The following
are the answers to the questions presented:
a. The joules of energy required to run a 100W light bulb for one day is 8640000J
b. The amount of coals that has to be burned to light that light bulb for one day is 0.96kg
The solution would
be like this for this specific problem:
<span>P=<span>W/s</span>→W=Pt=100W1day <span><span>24h/</span><span>1day </span></span><span><span>3600s/</span><span>1h</span></span>=8640000J</span>
<span>W=<span>30/100</span>wm→m=<span><span>100W/</span><span>30w</span></span>=<span><span>100×8640000J/</span><span>30×30×<span>10in thepowerof6 </span><span>J/<span>kg</span></span></span></span>=0.96kg</span>
<span>I am hoping that
these answers have satisfied your queries and it will be able to help you in
your endeavors, and if you would like, feel free to ask another question.</span>
Answer:
Incomplete question
The complete question is
A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.
At the bottom of the ride, what is the rate of change of the rider's momentum?
Explanation:
Radius of wheel is 6m
Rider mass=96kg
He completes one revolution in 9.6s
Let get angular velocity (w)
1 Revolution =2πrad
θ=2πrad
w= θ/t
w=2π/9.6
w=0.654rad/s
Linear speed is give as
v=wr
v=0.654×6
v=3.93m/s
Centripetal acceleration a
a=rw²
a=6×0.654²
a=2.57m/s²
Acceleration due to gravity g=9.81m/s²
According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider
ΣF = ma
N-W=ma
N-mg=ma
N=ma+mg
N=m(a+g)
N=96(2.57+9.81)
N=1188.48 N
Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.
Answer:
I'm not sure but I think it's 35-39
Answer:
Explanation:
We know that the gravity on the surface of the moon is,
<u>Gravity at a height h above the surface of the moon will be given as:</u>
..........................(1)
where:
G = universal gravitational constant
m = mass of the moon
r = radius of moon
We have:
is the distance between the surface of the earth and the moon.
Now put the respective values in eq. (1)
is the gravity on the moon the earth-surface.
Answer:
hope this was good for u and I believe it would be solid
