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seraphim [82]
3 years ago
6

Explain how earths lithosphere and asthenosphere work together

Physics
1 answer:
Aleonysh [2.5K]3 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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A 16 kg object cuts 4 km in 25 minutes, find the applied force on the object?
GenaCL600 [577]

Answer:

14min i think im not quite sure

Explanation:

7 0
2 years ago
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
Would u rather/content creator or a rap artist
BartSMP [9]

a content creator because if i was a rapper i probably wouldn't make good songs lol

7 0
2 years ago
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from
krek1111 [17]

Answer:

E=35921.96N/C

Explanation:

From the question we are told that:

Radius r=0.321mm

Charge Density \mu=0.100

Distance d= 5.00 cm

Generally the equation for electric field is mathematically given by

E=\frac{mu}{2\pi E_0r}

E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}

E=35921.96N/C

4 0
3 years ago
Applied force is the force of support exerted by an object that holds up another object.
kenny6666 [7]
False, applied force is when a person or an object pushes on another object 
6 0
3 years ago
Read 2 more answers
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