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yawa3891 [41]
3 years ago
5

HELP ASAP REAL ANSWERS PLEASE You push a box across the floor with a force of 30 N. You push it 15 meters in 8 seconds. How much

work did you do? How much power did you use?
Physics
2 answers:
dalvyx [7]3 years ago
6 0

When you lift a box up off the floor, you must exert a force at least equal to the weight of the box. If you lift the box at constant speed, you "do work" on the box that would be equal to the force you exert (the weight of the box) times the height through which you lifted the box. If you exert a force of 50 Newtons on the box and lift it 1 meter high, then you did (50 N)x(1 m) or 50 N-m of work on the object. 1 Newton-meter is called a Joule (J), pronounced "jewel." In lifting the 50 N box upward 1 meter at constant speed, you would do 50 Joules of work on the box. That means your body used 50 J of energy to lift the box.

<span>and here is the Equation for Work: Work = Force x distance, or W = F d.</span>


Leya [2.2K]3 years ago
6 0

When you lift a box up off the floor, you must exert a force at least equal to the weight of the box. If you lift the box at constant speed, you "do work" on the box that would be equal to the force you exert (the weight of the box) times the height through which you lifted the box. If you exert a force of 50 Newtons on the box and lift it 1 meter high, then you did (50 N)x(1 m) or 50 N-m of work on the object. 1 Newton-meter is called a Joule (J), pronounced "jewel." In lifting the 50 N box upward 1 meter at constant speed, you would do 50 Joules of work on the box. That means your body used 50 J of energy to lift the box.

<span>and here is the Equation for Work: Work = Force x distance, or W = F d.</span>

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Answer:

FB = 0.187 N

Explanation:

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the angle between B and L is given by:

\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N

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4 0
3 years ago
What is the value of x in the equation below 1+2e^x+1=9
GuDViN [60]
<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of x:

1+2e^x+1=9   (1)

Firstly, we have to clear x:

2e^x=9-1-1  

2e^x=7  

e^x=\frac{7}{2}     (2)

Applying<u> Natural Logarithm</u> on both sides of the equation (2):

ln(e^x)=ln(\frac{7}{2})     (3)

xln(e)=ln(\frac{7}{2})     (4)

According to the Natural Logarithm rules xln(e)=x, so (4) can be written as:

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Answer:

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\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2

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Salsk061 [2.6K]

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or, 29/19.6 = m

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6 0
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