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3 years ago

Explain how earths lithosphere and asthenosphere work together

Physics

1 answer:

Aleonysh [2.5K]2 years ago

3 0

Answer:

The lithosphere can affect the atmosphere when tectonic plates move and cause an eruption, where magma below spews up as lava above.

Explanation:

The lithosphere is broken into giant plates that fit around the globe like puzzle pieces. These puzzle pieces move a little bit each year as they slide on top of a somewhat fluid part of the mantle called the asthenosphere.

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ONLY ANSWER IF YOU KNOW FOR SURE PLEASE :)

sveticcg [70]

Answer:

Question one is b, Question two is b, and question three is b, im pretty positive that what it is

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2 years ago

How does an octopus or a squid uses Newton's third law of motion to escape from enemies

Sloan [31]

I think everytime they swim away, the water pushes the enemy back, which makes the octopus go faster. Every action has an equal and opposite reaction. Hope this helps.

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3 years ago

A crane lifts a 425 kg steel beam vertically upward a distance of 66 m. How much work does the crane do on the beam if the beam

iogann1982 [59]

Answer:

W = 311074.5 [J]

Explanation:

In order to solve this problem we must analyze two parts, in the first part by means of Newton's second law we can determine the acceleration of the beam, remembering that the sum of the forces is equal to the product of mass by acceleration.

∑F = m*a

F = forces acting on the beam [N]

m = mass = 425 [kg]

a = acceleration = 1.8 [m/s²]

The forces acting on the beam are the force of the crane up (positive) and the weight of the beam down (negative)

$F_{crane}-(425*9.81)= 425*1.8\\F_{crane}=4713.25 [N]$

Now in the second part, we use the definition of work, which is equal to the product of the force applied in the direction of displacement, that is, the product of force by distance.

$W=F*d$

where:

W = work [J]

F = force = 4713.25 [N]

d = distance = 66 [m]

$W=4713.25*66\\W=311074.5[J]$

5 0

3 years ago

In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

brainly.com/question/11422992

#SPJ1

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2 years ago

I need help with this question how to solve it for Brass and Cooper

Ksenya-84 [330]

Take into account that density and relative density are given by:

$\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}$

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

$\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}$

Cooper:

$\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}$

3 0

1 year ago