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Ahat [919]
3 years ago
8

If 3,600 J of work is done in 3.0 s, what is the power?

Physics
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

1,200 watts

Explanation:

1 watt = 1 Joule (J) of work / second

So, 3600 Joules of work / 3 seconds is:

3600 J / 3 seconds = 1,200 watts

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What is 1/12+7/9 hj​
Umnica [9.8K]

Explanation:

\frac{1}{12}  +  \frac{7}{9}  \\  \\  =  \frac{1 \times 3}{12 \times 3}  +  \frac{7 \times 4}{9 \times 4} \\  \\  =  \frac{3}{36}  +  \frac{28}{36}  \\  \\  =  \frac{3 + 28}{36}  \\  \\  =  \frac{31}{36}  \\  \\   \huge \purple{ \boxed{\therefore \:  \frac{1}{12}  +  \frac{7}{9}  =  \frac{31}{36} }} \\

6 0
3 years ago
Read 2 more answers
A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi
Rus_ich [418]
A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² =  1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² =  1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s


b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s² 
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
3 0
3 years ago
Two friends are carrying a crate of mass 200 kg up a flight of stairs. The crate has length 1.25 m and height 0.500 m, and its c
Marysya12 [62]

Ans; see attached file for calculation and answer

Explanation:

4 0
3 years ago
In an experiment, a researcher can make claims about causation
Yakvenalex [24]
<span>In an experiment, a researcher can make claims about causation if the independent variable changes because of changes made to the dependent variable. Causation works on cause and effect, so the changed independent variable is the cause and the changed dependent variable is the effect. In an experiment the independent variable is changed to determine the dependent variables value, so the two are directly related.</span>
8 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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