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telo118 [61]
2 years ago
8

Question If a force of 2N is used to move a box 1 meter, how much work was done to move the box?​

Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
7 0

Answer:

I assume that the force of 2N is applied along the direction of motion and was applied for the whole 1 meter, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula;

Explanation:

Work = 2N * 1m * cos 0º.

Work = 2N * 1m * 1

Work = 2Nm

Work = 2 joules

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Assume that a person bouncing a ball represents a closed system. Which statement best describes how the amounts of the ball's po
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Answer:

C.

Explanation:

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What is the relationship between and experiment and a hypothesis
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Answer:

an experiment is a way to test a hypothesis. A hypothesis is a prediction. You predict that if you change one thing (the independent variable) the other thing (the dependent variable) will change.

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2 years ago
From a height of 40.0 m, a 1.00 kg bird dives (from rest) into a small fish tank containing 50.5 kg of water. Part A What is the
Rom4ik [11]

Answer:

0.00185 °C

Explanation:

From the question,

The potential energy of the bird = heat gained by the water in the fish tank.

mgh = cm'(Δt)................... Equation 1

Where m = mass of the bird, g = acceleration due to gravity, h = height, c = specific heat capacity of water, m' = mass of water, Δt = rise in temperature of water.

make Δt the subject of the equation

Δt = mgh/cm'............... Equation 2

Given: m = 1 kg, h = 40 m, m' = 50.5 kg

constant: g = 9.8 m/s², c = 4200 J/kg.K

Substitute into equation 2

Δt = 1(40)(9.8)/(50.5×4200)

Δt = 392/212100

Δt = 0.00185 °C

4 0
2 years ago
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown
nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

v_x = 15.0 m/s

and with an initial vertical velocity of

v_{y0} = -20.0 m/s

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

y(t) = h + v_{0y} t + \frac{1}{2}gt^2 (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

v_y = -20 m/s

So the vertical position of the balloon after a time t is

y(t) = h + v_y t

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

v_x = 15.0 m/s

So the horizontal distance travelled in t = 6.00 s is

d_x = v_x t = (15.0 m/s)(6.00 s)=90 m

Considering also that the vertical height of the balloon after t=6.00 s is

d_y = 176.6 m

The distance between the balloon and the rock can be found by using Pythagorean theorem:

d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer at rest in the basket is

v_y (t) = gt

Substituting time t=6.00 s, we find

v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer on the ground is now given by

v_y (t) = v_{0y} + gt

Substituting time t=6.00 s, we find

v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s

6 0
2 years ago
The Special Olympics raises money through "plane pull" events in which teams of 25 people compete to see who can pull a 74,000 k
Murrr4er [49]

Answer:

28716.4740661 N

1.2131147541 m/s

51.2474965841%

Explanation:

m = Mass of plane = 74000 kg

s = Displacement = 3.7 m

f = Frictional force = 14000 N

t = Time taken = 6.1 s

u = Initial velocity = 0

v = Final velocity

s=ut+\frac{1}{2}at^2\\\Rightarrow 3.7=0\times 6.1+\frac{1}{2}\times a\times 6.1^2\\\Rightarrow a=\frac{3.7\times 2}{6.1^2}\\\Rightarrow a=0.198871271164\ m/s^2

Force is given by

F=ma+f\\\Rightarrow F=74000\times 0.198871271164+14000\\\Rightarrow F=28716.4740661\ N

The force with which the team pulls the plane is 28716.4740661 N

v=u+at\\\Rightarrow v=0+0.198871271164\times 6.1\\\Rightarrow v=1.2131147541\ m/s

The speed of the plane is 1.2131147541 m/s

Kinetic energy is given by

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}\times 74000\times 1.2131147541^2\\\Rightarrow K=54450.9540448\ J

Work done is given by

W=Fs\\\Rightarrow W=28716.4740661\times 3.7\\\Rightarrow W=106250.954045\ J

The fraction is given by

\dfrac{54450.9540448}{106250.954045}=0.512474965841

The teams 51.2474965841% of the work goes to kinetic energy of the plane.

3 0
3 years ago
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