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levacccp [35]
3 years ago
7

A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the

block and plane is 0.2. assuming that the block moves at a constant speed, how much work is required to move the block 100 ft considering the (a) the block and (b) the man as the system.
Physics
1 answer:
Talja [164]3 years ago
3 0
The first thing you should know is that the work is defined as:
 W = F * d
 Where
 F = force
 d = displacement
 We have then
 (a) the block
 F = (0.2) * (100) = 20
 d = 100
 W = (20) * (100) = 2000 ft.lbf
 (b) the man as the system.
 F = (0.2) * (100 + 180) = 56
 d = 100
 W = (56) * (100) = 5600 ft.lbf
 answer:
 (a) 2000 ft.lbf
 (b) 5600 ft.lbf
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3 0
4 years ago
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or
Margaret [11]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

\Delta U = 0

so from 1st law of thermodynamics we have

dQ = W + \Delta U

dQ = W

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

8 0
3 years ago
Gravity on earth is 9.8 m/s squared, and gravity on the moon is 1.6 m/s squared. So if the mass of an object on earth is 40 kilo
garik1379 [7]

The mass of an object on Earth is the same as its mass on the Moon. The weight is different.

Weight = m * g

Weight ( Moon ) = 40 kg * 1.6 m/s² = 64 N

If the mass of an object on Earth is 40 kg, its mass on the Moon is 40 kg and its weight on the Moon is 64 N. 

7 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
4 years ago
Please help i don't understand. will mark brainliest
soldier1979 [14.2K]

Answer:

marblebrainiest plz\c cmarble

Explanation:

8 0
3 years ago
Read 2 more answers
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