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Arada [10]
3 years ago
5

A ball rolls from point A to point B. The total energy of the ball at point A isn’t the same as the sum of its potential energy

and kinetic energy at point B. Which statement best explains this situation? it just goes downhill
A. The rolling ball doesn’t follow the law of conservation of energy.
B. Some of the ball’s energy is transformed to thermal energy.
C. The rolling ball loses some of its energy because of gravity.
D. The ball absorbs some of its own potential energy.
Physics
2 answers:
murzikaleks [220]3 years ago
3 0
B. Some of the ball’s energy is transformed to thermal energy.

Hope this helps you!
Licemer1 [7]3 years ago
3 0

B. Some of the ball’s energy is transformed to thermal energy.

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Explain how an alpine glacier can change the topography of a mountainous area
igor_vitrenko [27]
<span>An alpine glacier can change the topography of a mountainous area through Glacial Erosion and Glacial Deposition. Glaciers are agents of erosion, it can pick up and carry large rocks and sediments. In the process, a deep cavity or hole can form when the glacier plucks a big rock from where it passed. Glaciers have shaped many Mountain Ranges and have created distinct landforms by its erosion process.  In Glacial Deposition, as glaciers melt, it deposits all that it carried and a landform is developed.</span>
7 0
3 years ago
How is energy transferred between two objects at different temperatures?
mojhsa [17]

When calculated energy transferred between objects use the definition of heat as

4 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
The magnitude of the weight of a 3.0 kg object on the surface of the earth is 29 N. True False
madreJ [45]
True

In fact, the weight of an object on the surface of the Earth is given by:
F=mg
where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
F=mg=(3.0 kg)(9.81 m/s^2)=29 N
8 0
3 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
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