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Ivenika [448]
3 years ago
14

A calf stretch is a/an _______________________________.

Physics
1 answer:
swat323 years ago
5 0
<span>C anaerobic activity </span>
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z jaka maksymalna predkoscia porusza sie przez krotka chwile rowerzysta jezeli dysponuje moca maksymalna 420w i przy predkosci z
qwelly [4]

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!!!

6 0
3 years ago
Answer it fast and no spamming....! !
ExtremeBDS [4]

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

3 0
3 years ago
How do electromagnetic waves move through space?
lesantik [10]

Answer:through a small space through atoms

Explanation:

4 0
4 years ago
Clouds, wind, and rain are part of the _____.
Musya8 [376]
A its Stratosphere, Sorry I didn't see your answer, its bilogy I think not physics.. :)
8 0
3 years ago
A uniform, solid cylinder of radius 5.00 cm and mass 3.00 kg starts from rest at the top of an inclined plane that is 2.00 m lon
ch4aika [34]

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

KE_i = 0, since initially cylinder was at rest

PE_f = 0 since at the ground potential energy is zero

The mathematical values are,

mgh = \frac{1}{2} mV^2 + \frac{1}{2}I\omega^2

Here,

m = mass

g= Gravity

h = Height

V = Velocity

I = \frac{mr^2}{2} moment of Inertia in terms of its mass and radius

\omega = \frac{V}{r} 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 sin\theta, then

h = 2.00*sin 25

h = 0.845 m

Replacing,

v = \sqrt{\frac{4(9.8)(0.845)}{3}}

V = 3.32 m/s

Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s

6 0
3 years ago
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