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Lostsunrise [7]
2 years ago
7

Akito pushes a wheelbarrow with 800 W of power. How much work is required to get the wheelbarrow across the yard in 12 s? Round

your answer to the nearest whole number.
It takes blank J of work to get the wheelbarrow across the yard
Physics
2 answers:
masha68 [24]2 years ago
5 0

Answer:

9600 J

Explanation:

The power is given by the ratio between the work done and the time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, we know the power:

P = 800 W

and the time taken:

t = 12 s

So, we can re-arrange the equation to get the work required:

W=Pt=(800 W)(12 s)=9600 J

12345 [234]2 years ago
4 0
800 x 12 = 9,600
9,600 is your answer
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NemiM [27]
It depends what the volume of the plate is
4 0
3 years ago
What is the equation for the potential energy stored in a spring when it is stretched or compressed?
gregori [183]

Answer:

if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).

8 0
3 years ago
To apply Problem-Solving Strategy 12.2 Sound intensity. You are trying to overhear a most interesting conversation, but from you
Ivenika [448]

Answer:

r₂ = 0.316 m

Explanation:

The sound level is expressed in decibels, therefore let's find the intensity for the new location

            β = 10 log \frac{I}{I_o}

let's write this expression for our case

           β₁ = 10 log \frac{I_1}{I_o}

           β₂ = 10 log \frac{I_2}{I_o}

           

          β₂ -β₁ = 10 ( log \frac{I_2}{I_o} - log \frac{I_1}{I_o})

          β₂ - β₁ = 10 log \frac{I_2}{I_1}

          log \frac{I_2}{I_1} = \frac{60 - 20}{10} = 3

           \frac{I_2}{I_1} = 10³

           I₂ = 10³ I₁

having the relationship between the intensities, we can use the definition of intensity which is the power per unit area

           I = P / A

           P = I A

the area is of a sphere

          A = 4π r²

           

the power of the sound does not change, so we can write it for the two points

          P =  I₁ A₁ =  I₂ A₂

          I₁ r₁² = I₂ r₂²

we substitute the ratio of intensities

          I₁ r₁² = (10³ I₁ ) r₂²

         r₁² = 10³ r₂²

         

         r₂ = r₁ / √10³

         

we calculate

          r₂ = \frac{10.0}{\sqrt{10^3} }

          r₂ = 0.316 m

8 0
3 years ago
Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen
m_a_m_a [10]

Answer:

F=2.25\times 10^{-5}\ N

Explanation:

According to question,

Charge 1 and charge 2 are 10^{-7}\ C

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N

So, the electric force of repulsion is 2.25\times 10^{-5}\ N.

8 0
3 years ago
A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the sta
Volgvan
The equation for this is very simple you add then you subtract then you get the answer then you divide then it all works out for you
6 0
3 years ago
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