17- Find the magnitude of vector product \BxĀ| for A=– 23 +3Â and B = 2î – 3+ Å vectors.
1 answer:
You might be interested in
1) The law of motion of the projectile is
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
Answer:
5.15348 Beats/s
4.55 mm
Explanation:
= Velocity of sound = 342 m/s
= Velocity of sound = 346 m/s
= First frequency = 440 Hz
Frequency is given by
Beat frequency is given by
Beat frequency is 5.15348 Hz
Wavelength is given by
Relation between length of the flute and wavelength is
At v = 346 m/s
Difference in length is
It extends to 4.55 mm