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2 years ago

As a raindrop falls from a cloud to the surface of Earth,

1 answer:

6 0

The correct answer would be C. it moves at a constant speed. The troposphere(the layer our weather is in) is not nearly high enough for gravity to be different at different altitudes.

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According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 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

when a hockey stick strike a hockey puck and sends it towards the goalie, (thermal, mechanical) energy is being transferred from

velikii [3]

Answer:

Yes... definitely true.. can you tell me what is the question..

Explanation:

Please mark me brainliest :D

And also.. I would appreciate it if you give me 5 stars and a thanks.. :D

8 0

2 years ago

What is the mass of an atom that has 4 protons, 5 neutrons, and 4 electrons? Please explain. This is a final exam question. Plea

12345 [234]

Answer:

Explanation:

You get this answer by adding the protons and neutrons together.

5 0

3 years ago

A soccer player carries the ball for a distance of 40.0 m in the direction 42.0° west of south. find the westward component of t

Maksim231197 [3]

Actually what the problem meant about the westward component of the ball’s displacement is the horizontal component of the displacement. To help us better understand the problem, I attached a figure of the situation.

We can see from the figure that to solve for the value of the horizontal component, we have to make use of the sin function. That is:

sin θ = side opposite to the angle / hypotenuse of the triangle

sin 42 = x / 40 m

x = (40 m) sin 42

x = 26.77 m

Therefore the ball has a westward displacement of about 26.77 m

3 0

3 years ago