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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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a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha
mina [271]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

7 0
1 year ago
A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener
Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

   potential energy store has been transferred into its kinetic energy

   store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = \frac{1}{2}m(v^{2}-v_{0}^{2})

where v is the final speed, v_{0} in the initial speed and m

is the mass

→ v = 20 m/s and v_{0} = 0 m/s

→ K.E = \frac{1}{2}m(20^{2}-0^{2})

→ K.E = \frac{1}{2}m(400)

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0
2 years ago
Round your answers to one decimal place. This parallel circuit has two resistors at 15 and 40 ohms. What is the total resistance
goldenfox [79]
-- With two resistors in parallel, the total effective resistance is
the reciprocal of  (1/R₁ + 1/R₂).

1/R₁ + 1/R₂  = 1/15 + 1/40

= 8/120 + 3/120

= 11/120

So the total effective resistance is 120/11 = 10.9 ohms .

Current = (voltage) / (resistance)

= 12 / (120/11)

= (12 · 11) / 120

= 132/120  =  1.1 Amperes  
4 0
2 years ago
Read 2 more answers
The weight of a bucket is 186 N. The bucket is being raised by two ropes. The free-body diagram shows the forces acting on the b
amm1812
Fnet=(115+106)-186= 34 N

mass=Force/g= 186N/9.8m/s^2 = 18.98 kg

a=fnet/mass => 34N/18.98kg = 1.79 m/s^2

so A= 1.8m/s^2
4 0
2 years ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

Q is the charge on an object

Q=\dfrac{Er^2}{k}

Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

Q=1.56\times 10^{-8}\ C

So, the charge on the object is 1.56\times 10^{-8}\ C. Hence, this is the required solution.

7 0
2 years ago
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