Answer:
The time he can wait to pull the cord is 41.3 s
Explanation:
The equation for the height of the skydiver at a time "t" is as follows:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.
When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 m = 15000 m - 4.9 m/s² · t²
-15000 m / -4.9 m/s² = t²
t = 55.3 s
Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.
Answer:
The speed of the object is (
)m/s
The magnitude of the acceleration is 4.00m/s²
Explanation:
Given - position vector;
r = (2.0 + 3.00t)i + (3.0 - 2.00t²)j -------------------(i)
To get the speed vector (
), take the first derivative of equation (i) with respect to time t as follows;
= 
= ![\frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j] }{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5B%282.0%20%2B%203.00t%29i%20%2B%20%283.0%20-%202.00t%5E2%29j%5D%20%20%7D%7Bdt%7D)
= ------------------------(ii)
To get the acceleration vector (
), take the first derivative of the speed vector in equation(ii) as follows;
j
The magnitude of the acceleration |a| is therefore given by
|a| = |-4.00|
|a| = 4.00 m/s²
In conclusion;
the speed of the object is ()m/s
the magnitude of the acceleration is 4.00m/s²
The wavelength of the laser beam in the liquid is 517 nm
Brainliest?
Answer:
5.3 cm
Explanation:
This question is an illustration of real and apparent distance.
From the question, we have the following given parameters
Real Distance, R = 8.0cm
Refractive Index, μ = 1.5
Required
Determine the apparent distance (A)
The relationship between R, A and μ is:
μ = R/A
i.e.
Refractive Index = Real Distance ÷ Apparent Distance
Substitute values in the above formula
1.5 = 8/A
Multiply both sides by A
1.5 * A = A * 8/A
1.5A = 8
Divide both side by 1.5
1.5A/1.5 = 8/1.5
A = 8/1.5
A = 5.3cm
Hence, the letters would appear at a distance of 5.3cm
In general,
Power = (energy moved) / (time to move the energy) .
If it's mechanical power, then
Power = (work done) / (time to do the work) .
If it's electrical power, then it can be any one of these:
Power = (volts) x (amperes)
Power = (volts)² / (resistance, ohms)
Power = (amperes)² x (resistance, ohms) .
Whatever kind of energy you're dealing with, power always
turns out to be
(amount of energy produced, used, or moved)
divided by
(time taken to produce, use, or move the energy) .
