The correct answer is D: Watt. This unit was named after James Watt, and
is used to express the equivalent of one joule per second in energy. In
experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated
format: W.
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Answer:
It will be cut in half
Explanation:
The diffraction of a slit is given by the formula
a sin θ = m where
a = width of the slit,
λ = wavelength and
m = integer that determines the order of diffraction.
Next we divide both sides by a, we have
sin θ = m λ / a
Also, recall that
a’ = 2 a
Then we substitute in the previous equation
2asin θ' = m λ, if divide by 2a, we have
sin θ' = (m λ / 2a).
Now again, from the first equation, we said that sin θ = m λ / a, so we substitute
sin θ ’= sin θ / 2
Then we use trigonometry to find the width, we say
tan θ = y / L
Since the angle is small, we then have
tan θ = sin θ / cos θ
tan θ = sin θ, this then means that
sin θ = y / L
we will then substitute
y’ / L = y/L 1/2
y' = y / 2
this means that when the slit width is doubled the pattern width will then be halved
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
Answer:
n=2.053
Explanation:
We will use Snell's Law defined as:
Where n values are indexes of refraction and 
values are the angles in each medium. For vacuum, the index of refraction in n=1. With this we have enough information to state:
Solving for 
yields:
Remember to use degrees for trigonometric functions instead of radians!
D. 27 km/hr :) Just msg if you need any more help :)
