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Sergio039 [100]
3 years ago
14

How much work did the movers do (horizontally) pushing a 170-kg crate 10.2 m across a rough floor without acceleration, if the e

ffective coefficient of friction was 0.60?
Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

Answer:

Work done, W = 10195.92 Joules

Explanation:

Given that,

Mass of the crate, m = 170 kg

Distance, d = 10.2 m

The coefficient of friction, \mu=0.6

Let W is the work done by the mover. It is given by in terms of coefficient of friction as :

W=\mu mg\times d

W=0.6\times 170\times 9.8 \times 10.2

W = 10195.92 Joules

So, the work done by the mover is 10195.92 Joules. Hence, this is the required solution.

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Answer:

See the explanation below.

Explanation:

A lever is a simple machine that changes the magnitude and direction of the force applied to move an object. Minimizes the force needed to lift the object.

By means of the following image, we can see the principle of operation of a lever.

The load can be moved thanks to the force multiplied by the distance to the fulcrum.

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Which action might lead scientists to develop new explanations about the
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Answer:

The correct option is;

B. Designing experiments to replicate the conditions in which life may have first evolved on Earth

Explanation:

The proof to the hypothesis that life originated from inanimate inorganic, or non-living  molecules which is an explanation for the origin of life on Earth was provided by an experiment designed and performed in 1953 by Stanley L. Miller and Harold C. Urey which consisted of using chemicals proposed in the hypothesis and combining them through a specific design process to replicate expected atmospheric condition before the life began on Earth.

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Answer:

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Explanation:

Given that :

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Which components of an atom are found outside of the nucleus
butalik [34]
Electrons are found outside of the nucleus.
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Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
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