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frozen [14]
3 years ago
13

Need help on this please

Physics
1 answer:
Ratling [72]3 years ago
8 0

question 2 answer is ALL OF THE ABOVE

question 3 answer is WARM AND MOIST.

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I NEED HELP PLEASE, THANKS! Light passes from air into water at an angle of 40.0° to the normal. What is the angle of refraction
Vlad [161]

Answer:

Angle of Refraction = 28.9 degrees

Explanation:

<u><em>We'll use Snell's law for this. It's mathematical form is:</em></u>

=> n_1* sin(\alpha _1)=n_2 * sin(\alpha _2)

Where n_{1} = 1, n_{2} = 1.33, \alpha _{2} = 40^o

=> n_{1} and n_{2} are the refractive indexes of the air and water respectively.

<u><em>Solution:</em></u>

=> 1 * sin (40) = 1.33 * sin(\alpha _{2})

=> sin \alpha _{2} = \frac{sin 40^0}{1.33}

=> sin \alpha _{2} = 0.4821

=> \alpha _{2} = 28.9 degrees

8 0
2 years ago
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Helen [10]

Answer:

A

Explanation:

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6 0
2 years ago
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Box A has two moles of gas at a temperature of 400 K. Box B has one mole of gas at a temperature of 800 K. Which of the two gas
mina [271]
B I think if not sorry
8 0
3 years ago
A 0.10 newton spring toy with a spring constant of 160 newtons per meter is compressed 0.05 meter before it is launched. When re
forsale [732]

Answer:

(1) V = 0.2 J (2) 0.05J

Explanation:

Solution

Given that:

K = 160 N/m

x = 0.05 m

Now,

(1) we solve for the  initial potential energy stored

Thus,

V = 1/2 kx² = 0.5 * 160 * (0.05)²

Therefore V = 0.2 J

(2)Now, we solve for how much of the internal energy is produced as the toy springs up to its maximum height.

By using the energy conversion, we have the following

ΔV = mgh

=(0.1/9.8) * 9.8 * 1.5 = 0.15J

The internal energy = 0.2 -0.15

=0.05J

8 0
3 years ago
A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc
Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s  c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it  continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that  this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive  for the y-axis (perpendicular to the chosen  x-axis), we can write the vertical component of  the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ =\frac{-17.2}{16.6} =-1.036

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0
3 years ago
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