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3 years ago

Describes an increase in potential energy

1 answer:

7 0

I’m not quite sure what your asking, but if you need an example of an increase in potential energy, a good example would be when a ball begins to slow down after rolling down a hill. If there is another part of this question, that would be great so I could help a little more but for now I hope this helped! :-)

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How is the independent variable affected by the dependent variable

stiv31 [10]

Answer:

A dependent variable is a variable that is tested in an experiment. An independent variable is that can be modified. Depending on what you are testing, the dependent variable will change accordingly to the dependent variable.

- I'm reading this back and it doesn't make much sense, if you want me to reword this I can

5 0

3 years ago

a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&

Katen [24]

Answer:

a) T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

sin 30 = $\frac{T_x}{T}$

cos 30 = $\frac{T_y}{T}$

Tₓ = T sin 30

T_y = T cos 30

Y axis

T_y -W = 0

T cos 30 = mg (1)

X axis

Tₓ = m a

they relate it is centripetal

a = v² / r

we substitute

T sin 30 = m $\frac{v^2}{r}$ (2)

a) we substitute in 1

T = $\frac{mg }{cos 30}$

T = $\frac{ 0.2 \ 9.8}{cos \ 30}$

T = 2.26 N

b) from equation 2

v² = $\frac{T \ sin 30 \ r}{m}$

If we know the length of the string

sin 30 = r / L

r = L sin 30

we substitute

v² = $\frac{ T \ sin 30 \ L \ sin 30}{m}$

v² = $\frac{TL \ sin^2 30}{m}$

For the problem let us take L = 1 m

let's calculate

v = $\sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }$

v = 1.68 m / s

8 0

3 years ago

According to the picture, how would the gravitational potential energy of the car change if its weight was doubled? A) The gravi

Lorico [155]

Answer is A

Samuriguy125 on YT

3 0

3 years ago

Read 2 more answers

Figure 10.20 in your textbook shows an energy diagram for a system with total energy E1. Suppose the system's total energy is E2

wolverine [178]

The particles can undergo small oscillations around x₂.

The given parameters;

<em>initial energy of the particles = E₁</em>
<em>final energy of the particles, E₂ = 0.33E₁</em>

The movement of the particles depends on the kinetic energy of the particles.

When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.

However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.

The maximum position the particle can oscillate is x₅

The half position the particles can oscillate is x₃

Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.

Thus, we can conclude that the particles can undergo small oscillations around x₂.

Learn more here:brainly.com/question/23910777

3 0

3 years ago

If the distance d (in meters) traveled by an object in time t (in seconds) is given by the formula d = A + Bt^2, the SI units of

Yuliya22 [10]

Answer:

The SI units of the “A” is m (meters)

The SI units of the “B” is m/s^2

Explanation:

Given the distance = d meters.

Time taken to travel = t (seconds)

Function of the distance, d = A + Bt^2

Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.

Thus, the SI units of the “A” is = m (meters)

The SI units of the “B” is = m/s^2

6 0

4 years ago