Answer:
4.6 m
Explanation:
First of all, we can find the frequency of the wave in the string with the formula:
where we have
L = 2.00 m is the length of the string
T = 160.00 N is the tension
is the mass linear density
Solving the equation,
The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.
The n=1 mode (fundamental frequency) of an open-open tube is given by
where
v = 343 m/s is the speed of sound
Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:
8500 Hz and Longitudinal
Speed = frequency x wavelength
Speed of sound at 20 degrees Celsius is approximately 340 m/s
Answer:
yes it was a constant speed and the car traveled 10 meters in 20 seconds.
Explanation:
Ok, this is a 2d kinematics problem, the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.
If we know it goes 4m into the air, we know d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!
Answer:
Magnetic field, B = 0.23 T
Explanation:
Given that,
Length of the copper rod, L = 0.49 m
Mass of the copper rod, m = 0.15 kg
Current in rod, I = 13 A (in +ve y direction)
When the rod is placed in magnetic field, the magnetic force is balanced by its weight such that :
So, the magnitude of the minimum magnetic field needed to levitate the rod is 0.23 T.
