Answer:
(a) Workdone = -27601.9J
(b) Average required power = 1314.4W
Explanation:
Mass of hoop,m =40kg
Radius of hoop, r=0.810m
Initial angular velocity Winitial=438rev/min
Wfinal=0
t= 21.0s
Rotation inertia of the hoop around its central axis I= mr²
I= 40 ×0.810²
I=26.24kg.m²
The change in kinetic energy =K. E final - K. E initail
Change in K. E =1/2I(Wfinal² -Winitial²)
Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]
Change in K. E= -27601.9J
(a) Change in Kinetic energy = Workdone
W= 27601.9J( since work is done on hook)
(b) average required power = W/t
=27601.9/21 =1314.4W
Answer:
G M m / R^2 = m v^2 / R gravitational force = centripetal force
G M = v^2 R = constant
As v increases R will must decrease
Take the moon as an example
S = 2 pi R where R is about 240,000 miles for one orbit
S / 1 day = 54,000 miles/day for a 28 day circuit
S / 1 hr = 54000 / 24 = 2200 mph which is much less than a satellite in orbit
Answer:
a) a = 1,865 m / s² and b) t = 8.1 s
Explanation:
a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction
F = m .a
a = F / m
a = 8.02 10² /4.3 10²
a = 1,865 m / s²
b) We use kinematic relationships in one dimension
vf = vo + at
vf = 0 + a t
t = vf / a
t = 15.1 / 1.865
t = 8.1 s
The resistance of the lamp is apparently 50V/2A = 25 ohms.
When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.
In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.
The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.
The power delivered by the 200V mains is (200V) x (2A) = 400 watts.
The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.
The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.
Together, they add up to the 400 watts delivered by the mains.
CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
Answer : The mass of ice melted can be, 3.98 grams.
Explanation :
First we have to calculate the moles of ice.
where,
Q = energy absorbed = 27.2 kJ
= enthalpy of fusion of ice = 6.01 kJ/mol
n = moles = ?
Now put all the given values in the above expression, we get:
Now we have to calculate the mass of ice.
Molar mass of ice = 18.02 g/mol
Thus, the mass of ice melted can be, 3.98 grams.
