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cestrela7 [59]
3 years ago
14

What does a convergent boundary between oceanic and continental plate form?

Physics
1 answer:
Romashka [77]3 years ago
7 0

Answer:

a convergent plate boundary

Explanation:

when one plate is oceanic, there are large volcanoes. These volcanoes are found in lines that outline the subduction zone. Earthquakes also happen in these zones. The Aleutian Islands that border southern Alaska are an island arc

You might be interested in
Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele
Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0
2 years ago
A ball at the end of a string is swinging in a horizontal circle of radius 0.85 m. The ball makes exactly 3 revolutions per seco
Anettt [7]

Answer:

It's centripetal acceleration is 301.7 m/s²

Explanation:

The formula to be used here is that of the centripetal acceleration which is

ac = rω²

where ac is the centripetal acceleration = ?

ω is the angular velocity = 3 revolutions per second is to be converted to radian per second: 3 × 2π  = 3 × 2 × 3.14 = 18.84 rad/s

r is the radius = 0.85 m

ac = 0.85 × 18.84²

ac = 301.7 m/s²

It's centripetal acceleration is 301.7 m/s²

8 0
2 years ago
3) If an airplane is in flight cruising at constant velocity, what can be said about the net work
Crazy boy [7]

Explanation:

<h2>Yes!</h2>

<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>
6 0
3 years ago
A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

\rho (r) \  \alpha  \  \delta (r -R)

\rho (r) = k \  \delta (r -R) \ \  at \ \  (r = R)

\rho (r) = 0\ \ since \ r< R  \ \ or  \ \ r>R---- (1)

To find the constant k, we  examine the total charge Q which is:

Q = \int \rho (r) \ dV = \int \sigma \times dA

Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2

∴

\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta  \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2

\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2

Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

k * R^2= \sigma \times  R^2

k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

Thus, the units are verified.

The integrated charge Q

Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \  sin \theta  \ dr \ d\theta \  d \phi  \\ \\  Q = \int ^{2 \pi}_{0} \  d \phi  \int ^{\pi}_{0} \ sin \theta  \int ^R_{0} \rho (r) r^2 \ dr

Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  \int ^R_0  * \delta (r-R) r^2 \ dr

Q = 4 \pi  \sigma  *R^2    since  ( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )

\mathbf{Q = 4 \pi R^2  \sigma  }

6 0
3 years ago
A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

= 61.37 \ s

8 0
2 years ago
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