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frozen [14]
3 years ago
15

If you perform 40 J of work lifting a 10-N box from the floor to a shelf, how high is the shelf?

Physics
1 answer:
Kaylis [27]3 years ago
4 0
Before anything, if you are lifting the Work value have to be negative, reason is because the formula to the work of the weight is:

w = p \times d \times cos \:  \theta
*Photo to help you to visualize. (P = Weight Force, F = Force)

If you are lifting, you are exercising a force that opposes the Weight Force, Force up (lifting) and Weight Force down (gravity pulling).
This forms a 180 angle and the cos 180 = -1.

Well let's do the question:

Work (W) = -40 J (Joule = Kg x m^2/s^2)
(negative, explanation given)

Weight Force (P) = 10 N (Newton = kg x m/s^2)

Distance (D) = ?

W = P x d x cos theta

-> -40 = 10 x d x -1
-> -40 = -10 x d
-> d = -40/-10 (J/N)
-> d = 4 meters

d =j  \div n \\ d = ((kg \times  {m}^{2} ) \div  {s}^{2}) \div ((kg \times m) \div  {s}^{2}) \\ d = m \: (meters)

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A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to
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Answer:

The answer is below

Explanation:

a) The vertical displacement = Δy = 21.5 m - 1.5 m = 20 m

The horizontal displacement = Δx = 69 m wide

Using the formula:

\Delta y = u_yt+ \frac{1}{2}a_yt^2\\ \\u_y=initial\ velocity\ of \ car\ in\ y\ direction = 0,a_y=g=acceleration\ due\ to\ gravity\\=10m/s^2\\\\\Delta y =  \frac{1}{2}a_yt^2\\\\\Delta y=\frac{1}{2}a_yt^2\\\\t=\sqrt{\frac{2\Delta y}{a_y} }=\sqrt{\frac{2*20}{10} }  =2\ m/s

Also:

\Delta x = u_xt+ \frac{1}{2}a_xt^2\\ \\u_x=initial\ velocity\ of \ car\ in\ x\ direction = 0,a_x=acceleration=0\\\\\Delta x =  u_xt\\\\u_x=\frac{\Delta x}{t}=\frac{69}{2} =34.5\ m/s

b)The car is moving at a constant speed in the horizontal direction, hence the initial velocity = final velocity

v_x=u_x=34.5\ m/s\\\\v_y=u_y+a_yt\\\\v_y=0+gt\\\\v_y=10(2)=20\ m/s\\\\v=\sqrt{v_x^2+v_y^2}=\sqrt{34.5^2+20^2}=39.9\ m/s\\ v=39.9\ m/s

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Read 2 more answers
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height
rewona [7]

Answer:

W=17085KJ

Explanation:

From the question we are told that:

Height H=16m

Radius R=3

Height of water H_w=9m

Gravity g=9.8m/s

Density of water \rho=1000kg/m^3

Generally the equation for Volume of water is mathematically given by

 dv=\pi*r^2dy

 dv=\frac{\piR^2}{H^2}(H-y)^2dy

Where

   y is a random height taken to define dv

Generally the equation for Work done to pump water is mathematically given by

 dw=(pdv)g (H-y)

Substituting dv

 dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)

 dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy

Therefore

 W=\int dw

 W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy

 W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)

 W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0

 W=3420.84*0.25[2401-65536]

 W=17084965.5J

 W=17085KJ

 

'

'

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