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madam [21]
4 years ago
8

What is the lithosphere

Physics
2 answers:
hoa [83]4 years ago
8 0

Answer:

the rigid outer part of the earth, consisting of the crust and upper mantle.

Explanation:

UNO [17]4 years ago
3 0

Answer:

Lithosphere is the rigid outer part of the earth.

Explanation:

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A nonconducting sphere has radius R = 1.81 cm and uniformly distributed charge q = +2.08 fC. Take the electric potential at the
Sloan [31]

Answer:

a) V = -0.227 mV

b) V = -0.5169 mV

Explanation:

a)

Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude

E = (qr) / (4πε₀R³)

As we know that

V = -\int\limits^r_0 {E} \, dr

By solving above equation, we get

V = (-qr²) / (8πε₀R³)

When

R = 1.81 cm

r = 1.2 cm

q = +2.80 fC

ε₀ = 8.85 × 10⁻¹²

V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)

V = -2.27 × 10⁻⁴ V

V = -0.227 mV

b)

When

r = R

R = 1.81 cm

q = +2.80 fC

ε₀ = 8.85 × 10⁻¹²

V = (-qR²) / (8πε₀R³)

V = (-q) / (8πε₀R)

V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))

V = -5.169 × 10⁻⁴ V

V = -0.5169 mV

4 0
4 years ago
Which aspect of health is best illustrated by the following example? I work well in a group and
Yuri [45]
Social is the answer
7 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
Which vector has an x-component with a length of 4?
Masja [62]

Answer:ITS Y

Explanation:

6 0
4 years ago
A dart is thrown horizontally at a target's center that is 5.00 m, away. The dart hits the target 0.150 m below the targets cent
boyakko [2]

Answer:

  v₀ₓ = 28.6 m / s

Explanation:

This is a missile throwing exercise

          y = v_{oy} t - ½ g t²

as the dart is thrown horizontally the vertical velocity is zero (I go = 0)

          y = - ½ g t²

          t = \sqrt{- \frac{2y}{g}  }

let's calculate

         t = \sqrt{- \frac{2 \  (-0.150)}{9.8} }

         t = 0.175 s

the expression for horizontal displacement is

         x = v₀ₓ t

         v₀ₓ = x / t

         v₀ₓ = 5.00 / 0.175

         v₀ₓ = 28.6 m / s

7 0
3 years ago
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