Answer:
none of the answers is correct, the time is the same t₁ = t₂ = 0.600 s
Explanation:
This is a kinematics exercise, analyze the situation a bit. The vertical speed in both cases is the same is zero, the horizontal speed in the second case is double (vₓ₂ = 2 vₓ₁)
let's find the time to hit the ground
y = y₀ + I go t - ½ g t²
0 = y₀ - ½ g t²
t = √ 2y₀ / g
with the data from the first launch
y₀i = ½ g t²
y₀ = ½ 9.8 0.6²
y₀ = 1,764 m
with this is the same height the time to descend in the second case is the same
t₂ = 0.600 s
this is because the horizontal velocity change changes the offset on the x axis, but does not affect the offset on the y axis
Therefore, none of the answers is correct, the time is the same
t₁ = t₂ = 0.600 s
<h2>Answer:</h2>
<u>With radiocarbon dating scientists compare an object carbon 14 levels with </u><u>the fossil or rock for which the age measurement is required</u>
<h2>Explanation:</h2>
Radiocarbon, or carbon 14, is an isotope of the element carbon that is unstable and weakly radioactive. The carbon-14 method was developed by the American physicist Willard F. Libby about 1946. It can be used to determine the age of a rock or a fossil by comparing the specimen of the required or fossil and compared it with the carbon 14 sample. Carbon 14 decays at constant rate therefore an estimate of the date at which an organism died can be made by measuring the amount of its residual radiocarbon and comparing it with Carbon 14.
Answer:
A) True, B) False, C) False and D) false
Explanation:
Let's solve the problem using the law of conservation of energy to know if the statements are true or false
Let's look for mechanical energy
Initial
Emo = Ke = ½ k Dx2
Final
Em1= ½ m v12
Emo = Em1
½ k Δx2 = ½ m v₁²
v₁² = k / m Δx²
v₁ = √ k/m Δx
Now let's calculate the speed when it falls
Vfy² = Voy² - 2gy
Vfy² = - 2gy
Vf² = v₁² + vfy²
A) True v₁ = A Δx
.B) False. As there is no rubbing the mechanical energy conserves
.C) False the velocity is proportional to the square root of the height
v2y = v2 √2
. D) false promotional compression speed
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
The type of potential energy is elastic energy
