling at 2 meters per second?

1 answer:

Sedaia [141]2 years ago

7 0

KE=1/2m v2= 1/2 (3kg) (2m/s)2 = 6J (6joules)

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How is kinetic energy related to the distance

astra-53 [7]

Kinetic energy =

1/2 (mass) • (velocity squared).

I don't see distance in that formula anywhere, so they're not related.

8 0

3 years ago

Read 2 more answers

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg

n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

$\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N$

Thus, the net force over the body is:

$F=(913.14N)\hat{i}+(274.87N)\hat{j}$

Next, you calculate the magnitude of the force:

$F=\sqrt{(913.14N)+(274.87N)^2}=953.61N$

and the direction is:

$\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°$

7 0

3 years ago

Find the potential energy of a 40kg cart that is on a hill that is 10 meters high

Vilka [71]

Explanation:

P.E. = mgh

PE = 40 × 10 × 10

PE = 4000 Joule

3 0

1 year ago

Which of the following is an example of an irregular meter? A. Four beats per measure B. Five beats per measure C. Two beats per

Komok [63]

B. 5 beats per measure.

7 0

2 years ago