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Vesna [10]
4 years ago
14

On Jupiter a rover drops a ball down a 430 m high cliff and records a fall time of 24.5 sec. what is the gravitational accelerat

ion?
Physics
1 answer:
Dahasolnce [82]4 years ago
5 0
17.6 meters/ second squared
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Science<br> Write two examples of heating appliances
bearhunter [10]

Answer:

a} <u>Infrared radiant heaters.</u>

<u>b} Fan heaters.</u>

Explanation:

I hope this is okay, should I give more.

4 0
3 years ago
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
A student measures the speed of yellow light in water to be 2.00x10^8
max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or,  The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or,                                           = (4/3) × 2.00 x 10⁸ m/s

or,                                           = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.

7 0
3 years ago
The candle burning
vodka [1.7K]

Answer:

A

Explanation:

its oxygen I hope i am correct have a great day.

3 0
3 years ago
Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s
Marrrta [24]
F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m  a₂ ⇒ a₁ = 2 a₂
 
since v = at + v₀  with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m

p₁ = p₂

8 0
3 years ago
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