Ball 1 is launched with an initial vertical velocity v1 = 146 ft/sec. Ball 2 is launched 2.3 seconds later with an initial vertical velocity v2. Determine v2 if the balls are to collide at an altitude of 234 ft. At the instant of collision, is ball 1 ascending or descending?
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
At the time of collision velocity of ball one is descending.
Velocity of ball 1 = 146 ft/sec = 44.50m/s
The balls are to collide at an altitude of 234 ft
H = 234 ft = 71.32 m
We have equation of motion
v² = u² + 2as
v² = 44.50² + 2 x (-9.81) x 71.32
v = ±24.10 m/s.
Time for each velocity can be calculated using equation of motion
v = u + at
24.10 = 44.50 - 9.81 t , t = 2.07 s
-24.10 = 44.50 - 9.81 t , t = 6.99 s
Since the second ball throws after 2.3 seconds we can avoid case with t = 2.07 s.
So at the time of collision velocity of ball one is descending.
The collision occurs at t = 6.99 s.
Time of flight of ball 2 = 6.99 - 2.3 = 4.69 seconds.
Height traveled by ball 2 = 71.32 m
We need to find velocity
s = ut + 0.5 at²
71.32 = u x 4.69 - 0.5 x 9.81 x 4.69²
u = 38.21 m/s = 125.36 ft/s
Initial velocity of ball 2 = 38.21 m/s = 125.36 ft/s
First we just look at the motion of the first particle. It is moving vertically in a gravitational field so is decelerating with rate g = 9.81 m/s^2 = 32.18 ft/s^2. We can write it's vertical position as a function of time.
We set this equal to 234 ft to find when the body is passing that point, a solve the quadratic equation for t.
t_1,2=v_1±(√v_1^2-4*a/2*h_1)/a=2.57 s, 7.44 s
Since we know the second ball was launched after 2.3 seconds, we know that the time we are looking for is the second one, when the first ball is descending. The second ball will have 2.3 seconds less so the time we further use is t_c = 7.44 - 2.3 = 5.14 s. With this the speed of the second ball needed for collision at given height, can be found.
Solving a similar equation, but this time for v2 to obtain the result.
Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? A. Use a lens to focus the power into a smaller area. B. Increase the power output of the lamp. C. Either A or B.
the correct option is C
The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.
Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.
When we see the possibilities we see that the correct option is C
In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is recorded on a flat screen-like detector that is 2.00 m away from the slits. If the seventh bright fringe on the detector is 10.0 cm away from the central fringe, what is the wavelength of the light passing through the slits? (The central bright fringe is zeroth one).
The wavelength of the light passing through the slit is 214 nm.
What is the wavelength?
The wavelength is the distance between identical points in the adjacent cycles of a waveform.
Given that the separation between two slits d is 3.00 × 10^-5 m and the distance from the slit to screen r is 2 m. The distance from the central spot to fringe s is 10.0 m and the bright bands of the spectrum m are 7 for the seventh bright fringe.
The wavelength of the light passing through the slit is calculated as given below.
Hence we can conclude that the wavelength of the light passing through the slit is 214 nm.
To know more about the wavelength, follow the link given below.
To solve this problem, the concepts related to destructive and constructive Interference of light spot and dark spot are necessary.
By definition in the principle of superposition, light interference is defined as
d = Separation of the two slits
R = Distance from slit to screen
m= Any integer, which represents the repetition of the spectrum. The order of m equal to 1,2,3,4,5 represent bright bands and the order of m equal to 1.5,2.5,3.5 represent the dark bands.
Y = Distance from central spot to fringe.
Re-arrange the equation to find \lambda we have that
Our values are gives according the problem as,
m = 7 (The seventh bright fringe)
R = 2m
Therefore the wavelength of the light passing through the slits is 214nm
Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?
mass of the body,
mass of the tyre,
length of hanging of tyre,
distance run by the body,
acceleration of the body,
Using the equation of motion :
v=final velocity of the body
u=initial velocity of the body
here, since the body starts from rest state:
putting the values in eq. (1)
Now, the momentum of the body just before the jump onto the tyre will be:
Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.
Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.
above the normal hanging position.
A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?
According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.
As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.
The flux across the open surface can be expressed as follows:
As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².
⇒Flux = E*π*R²
A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?
ratio of the piccolo's length to the flute's length is 0.4916
frequency of piccolo = 522.5 Hz
frequency of flute = 256.9 Hz
to find out
ratio of the piccolo's length to the flute's length
we get here length of tube that is express as
length of tube = velocity of sound ÷ fundamental frequency .......................1
so here ratio of Piccolo length to flute that is
so ratio of the piccolo's length to the flute's length is 0.4916
A truck traveling with an initial velocity of 44.1 m/s comes to a stop in 15.91 secs. What is theacceleration of the truck?