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Artemon [7]
3 years ago
13

Which term is applied to an object through which light passes?

Physics
2 answers:
Zepler [3.9K]3 years ago
7 0

Answer:

The answer is D. Transparent

Explanation:

I took AP EX quiz.

Elena L [17]3 years ago
5 0

Answer:

D

Explanation:

transparent_objects that allows light to pass through and can you see through them

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
An object is accelerating at 2m/s^2. What is the net force if the mass is 125kg?
natita [175]

Answer:

250 N

Explanation:

Newton's second law:

∑F = ma

∑F = (125 kg) (2 m/s²)

∑F = 250 N

3 0
3 years ago
The figure shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulle
garik1379 [7]

Answer:

A) m = F/a = 91.7/9.81 = 9.35 kg

B) m = F/a = 59.2/9.81 = 6.03 kg

C) m = F/a = 33.4/9.81 = 3.40 kg

D) m = F/a = 9.65/9.81 = 0.984 kg

Explanation:

7 0
3 years ago
A hockey puck of mass m1=165 g slides from left to right with an initial velocity of 15.5 m/s. It collides head on with a second
Fynjy0 [20]

Answer:

it may be -6 m/s

Explanation:

3 0
3 years ago
Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
kirza4 [7]

Answer:

a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

= 0.8654 kg-m²

The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

5 0
3 years ago
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