Answer:
![I = -2.1 kg.m/s](https://tex.z-dn.net/?f=I%20%3D%20-2.1%20kg.m%2Fs)
Explanation:
Given,
mass of the object,m = 0.30 Kg
initial speed, v_i = 3 m/s
time of collision = 0.20 s
final speed, v_f = -4 m/s
Impulse = change in momentum
![I = m (v_f-v_i)](https://tex.z-dn.net/?f=I%20%3D%20m%20%28v_f-v_i%29)
![I = 0.30\times (-4-3)](https://tex.z-dn.net/?f=I%20%3D%200.30%5Ctimes%20%28-4-3%29)
![I = -2.1 kg.m/s](https://tex.z-dn.net/?f=I%20%3D%20-2.1%20kg.m%2Fs)
Hence, impulse of the object is equal to ![I = -2.1 kg.m/s](https://tex.z-dn.net/?f=I%20%3D%20-2.1%20kg.m%2Fs)
Answer:
False
Explanation:
According to the big bang theory, matter was an infinitely small and very high density point which at one point exploded and expanded in all directions, creating what we know as our Universe, which also includes space and time . This happened about 13.8 billion years ago. Theoretical physicists have managed to reconstruct this chronology of events from 1/100 of a second after the Big Bang. After the explosion, while the Universe expanded, it cooled sufficiently and the first subatomic particles were formed: Electrons, Positrons, Mesons, Barions, Neutrinos, Photons among others. Today more than 90 particles are known. This theory solves many unknowns and is very well received by the scientific community, however there is still much to solve, for example, one of the great unsolved scientific problems in the expanding Universe model is whether the Universe is open or closed.
An attempt to solve this problem is to determine if the average density of matter in the Universe is greater than the critical value in Friedmann's model. The mass of a galaxy can be measured by observing the movement of its stars; multiplying the mass of each galaxy by the number of galaxies, it is seen that the density is only 5 to 10% of the critical value.
Answer:
Potential Energy to Kenetic Energy
Explanation:
When holding a ball in the air, the ball has potential energy. Once you drop the ball, the ball gains Kenetic Energy
Answer
given,
Pressure on the top wing = 265 m/s
speed of underneath wings = 234 m/s
mass of the airplane = 7.2 × 10³ kg
density of air = 1.29 kg/m³
using Bernoulli's equation
![P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2](https://tex.z-dn.net/?f=P_1%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%20%3D%20P_2%20%2B%20%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2)
![\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%20%28v_2%5E2-v_1%5E2%29)
![\Delta P =\dfrac{1}{2}\times 1.29\times (265^2-234^2)](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%201.29%5Ctimes%20%28265%5E2-234%5E2%29)
![\Delta P =9977.5 Pa](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D9977.5%20Pa)
Applying newtons second law
2 Δ P x A - mg = 0
![A =\dfrac{mg}{2\Delta P}](https://tex.z-dn.net/?f=A%20%3D%5Cdfrac%7Bmg%7D%7B2%5CDelta%20P%7D)
![A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}](https://tex.z-dn.net/?f=A%20%3D%5Cdfrac%7B7.2%5Ctimes%2010%5E3%20%5Ctimes%209.8%7D%7B2%5Ctimes%209977.5%7D)
A = 3.53 m²
The question is incomplete. You dis not provide values for A and B. Here is the complete question
Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.
A = 12
B = 18
Answer:
18.5⁰
Explanation:
Angle of incidence i = 12.0 + A
A = 12
= 12.0 + 12
= 14
Refractive index u = 1.10 + B/100
= 1.10 + 18/100
= 1.10 + 0.18
= 1.28
We then find the angle of refraction index u
u = sine i / sin r
u = sine24/sinr
1.28 = sine 24 / sine r
1.28Sine r = sin24
1.28 sine r = 0.4067
Sine r = 0.4067/1.28
r = sine^-1(0.317)
r = 18.481
= 18.5⁰