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3 years ago

What occurs along a convergent plate boundary?

Physics

2 answers:

vova2212 [387]3 years ago

8 0

The answer is C.

Hope this helps

leonid [27]3 years ago

7 0

So first things first its c. because when two plates collide it causes a earthquake so u can rule those out and of course no not volcano so that's it u only have c left as your answer choice.
:/

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Compared to the sun, a star whose spectrum peaks in the infrared is:

Vladimir79 [104]

The answer is cooler. Hope this helps.

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3 years ago

What is direct current? In which direction does current go according to the electron flow convention?

KatRina [158]

Answer:

Answer to the question is:

Explanation:

Direct Current:

It is that current where electrons circulate in the same amount and sense in time, that is, flowing in the same direction. Its polarity is invariable and causes a current of relatively constant amplitude to flow through a load. This type of current is known as direct current (DC), and is generated by a battery.

the current of electrons will leave the negative terminal of the battery, (because they repel each other and also repel free electrons in the copper conductor), and go to the positive terminal where there is a lack of electrons, passing through the circuit to which it is connected. In this way the electric current is produced.

4 0

3 years ago

These two pls :)))) ill mark brainliest :)

Anna71 [15]

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: $F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}$

$F_{G}$ = gravitational force (Newtons/N)

G = gravitational constant, 6.67430 × 10¹¹ $\frac{N*m^{2}}{kg^{2}}$

m₁ and m₂ = masses of two objects (kilograms/kg)

r² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

4 0

3 years ago

A turntable must spin at 33.4 rpm (3.50 rad/s) to play an old-fashioned vinyl record. how much torque must the motor deliver if

Westkost [7]

In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration

For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds

α = (3.5 - 0) / 1.8
α = 1.94 rad/s²

The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002

τ = 1.94 * 0.002
τ = 0.004

The torque is 0.004

4 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:

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):

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

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:

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

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

7 0

3 years ago