Answer:
Her speed is 1.1 m/s, and her velocity is 0 m/s
Explanation:
Speed = Distance covered/Time
Given
Distance = 400m
Time = 6minutes = 6*60 = 360 secs
Substitute the given parameter into the formula;
Speed = 400/360
Speed = 1.1m/s
Since the track is a circular track, the displacement will be zero. She is only moving in a circular path (no direction)
Velocity = Displacement/Time
Velocity = 0/3600
Velocity = 0m/s
Hence her speed is 1.1 m/s, and her velocity is 0 m/s
in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s
Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js
now 1 Joule= 10⁷ ergs
so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)
h=6.626 x10⁻²⁷ erg s
Answer:
A copper wire current consists of electrons appropriately called conduction electrons.
Explanation:
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Answer:
a) ![(Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]](https://tex.z-dn.net/?f=%28Qa%2Ag%2AVb%29-%28Qh%2AVb%2Ag%29%3D%28Qh%2AVb%2Aa%29%5C%5Cwhere%20%5C%5Cg%3Dgravity%20%5Bm%2Fs%5E2%5D%5C%5Ca%3Dacceleration%20%5Bm%2Fs%5E2%5D)
b) a = 19.61[m/s^2]
Explanation:
The total mass of the balloon is:
![massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\](https://tex.z-dn.net/?f=massball%3Ddensityheli%2Avolumeheli%5C%5C%5C%5Cmassball%3D0.41%20%5Bkg%2Fm%5E3%5D%2A0.048%5Bm%5E3%5D%5C%5Cmassball%3D0.01968%5Bkg%5D%5C%5C%5C%5C)
The buoyancy force acting on the balloon is:
![Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]](https://tex.z-dn.net/?f=Fb%3Ddensityair%2Agravity%2Avolumeball%5C%5CFb%3D1.23%5Bkg%2Fm%5E3%5D%2A9.81%5Bm%2Fs%5E2%5D%2A0.048%5Bm%5E3%5D%5C%5CFb%3D0.579%5BN%5D)
Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.
In the attached image we can see the free body diagram and the equation deducted by Newton's second law
Answer:
<h2>
128.61 Watts</h2>
Explanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque =
= 
I is the rotational inertia = 16kgm²



To get the angular acceleration, we will use the formula;


Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts