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dimaraw [331]
3 years ago
7

Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude

of acceleration (c) Magnitude of displacement (d) Average Speed​
Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

c.

magnitude of displacement

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U=RI Ohm's law
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A car drives around a curve with a radius of 42 m at a velocity of 24 m/s. What is the centripetal acceleration of the car?
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Match the lithification processes.
butalik [34]

Explanation:

Recrystallization: contact pressure causing grains to "fuse" together

Cementation : precipitation of bonding agents between grains

Compaction : increase in density due to weight of overburden

Lithification is the process by which sediments are converted into sedimentary rocks. During this process, recrystallication, compaction and cementation of mineral grains occur.

The process starts with the compaction of sediments. The over burden weight of new sediments in the basin adds to the one originally deposited. This compresses the sediment. The volume of reduced and the density increases.

Recrystallization follows suit as the contact pressure of grains makes them fuse together. It is more like reworking of sediments. In this process, cementing materials can precipitate and cause sediments to be more fused together.

This is why most sediment are made up of clasts in a matrix of cementing materials.

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4 0
3 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
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