Joseph's experiment could be improved by using the same antenna at each part of the house during each trial instead of using different antenna. By doing so, he can obtain accurate results how is the signal in different part of the house under the same conditions (despite the location). So, he will see the dependence of the signal on the location. If he uses different antenna, than this antenna can also have influence of the signal.
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Answer:
fb = 240.35 Hz
Explanation:
In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.
Open tube:
(1)
vs: speed of sound = 343m/s
L: length of the open tube = 0.47328m
You replace in the equation (1):
Closed tube:
L': length of the closed tube = 0.702821m
Next, you use the following formula for the beat frequency:
The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz
Answer
given,
vertical speed of stone,v = 12 m/s
height of the cliff = 70 m
a) time taken by the stone to reach at the bottom of the cliff
We know that,
S = u t + 1/2 a t²
- 70 = 12 t - 0.5 x 9.8 t²
4.9 t² - 12 t - 70 = 0
solving the equation
t = 5.2 s (neglecting the negative value)
b) again using equation of motion
v = u + a t
v = 12 - 9.8 x 5.2
v = -38.96 m/s
ignoring the negative sign
magnitude of velocity is equal to 38.96 m/s
c) total distance travel by the stone
vertical distance covered by the stone
v² = u² + 2 g h
0 = 12² - 2 x 9.8 x h
h = 7.34 m
to reach the stone to the same level distance travel be doubled.
Total distance travel by the stone
H = h + h + 70
H = 7.34 x 2 + 70
H = 84.7 m.
