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yulyashka [42]
3 years ago
9

Earth has four motions in its movement through space, rotation, revolution, processional, and solar motion, which two are of any

importance in meteorology?
Physics
1 answer:
Anna007 [38]3 years ago
6 0

Answer:

rotation and revolution

Explanation:

out of the four motions the earth is subject to which are: rotation about its axis, revolution around  the Sun, processional motion (a slow conical movement ) of the axis, and the solar motion (this refers to the

movement of the whole solar system with space),  only two are of any

importance to meteorology as this two causes changes in weather and seasons. The first motion is rotation. Earth rotates on its axis

once every 24 hours. One-half of the Earth’s surface is

therefore facing the Sun at all times. The second motion of Earth is its revolution around  the Sun. The revolution around the Sun and the earth tilt on its axis are responsible for changes in seasons. The Earth

makes one complete revolution around the Sun in

approximately 365 1/4 days.

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A uniform sphere (I = 2/5 MR 2 ) rolls down an incline. (a) What must be the incline angle if the linear acceleration of the cen
Liono4ka [1.6K]

Answer:

Part a)

\theta = 8.05 degree

Part b)

a = 1.37 m/s^2

Explanation:

As the uniform sphere is rolling down the inclined plane then the net force on the sphere is given as

mg sin\theta - F_f = ma

also we have torque equation on it

F_f R = I\alpha

for pure rolling

a = R \alpha

F_f = \frac{Ia}{R^2}

now we have

mg sin\theta = ma + \frac{Ia}{R^2}

now we have

mg sin\theta = (m + \frac{2}{5}m)a

a = \frac{5}{7}g sin\theta

now given that

a = 0.10 g

so we have

0.10 g = \frac{5}{7} g sin\theta

sin\theta = 0.14

\theta = 8.05 degree

Part b)

If the inclined plane is frictionless then the acceleration is given as

a = g sin\theta

a = 9.8(0.14)

a = 1.37 m/s^2

5 0
3 years ago
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One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor
Bess [88]

Answer:

a) The final velocity is 20 m/s when the large-mass object is the one moving initially.

b) The final velocity is 9.0 m/s when the small-mass object is the one moving initially.

Explanation:

The momentum of the system is calculated as the sum of the momenta of each object. Each momentum is calculated as follows:

p = m · v

Where:

p =  momentum.

m =  mass.

v = velocity.

Then, the momentum of the system is the following:

m1 · v1 + m2 · v2 = (m1 + m2) · v

Where:

m1 = mass of the bigger object.

v1 = velocity of the bigger object.

m2 = mass of the smaller object.

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a) Let´s calculate the final velocity when the bigger object is moving:

(7.1 kg · 29 m/s + 3.2 kg · 0)/(7.1 kg + 3.2 kg) = v

<u>v = 20 m/s</u>

b) When the smaller object is moving:

(7.1 kg · 0 m/s + 3.2 kg · 29 m/s) / (7.1 kg + 3.2 kg) = v

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3 years ago
A ball is thrown horizontally from the top of a building 14.9 m high. The ball strikes the ground at a point 107 m from the base
umka2103 [35]

Answer:

1) t=1.743 sec

2)Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)Vf=17.08 m/s

Explanation:

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h=Vit+(1/2)gt^2

here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion

14.9=(0)*t+(1/2)(9.8)t^2

t^2=14.9/4.9

t^2=3.040 sec

t=1.743 sec

2) s=Vo*t

Putting values we get

107=Vo*1.743

Vo=61.388  m/sec

3)the x component of its velocity just be- fore it strikes the ground is the same as the  initial velocity of the ball that is=61.388  m/sec

4)From third equation of motion we know that

Vf^2-Vi^2=2gh

here Vi=0 m/s,h=14.9 m

Vf^2=Vi^2+2gh=0+2(9.8)(14.9)

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faust18 [17]
Set up a proportion

1 mile/1.6 km = 20,000miles/x

Cross Multiply
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6 0
4 years ago
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