Answer:
12 m/s
Explanation:
English Translation
At what maximum speed is the cyclist traveling for a short time if he has a maximum power of 420W and at this speed he must overcome the resistance 35N.
The Power, P, expended by a moving body with velocity, v, moving against a resistive force of F is given as
P = Fv
P = 420 W
F = 35 N
v = ?
420 = 35v
v = (420/35) = 12 m/s
In Polish/Po polsku
Moc P, wydana przez poruszające się ciało z prędkością v, poruszająca się przeciw sile oporu F, jest podana jako
P = Fv
P = 420 W
F = 35 N
v = ?
420 = 35v
v = (420/35) = 12 m/s
Hope this Helps!!!
Mam nadzieję że to pomoże!!!
Answer:
When the resistances are connected in the parallel the equivalent resistance will be always less than the value of the lowest resistance in the parallel circuit.
The equivalent resistance in the parallel circuit can be calculated using the following formula.
1/R=1/R1+1/R2+1/R3
1/R= 1/10+1/20+1/30
1/R=(6+3+2)/60
1/R=11/60
R=60/11
R=5.45 Ohms
Answer:through a small space through atoms
Explanation:
A its Stratosphere, Sorry I didn't see your answer, its bilogy I think not physics.. :)
To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is
KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f
, since initially cylinder was at rest
since at the ground potential energy is zero
The mathematical values are,
Here,
m = mass
g= Gravity
h = Height
V = Velocity
moment of Inertia in terms of its mass and radius
Angular velocity in terms of tangential velocity and its radius
Replacing the values we have that
mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2
gh = \frac{v^2}{2}+\frac{v^2}{4}
v = \sqrt{\frac{4gh}{3}}
From trigonometry the vertical height of inclined plane is the length of this plane for 
, then
Replacing,
Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s
