<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
Explanation:
The force of a spring is described by Hooke's law:
F = kx
where k is the spring stiffness in N/m, and x is the displacement in m.
A spring force vs displacement graph is a line passing through the origin with a slope of k.
The acceleration of the skier over the distance traveled is 10.38 m/s²
The given parameters;
- Initial velocity of the skier, u = 56 m/s
- Distance covered by the skier, s = 151 m
Acceleration is defined as the change in velocity per change in time of motion.
The magnitude of the acceleration of the skier is calculated as follows;
where;
<em>v is the final velocity at the end of 151 m = 0</em>
Thus, the acceleration of the skier over the distance traveled is 10.38 m/s²
Learn more here:brainly.com/question/20453233
The mistake here is that the total energy of the system does not change - generally, the total amount of energy in a system never changes unless the energy enters or leaves the system somehow (for example, if you see the Earth alone as a system, then energy from the sun enters the system and some energy is lost to space in the form of heat).
Answer:
Increases
Explanation:
In the Compton scattering experiment with x-rays,
The change in operation
![\Delta \lambda = \frac{h}{m_oc} [1-cos\theta]](https://tex.z-dn.net/?f=%5CDelta%20%5Clambda%20%3D%20%5Cfrac%7Bh%7D%7Bm_oc%7D%20%5B1-cos%5Ctheta%5D)
Now rest being constant, as \theta increases, cos\theta decreases
Hence, The change in wavelength will increase with the increase in \theta.
Hence, wavelength increases with an increase in the angle of scatttering.
