Answer: If you use a very small resistance AND the circuit design is such that the voltage drop is across the resistance of the heating-wire-circuit, [nothing else in the circuit is limiting current flow] you will get more heating. That's what we have in a heat anticipator internal-nichrome-wire-heater device.
Explanation:
Answer: I actually need the same answer
Explanation:
Answer:
(a) 2.34 s
(b) 6.71 m
(c) 38.35 m
(d) 20 m/s
Explanation:
u = 20 m/s, theta = 35 degree
(a) The formula for the time of flight is given by


T = 2.34 second
(b) The formula for the maximum height is given by


H = 6.71 m
(c) The formula for the range is given by


R = 38.35 m
(d) It hits with the same speed at the initial speed.
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
The work done by a rotating object can be calculated by the formula Work = Torque * angle.
This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.
An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.
The formula is Work = torque * angle.
Torque = 1000 N*m
Angle = [50 revolutions] * [2π radians/revolution] = 100π radians
=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.