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kompoz [17]
3 years ago
9

Will give brainliest

Physics
1 answer:
user100 [1]3 years ago
7 0

Answer:

I believe the answer is 2.5 N.......

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20 ohms 5 ohms
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Realiza la siguiente conversión de unidades: 340 N a Kgf
OleMash [197]

Answer:

The answer is 34.67 kilograms force

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A person pulls a bucket of water up from a well with a rope. Assume the initial and final speeds of the bucket are zero (Vi-Vf-0
n200080 [17]

Answer:

a

This a closed system because the mass of the system is conserved

The energy system that undergoes change is the Potential energy system

The energy system diagram is shown on the first uploaded image

b

Work done = Change in gravitational potential energy

So solving algebraically for work done would be

    Work done   = m*g*h

where m is mass

          g is acceleration due to gravity

          and h is the height

c

Work done in terms of force and distance is = mg

where  m is mass of bucket and

            g is acceleration due to gravity  

Explanation:

a) At the start, potential and kinetic energy were zero. so, energy is zero.

As the person pulls the bucket up, the potential energy becomes mgh.

so,final energy will be consisting of only potential energy.

B) Here work done is equal to change in gravitational potential energy.

W = \Delta P.E

W = m*g*h

where g = 9.9 m/s^2

C) Work = force * distance

mgh = force * h

force = mg

force = weight of bucket

6 0
3 years ago
What is the electron configuration of this element? *for ne
Serhud [2]

Answer:

[He]2s²2p⁶

Explanation:

8 0
3 years ago
Read 2 more answers
A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

m_{b}v_{b}\cos\theta= m_{c}v_{c}

v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0
3 years ago
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