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sergeinik [125]
2 years ago
11

Glass has a _____________ index of refraction than air

Physics
1 answer:
rewona [7]2 years ago
4 0

Glass has a <u>grater </u>index of refraction than air. The glass's and air's indexes of refraction will be 1.5 and 1, respectively.

<h3>What is an index of refraction?</h3>

The refractive index of a substance is a dimensionless quantity that specifies how quickly light passes through it in optics.

The index of refraction of the glass and air will be 1.5 and 1 respectively.

Hence,glass has a <u>grater </u>index of refraction than air.

To learn more about the index of refraction, refer to the link;

brainly.com/question/23750645

#SPJ4

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A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. Ho
Flauer [41]

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

3 0
3 years ago
Two planets have the same surface gravity, but planet b has twice the radius of planet
nevsk [136]
The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>
8 0
3 years ago
17.
avanturin [10]

Answer:

Gamma rays

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Explanation:

6 0
3 years ago
Objects falling through air experience a type Of friction called ____
DaniilM [7]

Answer: That's air resistance.

Explanation: Well, air resistance is an upward force exerted on falling objects.

( I hope this helped <3 )

6 0
3 years ago
Read 2 more answers
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
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