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3 years ago

What is the temperature of the lithosphere?

Physics

1 answer:

tester [92]3 years ago

5 0

The temperature of the lithosphere is around 300°C - 500°C

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A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago

A. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.

vovikov84 [41]

Answer:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

Explanation:

a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase

b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.

c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase

8 0

3 years ago

Read 2 more answers

Which orbit has the highest energy? n = 1 n = 2 n = 3

PIT_PIT [208]

Answer:

Explanation:

The closer an orbit is to the nucleus the fewer energy

4 0

3 years ago

A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.

sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

= 1.10 + 1.4(-0.57)

= 1.10 -0.798

= 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

7 0

3 years ago

An electric dipole is formed from two charges, ±q, spaced 0.800 cm apart. The dipole is at the origin, oriented along the y-axis

Simora [160]

Answer: q = 2.781e-9C = 2.781nC

E=200C

Explanation:

E = Qd/(2πEor^3)

Where

E=Electric field intensity

Q=Charge

d=distance between the dipole=0.008m

Eo=permitivitty

400 N/C = Q(0.80e-2 m)/(2πε*(10e-2 m)^3)

Q= (400* 2* 3.142 * 8.85 x 10-12 * 0.1^3)/0.008

q = 2.781e-9C = 2.781nC

Though the dipole are two separate charges. And since the point is on the x-axis, the electric field strengths are equivalent. The magnitude of the vector sum is:

E = kq*2sin θ/r^2

= 2(8.99e9 N*m^2/C^2)(2.781e-9 C)*sin(arctan(.4/10))/(10e-2 m)^2

= 2(8.99e9) * (2.781e-9) * sin(2.290)/(10e-2 m)^2

=200 C

3 0

3 years ago