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3 years ago

Describe the three major layers of the earth including temperature thickness of chemical composition

1 answer:

3 0

The three main layers are the core, the mantle, and the crust. The core is divided into two parts, the liquid outer core, and the solid inner core. Together it is 3450 km thick. The mantle is 2100 km thick, and the crust is 35-70 km thick. Hope I helped!

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An automobile rounds a curve of radius 50.0 m on a flat road.

bixtya [17]

Answer:

14m/s

Explanation:

Given parameters:

Radius of the curve = 50m

Centripetal acceleration = 3.92m/s²

Unknown:

Speed needed to keep the car on the curve = ?

Solution:

The centripetal acceleration is the inwardly directly acceleration needed to keep a body along a curved path.

It is given as;

a = $\frac{v^{2} }{r}$

a is the centripetal acceleration

v is the speed

r is the radius

Now insert the parameters and find v;

v² = ar

v² = 3.92 x 50 = 196

v = √196 = 14m/s

6 0

3 years ago

PLEASE HELP ME ASAP PLEASEEEEE

AleksAgata [21]

I believe its the law of inertia

6 0

3 years ago

if a stone is projected at an angle of 50 degrees to the horizontal with an initial velocity of 50m/s, what is the vertical comp

Vilka [71]

Answer:

38.3 m/s

Explanation:

To find vertical component of initial velocity, you'd have to use sine ratio:

$\displaystyle{\sin \theta = \dfrac{u_y}{u}}$

$\displaystyle{u_y}$ is vertical component of initial velocity and $\displaystyle{u}$ is initial velocity given which is 50 m/s.

A stone is projected at an angle of 50 degrees so $\displaystyle{\theta}$ = 50°. Substitute in the formula:

$\displaystyle{\sin 50^{\circ} = \dfrac{u_y}{50}}\\\\\displaystyle{50 \sin 50^{\circ} = u_y}\\\\\displaystyle{u_y = 38.3 \ \, \sf{m/s}}$

Therefore, the vertical component of initial velocity is approximately 38.3 m/s

(The picture is also attached for visual reference!)

3 0

2 years ago

car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car

Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0

3 years ago

PLEASEEE HELPPP!!!

Alja [10]

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is :

$W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m$

Hence, this is the required solution.

3 0

3 years ago