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3 years ago

What is a Rift Valley ?

2 answers:

7 0

a steep-sided valley formed by the downward displacement of a block of the earth's surface between nearly parallel faults or fault systems.

Naddika [18.5K]3 years ago

6 0

a large elongated depression with steep walls formed by the downward displacement of a block of the earth's surface between nearly parallel faults or fault systems.

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Help me pls, I need it now

posledela

6.) a

7.) d

8.) b

9.) b

10.) a

3 0

2 years ago

Which reaction takes place in a nuclear fission reactor?

anzhelika [568]

Answer:

Last option in the list of possible answers, with U235 and n (neutron) in the left (originators) of the reaction diagram.

Explanation:

Uranium 235 (which is a fissile isotope of uranium) plus slow neutrons is what produce the chain reaction that feeds nuclear reactors.

3 0

2 years ago

What is the magnitude of the displacement of a car with an acceleration of 3.75 m/s2 as it increases its speed from 5.20m/s to 1

NeX [460]

Answer:

20.7 m

Explanation:

using v^2=u^2+2as

s=(v^2-u^2)/2a

v=13.5

u=5.2

a=3.75

7 0

2 years ago

An object of mass 30KG is falling in air and experiences a force due to air resistance of 50 newtons determine the net force act

ryzh [129]

Answer:

Assume that $g = 10\; \rm N \cdot kg^{-1}$ . The net force on this object will be $250\; \rm N$ (downwards.) The acceleration of this object will be approximately $8.3\; \rm m \cdot s^{-2}$ (also downwards.)

Explanation:

<h3>Net force</h3>

The object is falling towards the ground because of gravity. The size of the gravitational force on this object depends on its mass and the strength of the gravitational field at its location.

Near the surface of the earth, the gravitational field strength is approximately $10\; \rm N \cdot kg^{-1}$ . In other words, approximately $10\; \rm N$ of gravitational force acts on each kilogram of mass near the surface of the earth.

The mass of this object is given as $m = 30\; \rm kg$ . Therefore, the size of the gravitational force on it will be:

$W = m \cdot g \approx 30 \; \rm kg \cdot 10\; N \cdot kg^{-1} = 300\; \rm N$ .

Near the surface of the earth, gravitational forces point towards the ground. On the other hand, the direction of air resistance on this object will be opposite to its direction of motion. Since this objects is moving towards the ground, the air resistance on it will be directed in the opposite direction. That's exactly the opposite of the direction of the gravitational force on this object. The net force on this object will be:

$300\; \rm N - 50\; \rm N =250\; \rm N$ .