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lawyer [7]
2 years ago
7

A soap box derby car at the top of a ramp has more _________ energy and at the finish line at the bottom of the ramp it has more

_________ energy.
A) kinetic, kinetic
B) potential, kinetic
C) kinetic, potential
D) potential, potential
Physics
1 answer:
astra-53 [7]2 years ago
5 0
Kinetic, potential because, at the top of the ramp it’s going faster. Potential at the bottom of the ramp is potential because, it’s not doing any motion.
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When two atoms of the same nonmetal react, they often form a(an) A. ionic bond. B. polyatomic ion. C. diatomic molecule. D. pola
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<span>When two atoms of the same nonmetal react,they form what we know today as a diatomic molecule.</span>
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How are mass and inertia related?
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The greater the mass the greater is inertia.
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A ball is falling after rolling off a tall roof<br> The ball has what type of energy.
CaHeK987 [17]

Answer:

Linear and rotational Kinetic Energy + Gravitational potential energy

Explanation:

The ball rolls off a tall roof and starts falling.

Let us first consider the potential energy or more specifically gravitational potential energy (mgh; m = mass of the ball, g = acceleration due to gravity, h = height of the roof). This energy comes because someone or something had to do work to take the ball to the top of the roof against the force of gravity. The potential energy is naturally maximum at the top and minimum when the ball finally reaches the ground.

Now, the ball starts to roll and falls off the roof. It shall continue rotating because of inertia (Newton's first law). This contributes to the rotational kinetic energy (\frac{1}{2}I\omega^2; I=moment of inertia of the ball & \omega = angular velocity).

Finally comes the linear kinetic energy or simply, kinetic energy (\frac{1}{2}mv^2) which is caused due to the velocity v of the ball.

3 0
3 years ago
find the rate of positive acceleration of an automobile which went from a complete stop to a velocity of 30 meters per second in
maks197457 [2]

Answer:

3 m/s^2

Explanation:

acceleration= Change in velocity/time

= 30-0 / 10

= 30/10

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3 0
3 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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