physics
How can there be acceleration if the body is moving at constant velocity?
physics
(PE+KE)initial = (PE+KE)final mgh + mv0^2/2 = 0 + mv^2/2 => h = (v^2-v0^2)/2g
PHYSICS
The current through each bulb is the voltage across it divided by its resistance. In this case the bulbs are in parallel and voltage across them is 2.4V.
PHYSICS
The series resistor R should be chosen so as to have a voltage drop of 7.6V(10-2.4)across it. The currents in the two bulbs: i1=2.4/3.0 = 0.8A i2=2.4/2.5 = 0.96A Total current in the circuit = i1+i2 = 1.76A R = 7.6V/1.76A = 4.3 Ohms
physics
First find K of the spring: Mg = Kx here, x = elongation of the spring = 13.40 - 12.00 = 1.40 cm = 0.014 m K = Mg/x = 3.15*9.8/0.014 = ?? Having got K, find X from: KX^2/2 = 10.0J where X is the elongation when stored PE is 10.0 Joules. The total length of spring then would be...
physics
use the formula: V^2 = U^2 + 2*a*s here, V=0(final speed); U= Initial speed (convert units mi/hr to m/s) a = acceleration s = 108 m Find a in m/s^2 - your answer will be a negative value indicating deceleration.
Physics
Initial momentum of boat+Sally system is=(125+65)*5= 190*5=950 Kg m/s Sally's speed is 2.0m/s relative to boat i.e. w.r.t.a stationary observer it is (2+V)m/s where V is the speed of the boat when Sally walks. Since there is no external force acting on our system, the line...
physics
F = k q^2/r^2 = 9x10^9*1x10^-12/16x10^-4 = 5.6 N 5.6=m*9.8 => m=5.6/9.8=5.7 Kg
Physics
Yes, you did it correctly. In order to understand phasor diagram, read your text book. In a purely inductive load, the current lags voltage by 90 deg. In this case (R+L load)the phase angle will be less than 90 - you can find it out once you read the text.
physics
The string closest to the axis of rotation would have the maximum tension. Draw the free body diagram and it will be clear that the tension in the closest string balances not only the centrifugal force on the attached ball but also the tension in the string next in the chain.
Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | Next>>
For Further Reading
