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Tanya [424]
3 years ago
13

the law of conversation of energy and describe the energy transformations that occur as you coast down a long hill on a bicycle

and then apply the brakes to make the bike stop at the bottom.
Physics
1 answer:
deff fn [24]3 years ago
7 0

Answer:

Potential energy is transformed into kinetic energy

friction work decreases kinetic energy

Explanation:

The law of conservation of  the mechanical energy is the sum of kinetic energy plus the different forms of potential energy, this energy is constant throughout the trajectory if the dissipative force (friction) is zero.

Let us apply this to our case, in the upper part of the trajectory almost all the mechanical energy is potential, and a very small part is kinetic, the bicycle goes very slowly, as it descends without pedaling the speed increases so that the kinetic energy it increases and the height decreases therefore the potential energy decreases, but the sum of the two energies remains constant.

Potential energy is transformed into kinetic energy

When the brakes are applied, a dissipative force enters the system that causes part of the energy to be transformed into heat and part into work of this dissipative force against the wheel, two resulting in a net decrease in mechanical energy and therefore a decrease in the speed of the bicycle, the value of this decrease is given by

                  W = DK

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a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi
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The radius of the sphere in meters is ,r = 10\sqrt{5} -20

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

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1 year ago
A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

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