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STatiana [176]
4 years ago
11

The mechanical energy of a bicycle at the top of a hill is 6,000 J. The bicycle stops at the bottom of the hill by applying the

brakes. If the potential energy of the bicycle is 2,000 J at the bottom of the hill, how much thermal energy was produced?
Physics
1 answer:
Semmy [17]4 years ago
7 0

Answer:

Thermal energy produce =4,000 J.

Explanation:

Given that

Mechanical  energy at the top of hill = 6,000 J

Mechanical  energy at the bottom of hill = 2,000 J

We know that energy is conserve

Energy at top of hill = energy at bottom of hill + Thermal energy produce

So now by putting the values

Energy at top of hill = energy at bottom of hill + Thermal energy produce

6,000 = 2,000+ Thermal energy produce

Thermal energy produce =6,000-2,000 J

Thermal energy produce =4,000 J.

As we know that thermal energy produce due to friction when one mechanical component slides on the other mechanical component then always heat is generated and this heat is known as thermal energy.

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JulsSmile [24]

This is true. The break handles and brakes are levers and the seat adjustment (raise or lower) is a screw.

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3 years ago
Read 2 more answers
Please help Math Phys
Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land.  Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0
3 years ago
Objects 1 and 2 attract each other with a electrostatic force of 18.0 units. If the charge of Object 2 is tripled, then the new
Leya [2.2K]

Answer:

The force will be 54.0 units

Explanation:

The magnitude of the electrostatic force between two charged objects is given by Coulomb's Law:

F=\frac{kq_1 q_2}{r^2}

where

k is Coulomb's constant

q1, q2 are the magnitude of the two charges

r is the separation between the two charges

From the equation, we see that the magnitude of the force is directly proportional to the charge of object 2:

F\propto q_2

In this problem, the initial force between the two objects is

F = 18.0 N

And so, when the charge on object 2 is tripled,

q_2'=3q_2

The new electrostatic force will be

F'\propto q_2' = (3q_2) = 3F

So, the force will also triple: since the original force was 18.0 units, the new force will be

F'=3F=3(18.0)=54.0

5 0
3 years ago
Similarity of gravity and electromagnetic force
Mashutka [201]

Answer:

Gravity is an attractive force as well as electromagnetic, but electromagnetic attracts and repels.

Explanation:

5 0
3 years ago
A truck slows from a velocity of 25 m/s to a stop in 70 m. What was the truck’s acceleration
igomit [66]

v^2-u^2=2 x a x d

25^2-0^2=2 x a x 70

625-0=140 x a

625=140a

a=625/140

a=4.46 m/s^2

im not very sure but i think this is how you do this

4 0
3 years ago
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