Answer:
1.53 seconds
Explanation:
Applying,
T = 2usin∅/g................ Equation 1
Where, T = time of flight, u = initial velocity, ∅ = angle of projectile to the horizontal, g = acceleration due to gravity
From the question,
Given: u = 15 m/s, ∅ = 30°
Constant: g = 9.8 m/s²
Substitute these values in equation 1
T = 2(15)(sin30°)/9.8
T = 15/9.8
T = 1.53 seconds
Hence the time rate of flight is 1.53 seconds
Answer:
The SI units of the “A” is m (meters)
The SI units of the “B” is m/s^2
Explanation:
Given the distance = d meters.
Time taken to travel = t (seconds)
Function of the distance, d = A + Bt^2
Now we have given the above information and from the given distance function, we have to find the SI units of the A and B. Here, below are the SI units.
Thus, the SI units of the “A” is = m (meters)
The SI units of the “B” is = m/s^2
-- The string is 1 m long. That's the radius of the circle that the mass is
traveling in. The circumference of the circle is (π) x (2R) = 2π meters .
-- The speed of the mass is (2π meters) / (0.25 sec) = 8π m/s .
-- Centripetal acceleration is V²/R = (8π m/s)² / (1 m) = 64π^2 m/s²
-- Force = (mass) x (acceleration) = (1kg) x (64π^2 m/s²) =
64π^2 kg-m/s² = 64π^2 N = about <span>631.7 N .
</span>That's it. It takes roughly a 142-pound pull on the string to keep
1 kilogram revolving at a 1-meter radius 4 times a second !<span>
</span>If you eased up on the string, the kilogram could keep revolving
in the same circle, but not as fast.
You also need to be very careful with this experiment, and use a string
that can hold up to a couple hundred pounds of tension without snapping.
If you've got that thing spinning at 4 times per second and the string breaks,
you've suddenly got a wild kilogram flying away from the circle in a straight
line, at 8π meters per second ... about 56 miles per hour ! This could definitely
be hazardous to the health of anybody who's been watching you and wondering
what you're doing.
Answer:
10.53m/s²
Explanation:
Centripetal acceleration is the acceleration of an object about a circle. The formula for calculating centripetal acceleration is expressed by:
v is the velocity of the car = 24.5m/s
r is the radius of the track = 57.0m
Substitute the given values into the formula:
Hence the centripetal acceleration of the race car is 10.53m/s²
Answer:
A.) 355 m/s
B.) 0.71 m
C.) 500Hz
Explanation:
Given that a police car is traveling due east at a speed of 15.0 m/s relative to the earth. You are in a convertible following behind the police car. Your car is also moving due east at 15.0 m/s relative to the earth, so the speed of the police car relative to you is zero. The siren of the police car is emitting sound of frequency 500 Hz. The speed of sound in the still air is 340 m/s
a.) What is the speed of the sound waves relative to you?
Since the car is moving away from the observer, the relative velocity will be:
Relative velocity = 340 + 15
Relative velocity = 355 m/s
b.) What is the wavelength of the sound waves at your location?
Using the wave speed formula
V = frequency × wavelength
Make wavelength the subject of formula.
Wavelength = Velocity / frequency
Wavelength = 355/500
Wavelength = 0.71 m
c.) What frequency do you detect?
Fo = Fs ( C + V ) / ( C + v )
Fo = Fs
That is, the frequency of the observer will be equal to the frequency of the source.
Therefore, Fo = 500Hz
