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vovangra [49]
3 years ago
10

What is the result of seafloor spreading?

Physics
1 answer:
Over [174]3 years ago
6 0

Answer:

Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.

Explanation:

eventually the crust cracks.

You might be interested in
Can anyone help me please​
ArbitrLikvidat [17]

Answer:

Gravity.

Rocket ships.

Ball.

Basketball.

Explanation:

Gravity has to do a lot with air. It puts the planets in there area.

Rocket Ship has to do a lot with air. If i'm right, they calculate the area, weather, about the air.

A ball gets throwed in the air, which gravity comes into place.

Basketball is also a similar example to a ball.

7 0
3 years ago
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she tra
Ad libitum [116K]

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m

7 0
3 years ago
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
RSB [31]

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

6 0
3 years ago
A water wheel rotates with a period of 2.26 s. If the water wheel has a radius of l.94 m,
Dafna11 [192]

Answer:

The correct answer is B)

Explanation:

When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel.  So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.

The formula for calculating the velocity of a point on the edge of the wheel is given as

V_{r} = 2π r / T

Where

π is Pi which mathematically is approximately 3.14159

T is period of time

Vr is Velocity of the point on the edge of the wheel

The answer is left in Meters/Seconds so we will work with our information as is given in the question.

Vr = (2 x 3.14159 x 1.94m)/2.26

Vr = 12.1893692/2.26

Vr = 5.39352619469

Which is approximately 5.39

Cheers!

7 0
3 years ago
Consider a simple pendulum with a period
lisabon 2012 [21]

Answer:

1. The length is 8.35m

2. The period on the moon is 14.05 secs

Explanation:

1. Data obtained from the question. This includes the following:

Period (T) = 5.8 secs

Acceleration due to gravity (g) = 9.8 m/s2

Length (L) =...?

The length can be obtained by using the formula given below:

T = 2π√(L/g)

5.8 = 2π√(L/9.8)

Take the square of both side

(5.8)^2 = 4π^2 x L/ 9.8

Cross multiply

4π^2 x L = (5.8)^2 x 9.8

Divide both side by 4π^2

L = (5.8)^2 x 9.8 / 4π^2

L= 8.35 m

2. Data obtained from the question. This includes the following:

Acceleration due to gravity (g) = 1.67 m/s2

Length (L) = 8.35m (the length remains the same)

Period (T) =?

The period can be obtained as follow:

T = 2π√(L/g)

T = 2π√(8.35/1.67)

T = 14.05 secs

Therefore, the period on the moon is 14.05 secs

4 0
3 years ago
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