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Brilliant_brown [7]
3 years ago
6

The critical angle for diamond (n-2.42) surrounded by air is approximately 24 35 48 66

Physics
1 answer:
Veseljchak [2.6K]3 years ago
5 0

Answer:

sin(C) = 1/n = 1/2.42

C = 24.4 deg

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Leni [432]
Then the force will also be doubled
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3 years ago
evaluate the force of a skateboard exerts on you if your mass is 60kg and you push on the skateboard with a force of 60N
Sav [38]
Every action has an equal or opposite reaction.
You weigh 60kg 
<span>So your acceleration is 6N / 60 kg = 0.1m/s^2</span>
6 0
3 years ago
Class characteristics serve as corroboration for other, more subjective pieces of evidence in a courtroom (like witness testimon
Amiraneli [1.4K]

Answer: True

Explanation:

Class characteristics can be define as the features which are common to the group of objects. Like the make, model, label of the manufacturing company, design, shape and form. The individual characteristics can be define as the features which develop on the object or any other article with it's wear and use. Like tear, cuts, malformation and deposition of dust, dirt, and mud. The individual characteristic indicate towards the ownership of article or evidence to a particular person.

The class characteristics can only support the possibility of the evidence exactly alike that of the evidence found at the scene of crime. But the individual characteristics can directly link the evidence with the cause of crime. Hence, will be useful to prove that a crime has taken place in the court of law.

8 0
4 years ago
Read 2 more answers
A guy wire 1034 feet long is attached to the top of a tower. When pulled taut, it touches level ground 699 feet from the base of
kolezko [41]

Answer:

80.386 degrees

Explanation:

We use the cosine equation here (which is the adjacent side of the unknown angle divided by the hypotenuse

The adjacent side = 699ft

The hypotenuse = 1034ft

using cos∅ = Adjacent/hypotenuse

where ∅ is the unknown angle

cos ∅ = 699/1034 = 0.167

∅ = arccos 0.167 = 80.368°

As easy as one can imagine

8 0
3 years ago
A 460 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s. The rocket engine, when it is fired, exert
ollegr [7]

Answer:

The rocket should be launched at a horizontal distance of <u>6.45 m</u> left of the loop.

Explanation:

Given:

Mass of the rocket model (m) = 460 g = 0.460 kg [1 g = 0.001 kg]

Speed of the cart (v) = 3.0 m/s

Thrust force by the rocket engine (F) = 8.5 N

Vertical height of the loop (y) = 20 m

Let the horizontal distance left to the loop for launch be 'x'. Also, let 't' be the time taken by the rocket to reach the loop.

Now, there are two types of motion associated with the rocket- one is horizontal and the other vertical.

So, we will apply kinematics of motion in the two directions separately.

Vertical motion:

Given:

Force acting in the vertical direction is given as:

F_y=F-mg=8.5-0.46\times 9.8=3.992\ N

So, acceleration in the vertical direction is given as:

Acceleration = Force ÷ mass

a_y=\frac{F_y}{m}=\frac{3.992\ N}{0.46\ kg}=8.678\ m/s^2

Vertical displacement of rocket is same as the height of loop. So, y=20\ m

There is no initial velocity in the vertical direction. So, u_y=0\ m/s

Now, applying equation of motion in vertical direction. we have:

y=u_yt+\frac{1}{2}a_yt^2\\\\20=0+\frac{1}{2}\times 8.678t^2\\\\20=4.339t^2\\\\t^2=\frac{20}{4.339}\\\\t=\sqrt{\frac{20}{4.339}}=2.15\ s

Now, time taken to reach the loop is 2.15 s.

Horizontal motion:

There is no acceleration in the horizontal motion. So, displacement in the horizontal direction is equal to the product of horizontal speed and time.

Also, displacement of the rocket in the horizontal direction is nothing but the horizontal distance of its launch left of the loop. So,

x=vt\\\\x=3.0\ m/s\times 2.15\ s\\\\x=6.45\ m

Therefore, the rocket should be launched at a horizontal distance of 6.45 m left of the loop.

5 0
4 years ago
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