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Sholpan [36]
3 years ago
14

Many processes help to shape Earth’s surface. Starting with a rock, in what order would you expect the following Earth processes

to occur?
Physics
1 answer:
GuDViN [60]3 years ago
3 0
<span>Like most Earth materials, rocks are created and destroyed in cycles. The rock cycle is a model that describes the formation, breakdown, and reformation of a rock as a result of sedimentary, igneous, and metamorphic processes. </span><span>All rocks are made up of minerals. A mineral is defined as a naturally occurring, crystalline solid of definite chemical composition and a characteristic crystal structure. A rock is any naturally formed, nonliving, firm, and coherent aggregate mass of solid matter that constitutes part of a planet. i don't know if this is right but i hope it helps</span>
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How many atoms are in the formula CH4? <br><br> A ) 2 <br><br> B ) 4 <br><br> C ) 5 <br><br> D ) 6
Rufina [12.5K]
B) 4 is the answer for atoms in CH4
6 0
3 years ago
Read 2 more answers
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
I need help ASAP pleasee!!!
Vera_Pavlovna [14]

Answer: when in doubt go with B

Explanation:

3 0
3 years ago
Read 2 more answers
011 10.0 points
Sedbober [7]
<h2>The temperature of the air is 66.8° C</h2>

Explanation:

From the Newton's  velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .

In this case The velocity of sound = frequency x wavelength

= 798 x 0.48 = 383 m/sec

Suppose the temperature at this time = T K

Thus 383 ∝ \sqrt{T}                       I

The velocity of sound is 329 m/s at 273 K ( given )

Thus 329 ∝ \sqrt{273}                   II

Dividing I by II , we have

\frac{383}{329} = \sqrt{\frac{T}{273} }

or \frac{T}{273} = 1.25

and T = 339.8 K  = 66.8° C

4 0
4 years ago
A 29 foot ladder leans against a building so that the angle between the ground and the ladder is 75 ∘ . How high does the ladder
Aloiza [94]

Answer:

28.01m

Explanation:

Opp/Hyp = Sin

Sin 75 = x/29

x = 29 sin 75

x = 28.01m

3 0
3 years ago
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