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Igoryamba
2 years ago
9

A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pu

ll the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg.
-Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well.
-Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.
Physics
1 answer:
AlladinOne [14]2 years ago
4 0
The first problem cannot be solve because you did give the distance or length of the rope, because work = distance x force. i can only solve the the second problem. since the bucket is moving up then force due to gravity is going down, then the net force is:
Fnet = F1 - Fg
where Fg = mg
g is the accelaration due to gravity ( 9.81 m/s^2)
Fnet = 57.5 N - (3.9 kg)(9.81) N
Fnet = 19.24 N
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Mr. Chang's class participated in a science experiment where they placed three substances outside in the sunlight to determine w
Montano1993 [528]

Answer:

B or D but im pretty sure it is D

Explanation:

When molecules are left in the sun, it heats up. When molecules heat up, the begin to vibrate rapidly. The sun is not constant as it could get blocked by clouds, so it would, at times, slow down the movement of the molecules. The answer is most likely D.

7 0
2 years ago
A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0
2 years ago
A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

a) t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s

  Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

                  H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m

      Maximum altitude = 490.5 m

c) Range of projectile is given by

                              R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m

    Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

   Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

   We have equation of motion s = ut + 0.5 at²

   Horizontal motion

                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

                Substituting

                          s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

      Vertical motion

                         u = 98.1 m/s

                         a = -9.81 m/s²

                          t = 15 s

                Substituting

                          s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

   \texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m

   Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0
2 years ago
A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s
Furkat [3]

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

p_{rad} = \frac{I}{c}

<u>Where:</u>

p_{rad}: is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

3 0
2 years ago
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How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?
noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

U=mgh

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

U=(50 kg)(9.8 m/s^2)(15 m)=7350 J

3 0
2 years ago
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