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cupoosta [38]
3 years ago
5

A 45 N girl sits on a bench 0.6 meters off the ground. How much work is done on the bench?

Physics
1 answer:
ycow [4]3 years ago
6 0

Answer: 27 joules

Explanation:

Work is done when force is applied on the bench over a distance. it is measured in joules.

Workdone = force x distance

= 45 N x 0.6 metres

= 27 joules

Thus, 27 joules of work is done on the bench.

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Answer:

When the velocity of an object changes it is said to be accelerating. Acceleration is the rate of change of velocity with time. ... Acceleration occurs anytime an object's speed increases or decreases, or it changes direction.

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What effect would this have on the momentum?
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Momentum describes an object in motion and is determined by the product of two variables: mass and velocity. Mass -- the weight of an object -- is usually measured in kilograms or grams for momentum problems. Velocity is the measure of distance traveled over time and is normally reported in meters per second. Examining the possible changes in these two variables identifies the different effects momentum can have on an object in motion.
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A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder
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Answer:

70509.8039216 N/C

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

\lambda=\dfrac{q}{l}

Electric field is given by

E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C

The electric field at the surface of the cylinder is 70509.8039216 N/C

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