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sashaice [31]
3 years ago
6

A total of 750 j of work was done when a force of 125 n was exerted on a box to move it. how far was the box moved?

Physics
1 answer:
Slav-nsk [51]3 years ago
3 0
Work = Force times Distance
W = Fd

Given W = 750J, F = 125N;

750 = 125d

Solving for d:
d = 750/125
d = 6

The box moved a distance of 6 meters.
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Which of the following best describes why tidal energy is considered a renewable energy resource? A. The Moon receives energy fr
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Answer:

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Explanation:

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Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains
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Given that the density of heptane is

d_h=\frac{0.684g}{mL}

The mass of heptane is

m_h=31\text{ g}

The density of water is

d_w=\frac{1g}{mL}

The mass of water is

m_w=37\text{ g}

The volume of heptane will be

\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}

The volume of water will be

\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}

The total volume of liquid in the cylinder will be 82.32 mL.

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If a reaction is exothermic and produces large quantities of heat reaching equilibrium, its equilibrium constant will be
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Read 2 more answers
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

5 0
3 years ago
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