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3 years ago

What is one way that continental rifts are similar to mid-ocean ridges and what is one that they are different?

2 answers:

7 0

Answer: Both mid-ocean ridges and continental rifts form where two plates are moving apart from each other. However, mid ocean ridges occur in the ocean, while continental rifts occur on land.

Explanation:

This is word for word on the answer so i would change it up a little

adell [148]3 years ago

3 0

Explanation:

Continental rifts and mid-ocean ridges are both features of a divergent plate margin.

In both cases plates are moving away from one another. Therefore they are creating new land masses.

A continental drift like the east African rift valley is where a continent begins to pull apart or diverges.
A mid-ocean ridge is divergent margin in the ocean.

They are different in that, continental rift occurs within the continental plate that are on land.

But:

Mid-ocean ridges are in the oceanic crust in the ocean . They form the largest physiographic structure on the earth surface called the mid-ocean ridge.

learn more:

Descending lithosphere brainly.com/question/9582362

#learnwithBrainly

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A 1,200 kg dragster, starting from rest, reaches a maximum velocity of 140m/s in 5 seconds. At the 5 second mark, the dragster d

SSSSS [86.1K]

Answer:

Drag or air resistance

Explanation:

The force of friction caused by a moving fluid is called drag. When that fluid is air, it's also known as air resistance.

8 0

3 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

3 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance during a time t, and object B falls th

UNO [17]

Answer:

Distance covered by B is 4 times distance covered by A

Explanation:

For an object in free fall starting from rest, the distance covered by the object in a time t is

$s=\frac{1}{2}gt^2$

where

s is the distance covered

g is the acceleration due to gravity

t is the time elapsed

In this problem:

- Object A falls through a distance $s_A$ during a time t, so the distance covered by object A is

$s_A=\frac{1}{2}gt^2$

- Object B falls through a distance $s_B$ during a time 2t, so the distance covered by object B is

$s_B=\frac{1}{2}g(2t)^2 = 4(\frac{1}{2}gt^2)=4s_A$

So, the distance covered by object B is 4 times the distance covered by object A.

5 0

3 years ago

The type of energy that depends on position is called

attashe74 [19]

The type of energy that depends on position is called
kinetic energy

5 0

3 years ago

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .

5 0

3 years ago