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3 years ago

A force of 500 N is used to slide a box across a smooth surface; the box moved 5 m in 1.2 seconds. What power is used?

2 answers:

8 0

F = 500N
S = 5m
T = 1.2s
----------------------------------
P = ?

We can use formula P = F * V, just because it was a smooth slide we can assume that average speed was V = S/T = 5 / 1.2 = 50/12

So the final answer would be:

P = F * V = 500 * 50/12

Are you sure those are correct numbers? The answer don't look nice :D

inysia [295]3 years ago

6 0

$Given:\\F=500N\\s=5m\\t=1.2s\\\\Find:\\P=?\\\\Solution:\\\\P= \frac{W}{t} \\\\W=Fs\\\\P= \frac{Fs}{t} \\\\P= \frac{500N\cdot 5m}{1.2s} \approx 2083.33W$

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A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

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3 years ago

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muminat

Answer:

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Therefore, the electrons are negatively charged and want to get away from the Negative Terminal and go to the Positive Terminal, Hence the electrons move from left to right and current flows from right to left.

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Ivanshal [37]

Answer:

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