Answer:
The speed in the first point is: 4.98m/s
The acceleration is: 1.67m/s^2
The prior distance from the first point is: 7.42m
Explanation:
For part a and b:
We have a system with two equations and two variables.
We have these data:
X = distance = 60m
t = time = 6.0s
Sf = Final speed = 15m/s
And We need to find:
So = Inicial speed
a = aceleration
We are going to use these equation:
We are going to put our data:
With these equation, you can decide a method for solve. In this case, We are going to use an egualiazation method.
![[\sqrt{(15m/s)^2-(2*a*60m)}]^{2}=[15m/s-(a*6s)]^{2}](https://tex.z-dn.net/?f=%5B%5Csqrt%7B%2815m%2Fs%29%5E2-%282%2Aa%2A60m%29%7D%5D%5E%7B2%7D%3D%5B15m%2Fs-%28a%2A6s%29%5D%5E%7B2%7D)
If we analyze the situation, we need to have an aceleretarion greater than cero. We are going to choose a = 1.67m/s^2
After, we are going to determine the speed in the first point:
For part c:
We are going to use:
Answer: the speed at which it falls toward the Earth.
Explanation:
A bullet travelling across Earth's surface with some horizontal velocity is classical example of projectile motion.
Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).
This kind of motion is the result of two independent motions: vertical motion and horizontal motion.
The observed net velocity is the vectorial sum of the vertical and horizontal velocities.
The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).
The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.
Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.
Answer:
The separation between the first two minima on either side is 0.63 degrees.
Explanation:
A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:
with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:
for the first minimum
solving for θ1:
for the second minimum:
So, the angular separation between them is the rest:
I think the correct answer from the choices listed above is option A. The kinetic energy after the perfectly inelastic collision would be zero Joules. <span>A </span>perfectly inelastic collision<span> occurs when the maximum amount of kinetic energy of a system is lost. Hope this answers the question.</span>
The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
(1)
where
P is the power
A is the area
In our problem, the intensity is
. At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
And so if we re-arrange (1) we find the power emitted by the source:
