Answer:
Explanation:
Given that on the tree the gravitational energy stored is 8J
Then, mgh = 8J.
The apple begins to fall and hit the ground, what is the maximum kinetic energy?
Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)
So,
K.E(max) = 8J
Answer:
Explanation:
STEP 1
<u>Given</u>
Radius of cylinder = r = 25cm, 2.5m
mass = 27kg
cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder
height = 0.6m
<u>part 1</u>
The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by
<em>I = Icm + mh²</em>
<em>∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20 x (0.6)²</em>
<em>I=13.09kg.m²</em>
where
<em>I</em>cm is the rotational inertia of the cylinder about its central axis
m is the mass of the cylinder
h is the distance between the axis of the rotation and the central axis of the cylinder
r is the radius of the cylinder
<em> </em><em> I=13.09kg.m²</em>
<em>part2</em>
<em>from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero</em>
<em>Δk + Δu = 0</em>
<em>1/2 </em>I(w²-w²) = Ui-Uf
1/2 x 13.09w² = mgh
∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5
w=18.09rad/s
Answer:primary current is 3A
Explanation:
primary voltage Vp=240V
secondary voltage Vs=60V
resistance 5Ω
primary current in the circuit is
Answer:
From the narrative in the question, there seem to have been a break failure and the ordered step of response to this problem is to
1) Put on the hazard light to inform other road users of a problem or potential fault with your car and so they should continue their journey with caution.
2) Avoid pressing on the acceleration pedal as this might cause the car to gradually slow down due to friction and gravity
3)Try navigate the car to the service lane. This is the less busy lane where cars are sometimes parked briefly.
4) Continuously pump the breaks to try stop the car. Continuously pumping the breaks might just help you build enough pressure to stop the car because often time, there are some pressure left in the break.
5) At this point, the speed of the car should be relatively slow. So at this point, you could try apply the emergency hand break. Do not pull the emergency hand breaks if the car is on high speed. Doing this may cause the car to skid off the road.
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