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kari74 [83]
3 years ago
9

In this graph, what is the displacement of the particle in the last teo seconds?

Physics
1 answer:
White raven [17]3 years ago
6 0

Answer: B

Explanation:

Displacement is a vector quantity. it is the distance covered in a specific direction.

Whereas, Velocity = displacement/time

Make displacement the subject of formula

Displacement = velocity × time

From the graph, velocity is constant which is equal to 1 m/s

Displacement in the last two seconds will be velocity multiply by 2. Since time = 2s. Therefore,

Displacement = 1 × 2 = 2 m

So, the particle displacement in the last two seconds will be 2 metres

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4 0
3 years ago
How do the dark lines of an atom''s absorption spectrum relate to the bright lines of its emission spectrum?
tangare [24]

Wouldn't it be neat if an electron falling closer to the nucleus ... emitting a
photon ... actually gave out more energy than it needed to climb to its original
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All we'd need is a pile of it in our basement, with a bright light bulb over the pile,
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The atoms in the pile would absorb some photons, raising their electrons to higher
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releasing more energy than they needed to climb up.  We could take that energy,
use some of it to keep the light bulb shining on the pile, and use the extra to heat
the house or run the dishwasher.

The energy an electron absorbs when it climbs to a higher energy level (forming
the atom's absorption spectrum) is precisely identical to the energy it emits when
it falls back to its original level (creating the atom's emission spectrum).

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into it from somewhere can't be created there.

You get what you pay for, or, as my grandfather used to say, "For nothing
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3 0
3 years ago
5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?
GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

Learn more:

brainly.com/question/17506203?referrer=searchResults

3 0
2 years ago
Does the under water pressure depend on the molecules of the air or what does it depend on
miskamm [114]

Answer:the pressure depends on gas and it will be half as much underwater

Explanation:

Water pressure increases with the depth of the water. This is because the weight of the column of water above the object increases. But a large, shallow pond may have more water in it than a small, deep pond.

This is due to an increase in hydrostatic pressure, the force per unit area exerted by a liquid on an object. The deeper you go under the sea, the greater the pressure of the water pushing down on you. For every 33 feet (10.06 meters) you go down, the pressure increases by one atmosphere .

4 0
3 years ago
A 0.520-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of
Veseljchak [2.6K]

Answer:

Explanation:

Given that,

Mass attached to spring

M = 0.52kg

Force constant K = 8N/m

Amplitude A = 11.6 cm

a. Maximum speed?

Angular velocity is calculated using

w = √k/m

w = √8/0.52

w = √15.385

w = 3.922rad/s

Then, the relation ship between angular velocity and linear velocity is given as

v = - w•A

v = - 3.922 × 11.6

v = - 45.5 cm/s

Then, the maximum velocity is

vmax = |v|= 45.5cm/s

b. Acceleration a?

Acceleration can be determine using the formula

a = -w²• A

a = -3.922² × 11.6

a = -178.46 cm/s²

Magnitude of the acceleration is 178.46cm/s²

c. Speed when the object is at 9.6cm from equilibrium position?

Generally,

The position of the object at equilibrium is

x(t) = A•Cos(wt)

x(t) = 11.6 Cos (3.922t)

Then, when x(t) = 9.6cm

9.6 = 11.6 Cos(3.92t)

Cos(3.922t) = 9.6/11.6

Cos(3.922t) = 0.8276

3.922t = ArcCos(0.8276)

Note: the angle is in radiant

3.922t = 0.596

t = 0.596/3.922

t = 0.152 second

Then, v(t) at that time is

v(t) = x'(t) = -11.6×3.92Sin(3.922t)

v(t) = -45.5Sin(3.922t)

Now, when t =0.152

v(t) = -45.5 Sin(3.922×0.152)

v(t) = -45.5Sin(0.596)

v(t) = -25.5 cm/s

Then, it's magnitude is 25.5cm/s

d. Acceleration at same position

t = 0.152s

a(t) = v'(t) = - 45.5×3.922Cos(3.922t)

a(t) = -178.46Cos(3.92t)

a(t) = -178.46 Cos(3.92×0.152)

a(t) = -178.46 Cos(0.596)

a(t) = -147.68 cm/s²

Magnitude of the acceleration is 147.68 cm/s²

5 0
3 years ago
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