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AVprozaik [17]
2 years ago
9

An arrow is launched from a bow with an initial horizontal velocity of 40

Physics
1 answer:
Ivan2 years ago
4 0

Answer:

V = (Vx^2 + Vy^2)^1/2 = (40^2 + 62^2)^1/2

V = 73.8 m/s

tan theta = Vy / Vx = 62/40 = 1.55

theta = 57.2 deg

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When will heat transfer within a liquid stop? (25 points)
exis [7]
Choice C.
That's when convection stops.
3 0
3 years ago
Which of the following does NOT support the geologic evidence used to date the earth?
Zanzabum

Answer:

comparing the anatomy of organisms

Explanation:

Basically, Rocks are the solid evidence used to date the earth.

Due to the changing in the genetic make up of organisms,the anatomy of organisms is not a reliable evidence to date the earth.

4 0
3 years ago
Introduction to Symbolic Answers, Part B, Enter the expression 2cos2(θ)−1, where θ is the lowercase Greek letter theta.
lesantik [10]
2cos2(o)-1    is the answer
4 0
3 years ago
You yell into a canyon. You hear the echo in 3 seconds. How long did the sound of your voice travel before bouncing off a cliff?
Lilit [14]

Answer:

The sound travelled 516 meters before bouncing off a cliff.

Explanation:

The sound is an example of mechanical wave, which means that it needs a medium to propagate itself at constant speed. The time needed to hear the echo is equal to twice the height of the canyon divided by the velocity of sound. In addition, the speed of sound through the air at a temperature of 20 ºC is approximately 344 meters per second. Then, the height of the canyon can be derived from the following kinematic formula:

2\cdot h = v\cdot t (1)

Where:

h - Height, measured in meters.

v - Velocity of sound, measured in meters per second.

t - Time, measured in seconds.

If we know that  v = 344\,\frac{m}{s} and t = 3\,s, then the height of the canyon is:

h = \frac{v\cdot t}{2}

h = \frac{\left(344\,\frac{m}{s} \right)\cdot (3\,s)}{2}

h = 516\,m

The sound travelled 516 meters before bouncing off a cliff.

7 0
2 years ago
At what speed does a typical electron pass by any given point in the wire?
attashe74 [19]
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

</span>a) 5.0 A = 5.0 C/s 
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
7 0
3 years ago
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