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2 years ago

15

#2 Most of Earth's weather takes place in the O troposphere O stratosphere O mesosphere thermosphere

1 answer:

Stella [2.4K]2 years ago

6 0

Answer:

troposphere

hope this helps

have a good day :)

Explanation:

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___________is the tendency of an object to resist change in its state of motion

Mademuasel [1]

Inertia
<span> An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.this tendency to retain its motion is referred to as Inertia </span>

5 0

3 years ago

As you are cycling to the Q center one spring day, you pass the construction site at Gant and stop to watch a few minutes. A cra

Rama09 [41]

Answer:

30.63 m

Explanation:

Using y = ut + 1/2gt² where u = initial speed of block = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time of fall = 2.5 s and y = height of fall.

So, substituting the values of the variables into the equation, we have

y = ut + 1/2gt²

y = 0 m/s × 2.5 s + 1/2 × 9.8 m/s² × (2.5 s)²

y = 0 m + 4.9 m/s² × 6.25 s²

y = 0 m + 30.625 m

y = 30.625 m

y ≅ 30.63 m

So, the brick fell 30.63 m

7 0

3 years ago

1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as

telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

$v=r\omega$

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

$v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v$

New angular speed is twice that of the Chuck's speed.

8 0

3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

= $\frac{vFx}{Epir^{2} }$ = $\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}$

Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

5 0

3 years ago

If a ball is 10m high with what velocity will it fall?

Semmy [17]

14m/s

Explanation:

Given parameters:

Height of the ball = 10m

Unknown:

Velocity of fall or final velocity = ?

Solution:

We are going to use the appropriate equation of motion to solve this problem.

The object is falling with respect to gravity.

V² = U² + 2gH

where V is the final velocity

U is the initial velocity

g is the acceleration due to gravity 9.8m/s²

H is the height of fall

The initial velocity here is zero and

V² = 2 x 9.8 x 10 = 196

V = 14m/s

learn more:

Motion problems brainly.com/question/5248528

#learnwithBrainly

6 0

3 years ago