Answer:
1.59 seconds
12.3 meters
but if you are wise you will read the entire answer.
Explanation:
This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.
Time
<u>Given</u>
a = 0 This is the critical point. There is no horizontal acceleration.
d = 20 m
v = 12.6 m/s
<u>Formula</u>
d = vi * t + 1/2at^2
<u>Solution</u>
Since the acceleration is 0, the formula reduces to
d = vi * t
20 = 12.6 * t
t = 20 / 12.6
t = 1.59 seconds.
It takes 1.59 seconds to hit the ground
Height of the building
<u>Givens</u>
t = 1.59 sec
vi = 0 Another critical point. The beginning speed vertically is 0
a = 9.8 m/s^2 The acceleration is vertical.
<u>Formula</u>
d = vi*t + 1/2 a t^2
<u>Solution</u>
d = 1/2 a*t^2
d = 1/2 * 9.8 * 1.59^2
d = 12.3 meters.
The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)
The other vi is a vertical speed. It is 0.
Answer:
a = -0.33 m/s² k^
Direction: negative
Explanation:
From Newton's law of motion, we know that;
F = ma
Now, from magnetic fields, we know that;. F = qVB
Thus;
ma = qVB
Where;
m is mass
a is acceleration
q is charge
V is velocity
B is magnetic field
We are given;
m = 1.81 × 10^(−3) kg
q = 1.22 × 10 ^(−8) C
V = (3.00 × 10⁴ m/s) ȷ^.
B = (1.63T) ı^ + (0.980T) ȷ^
Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;
a = qVB/m
a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))
From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^
Thus;
a = -0.33 m/s² k^
Explanation:
Suppose you want to shine a flashlight beam down a long, straight hallway. Just point the beam straight down the hallway -- light travels in straight lines, so it is no problem. What if the hallway has a bend in it? You could place a mirror at the bend to reflect the light beam around the corner. What if the hallway is very winding with multiple bends? You might line the walls with mirrors and angle the beam so that it bounces from side-to-side all along the hallway. This is exactly what happens in an optical fiber.
The light in a fiber-optic cable travels through the core (hallway) by constantly bouncing from the cladding (mirror-lined walls), a principle called total internal reflection. Because the cladding does not absorb any light from the core, the light wave can travel great distances.
However, some of the light signal degrades within the fiber, mostly due to impurities in the glass. The extent that the signal degrades depends on the purity of the glass and the wavelength of the transmitted light (for example, 850 nm = 60 to 75 percent/km; 1,300 nm = 50 to 60 percent/km; 1,550 nm is greater than 50 percent/km). Some premium optical fibers show much less signal degradation -- less than 10 percent/km at 1,550 nm.
1
Answer:
The answer is B
Explanation:
It's past tense and dad is singular
