Answer:
<h2>0.39m/s^2</h2>
Explanation:
Step one:
given data
mass m= 300kg
applied force F= 1000N
coefficient of friction μ= 0.3
Step two:
The net force Fn= applied force-friction force
Fn=F-F1
F1= limiting force
F1=μ*m*g
F1=0.3*300*9.81
F1=882.9N
the Net force= 1000-882.9
Fn=117.1N
Step three:
we know that
F=ma
Fnet=ma
a= Fnet/m
a=117.1/300
a=0.39m/s^2
On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.
You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is
(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1
while during the return trip, it is
(0 - 5) / (12.5 - 10) = -5/2.5 = -2
Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.
Answer:
49.6°
Explanation:
= Unpolarized light
= Light after passing though second filter = 
Polarized light passing through first filter
Polarized light passing through second filter
The angle between the two filters is 49.6°
Answer:
Given that,
- Power = 2000 W
- time = 60 seconds
- distance= 10m
Power = work done ÷ time
Here, since the movement is vertical, w = mgh
So,
Power = mgh÷t
2000 = (m × 9.8 ×10) ÷ 60
m = (2000 ×60) ÷98
m = 1224.5kg
Answer:
the one above the surface of earth
Explanation:
earth has gravity the ball of the moon would float away
