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irinina [24]
2 years ago
7

A ray in flint glass (n = 1.61)

Physics
2 answers:
Bad White [126]2 years ago
7 0

Answer:

The answer to this would be 0

Explanation:

Got it right on acellus ✅

Scilla [17]2 years ago
4 0

Answer:0

Explanation:

0

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Consider this situation: A baseball player dives head-first
siniylev [52]
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.

I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
7 0
2 years ago
Which part of the eye holds eye color?
Anna [14]

Hi, it’s the iris. Hope this helped u

6 0
3 years ago
Two dimensional motion is not always projectile motion. In two dimensional motion that is not projectile motion, what is not acc
MatroZZZ [7]

gravity is the correct answer.

4 0
3 years ago
Read 2 more answers
Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the
Assoli18 [71]

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

7 0
2 years ago
. A bathysphere used for deep-sea exploration has a radius of 1.50 m and a mass of 1.20 104 kg. To dive, this submarine takes on
dangina [55]

Answer:

Explanation:

Given that,

Bathysphere radius

r = 1.5m

Mass of bathysphere

M = 1.2 × 10⁴ kg

Constant speed of descending.

v = 1.2m/s

Resistive force

Fr = 1100N upward direction

Density of water

ρ = 1.03 × 10³kg/m³

The volume of the bathysphere can be calculated using

V = 4πr³ / 3

V = 4π × 1.5³ / 3

V = 14.14 m³

The Bouyant force can be calculated using

Fb = ρgV

Fb = 1.03 × 10³ × 9.81 × 14.14

Fb = 142,846.18 N

Buoyant force is acting upward

Weight of the bathysphere

W = mg

W = 1.2 × 10⁴ × 9.81

W = 117,720 N

Weight is acting downward

The net positive buoyant using resolving

Fb+ = Fb — W

Fb+ = 142,846.18 — 117,720

Fb+ = 25,126.18 N

The force acting downward is the weight of the submarine and it is equal to the positive buoyant force and the resistive force

W = Fb+ + Fr

W = 25,126.18 + 1100

W = 26,226.18

mg = 26,226.18

m = 26,226.18 / 9.81

m = 2673.4kg

Mass of submarine is 2673.4kg

8 0
2 years ago
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