Answer:
975 m/s^2
Explanation:
The formula for rate of acceleration is
(Final velocity - intial velocity)/ time taken
If you plug in your data you will get 975 m/s^2
(1600 m/s -1210 m/s )/.4 s = 975 m^2
90 J / 2.2 s = 41 W. 41 Watts is the power <span>required to give a brick 90 J of potential energy in a time of 2.2 s</span>
I would say Anthony has more power than Angel. If they are both exerting the same force on the box, which isnt really mentioned here, which is why I believe this is a bit vague, then the both do the same work. So, if the work is equal then the person with the lower time period would have more power. Since Anthony only took 38 seconds, compared to Angel's 42 I would say that Anthony has more power than Angel.
Answer:
30 miliAmps
Explanation:
Step 1:
Obtaining an expression to solve the question. This is illustrated below:
From ohm's law,
V = IR
Were:
V is the voltage.
I is the current.
R is resistance.
From the question given, we were told that the resistance is constant. Therefore the above equation can be written as shown below:
V = IR
V/I = constant
V1/I1 = V2/I2
V1 is initial voltage.
V2 final voltage.
I1 is initial current.
I2 final current.
Step 2:
Data obtained from the question. This include the following:
Initial voltage (V1) = V
Initial current (I1) = 60 miliAmps
Final voltage (V2) = one-half of the original voltage = 1/2V = V/2
Final current (I2) =..?
Step 3:
Determination of the new current. This can be obtained as follow:
V1/I1 = V2/I2
V/60 = (V/2) / I2
Cross multiply to express in linear form
V x I2 = V/2 x 60
V x I2 = V x 30
Divide both side by V
I2 = (V x 30)/V
I2 = 30mA.
Therefore, the new current is 30miliAmps
Assuming our "closed tube" is closed at only one end, then
<span> v = fλ = f*4L/n
</span><span> where "n" is the harmonic number. So
</span><span> L = nv / 4f = n*346m/s / 4*256Hz = n*0.38 m
</span> <span>Since the only option in your list that is an integer multiple of 0.38 m is 1.35 m
</span><span> I'd say that we're hearing the fourth harmonic.
answer is
</span><span>A. 1.35 m</span><span>
</span>
