Answer:
The area of the water stream will be 1.74 cm^2
Explanation:
initial velocity of water u = 33.2 cm/s
initial area = 6.4 cm^2
height of fall = 7.05 cm
final area before hitting the sink = ?
as the water falls down the height, it accelerates under gravity; causing the speed to increase, and the area to decrease.
first we find the velocity before hitting the sink
using
-----Newton's equation of motion
where v is the velocity of the water stream at the sink
u is the initial speed of the water at the spout
h is the height of fall
g is acceleration due to gravity, and it is positive downwards.
g = 981 cm/s^2
imputing relevant values, we have
= 122.206 cm/s
according to continuity equation,
A1v1 = A2v2
where A1 is the initial area
V1 = initial velocity
A2 = final area
V2 = final velocity
6.4 x 33.2 = 122.206 x A2
212.48 = 122.206 x A2
A2 = 212.48 ÷ 122.206 ≅ 1.74 cm^2
Answer:
51 mph
Explanation:
Answer:
for the body to float, the density of the body must be less than or equal to the density of the liquid.
Explanation:
For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.
Weight is
W = mg
let's use the concept of density
ρ_body = m / V
m = ρ_body V
W = ρ_body V g
The thrust of the body is given by Archimedes' law
B = ρ_liquid g V_liquid
as the body floats the submerged volume of the liquid is less than or equal to the volume of the block
ρ_body V g = ρ_liquid g V_liquid
ρ_body = ρ liquid Vliquido / V_body
As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.
b. If the frequency is then increased while you remain 0.25 m from the center, what is the first frequency for which that location will be a maximum of sound intensity?
Answer:
a) 343.0 Hz b) 686.0 Hz
Explanation:
a) First, we need to know the distance to both speakers.
If the person is at halfway between the two speakers, and they are 4.0 m apart, this means that he is at 2.0 m from each speaker.
So, if he moves 0.25 m towards one of them, the distance from any speaker will be as follows:
d₁ = 2.0 m-0.25 m= 1.75 m
d₂ = 2.0 m + 0.25 m = 2.25 m
The difference between these distances is the path difference between the sound from both speakers:
d = d₂ - d₁ = 2.25 m - 1.75 m = 0.5 m
If the person encounters at this path difference a minimum of sound intensity, this means that this distance must be an odd multiple of the semi-wavelength:
d = (2*n-1)*(λ/2) = 0.5 m
The minimum distance is for n=1:
⇒ λ = 2* 0.5 m = 1 m
In any wave, there exists a fixed relationship between the speed (in this case the speed of sound), the wavelength and the frequency, as follows:
v = λ*f, where v= 343 m/s and λ=1 m.
Solving for f, we have:
b) If the person remains at the same point, for this point be a maximum of sound intensity, now the path difference (that it has not changed) must be equal to an even multiple of the semi-wavelength, which means that it must be met the following condition:
d = 0.5 m = 2n*(λ/2) = λ (for n=1)
if the speed remains the same (343 m/s) we can find the new frequency as follows:
⇒ f = 686.0 Hz
Two speakers create peaks and troughs of sound intensity due to constructive and destructive interference of waves. Using wave properties, the frequency of the sound when a minimum intensity is experienced 0.25m from the center is 680Hz. Increasing the frequency, the first to produce maximum intensity at the same position is about 2720Hz.
The behavior of sound intensity in this question is due to wave interference, specifically, constructive and destructive interference of sound waves. When you stand halfway between the speakers, the sound waves from each speaker are in phase, which means the pressure variations combine to create an intensified sound, known as constructive interference.
When you move towards one of the speakers and detect a minimum in sound intensity, this is due to destructive interference, which occurs when the crest of one wave overlaps with the trough of another, canceling each other and producing a minimum sound level.
a. The frequency of the sound can be calculated using the formula for wave speed, v = f.lambda, where v is the speed of sound (340 m/s under normal conditions), f is the frequency, and lambda is the wavelength. In this case, a minimum sound intensity indicates one-half wavelength. So, lambda = 0.5 m. Thus, frequency, f = v/lambda = 340/0.5 ~ 680 Hz.
b. When you increase the frequency while remaining 0.25m from the center, the first frequency for which the location will be a maximum of sound intensity will be when you are an integral multiple of the wavelength away from the source. Thus if we let this be 2λ, we can calculate the frequency as f = v / λ = v / (0.25m / 2) = 340 / 0.125 ~ 2720 Hz.
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The simple machines can help make work easier by working quicker than the people and making other peoples jobs easier.
Explanation:
I hope this helped.
The linear speed of the ball for the circular motion is determined as 12 m/s.
The given parameters;
The linear speed of the ball is calculated as follows;
v = ωr
where;
The linear speed of the ball is calculated as follows;
v = ωr
v = 120 x 0.1
v = 12 m/s
Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.
Learn more here:brainly.com/question/14404053
Answer:
The speed of ball is 12
Explanation:
Given:
Mass of ball kg
Radius of rotation m
Angular speed
Here barbell spins around a pivot at its center and barbell consists of two small balls,
From the formula of speed in terms of angular speed,
Where speed of ball
Therefore, the speed of ball is 12
Answer:
True
Explanation:
The normal line is defined as the line which is perpendicular to the reflecting surface at the point where the incident ray meet with the reflecting surface.
The angle of incident is defined as the angle which is subtended by the incident ray with respect to the normal ray by consider the normal ray as the base line and angle is measured from the point where incident ray is incident on the reflecting surface of the mirror.
Similarly reflecting ray can be defined as the ray which is reflected after the incident of a ray and the angle subtended by the reflecting ray is measure with respect to normal ray by considering normal ray as a base line.
Therefore, the normal ray is the perpendicular line to the reflecting surface at the point of incidence.