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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 NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

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5 0
3 years ago
The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t
Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of 8.1\times 10^5 N

Mass of train=1.9\times 10^7

Final speed (v)=68 km/h \approx 18.88 m/s

F=ma

so acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}

a=0.042631 m/s^2

and acceleration is

a=\frac{v-u}{t}

0.042631=\frac{18.88-0}{t}

t=443.089 \approx 443.1 s

8 0
3 years ago
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siniylev [52]
2m/s is the answer ! hopefully this helps
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3 years ago
A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?
givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

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8 0
3 years ago
In which condition the acceleration of a moving vehicle become zero​
Alekssandra [29.7K]

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When,the vehicle has uniform velocity, it's acceleration becomes zero

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