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2 years ago

What layer is just below the Earth’s crust?

1 answer:

4 0

I believe the answer you’re looking for is the mantle.

It goes inner core, outer core, mantle, crust,
from the very inside to the outside. I hope this is enough of a detailed reply to completely answer your question! Have a great night.

~Brooke❤️

You might be interested in

A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco

Alona [7]

Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.

So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal

here it can be said that medium 2 will be rarer then medium 1

So here the possible options are

1. Water

Air

2. Diamond

Air

So in above two options medium 1 is denser and medium 2 is rarer

8 0

2 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

The micrometer (1μm) is often called the micron. (a) how many microns make up 1 Km? (b) what fraction of a centimeter equals 1 μ

Vera_Pavlovna [14]

A) $1 \ km = 10^9 \ microns$
b) $1/10,000 \ cm = 1 \ micron$
c) $1 \ yd = 914,400 \ microns$

7 0

2 years ago

What is the final speed?

EastWind [94]

U=6.9ms-¹
a=0.62ms-²
t=3.4s
V=?
using
v=u+at
v=6.9+(0.62×3.4)
v= 6.9+2.108
v=9.008ms-¹

6 0

2 years ago

A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 60.0 N for 2.0s. What impulse acts on the b

NARA [144]

Impulse<span> is the change of momentum of an object when the object is acted upon by a force for an interval of time. It is simply calculated by the expression:
</span>
Impulse = F(change in time)
Impulse = 60(2)
Impulse = 120 Ns

Hope this answers the question. Have a nice day.

6 0

3 years ago