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kenny6666 [7]
2 years ago
15

Where is there kinetic energy in this system?

Physics
1 answer:
Alika [10]2 years ago
6 0

Answer:

So kinetic means to move, something like that right, so the two balls that go in the air are where the kinetic energy is.

Explanation:

Hope it helps.

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The energy stored in a wooden log transforms when the log is burned. Which of the following explanations BEST describes how the
Makovka662 [10]

Answer:

D. the amount of chemical energy equals the amount of heat and light energy.

Explanation:

Given that the first law of thermodynamics affirmed that energy is neither created nor destroyed however, it can be transformed from one form to another. In other words, while, during the transformation of energy, no energy is lost, the input energy is also equal to output energy.

Hence, the chemical energy stored in the log is EQUAL to the heat and light energy produced by burning.

5 0
2 years ago
A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the
antoniya [11.8K]
Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0
2 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
kipiarov [429]

Answer:

I=68.31\times 10^{-6}\ A

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}

By putting the values

I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A

I=68.31\times 10^{-6}\ A

7 0
2 years ago
Read 2 more answers
A conducting sphere of radius 5.0 cm carries a net charge of 7.5 µC. What is the surface charge density on the sphere?
11111nata11111 [884]

Answer:

\sigma=0.014\ C/m^2

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2

So, the surface charge density on the sphere is 0.014\ C/m^2.

7 0
2 years ago
What is the net force required to give an automobile with a mass of 1,600 kg an acceleration of 4.5 m/s2
Paladinen [302]

Answer:

Fnet=7200 N

Explanation:

Fnet=mass x acceleration

mass= 1600kg acceleration=4.5m/s^2

8 0
2 years ago
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