The net force on the boy is zero.
Based on this analysis, I would expect the boy to <u>not accelerate</u>.
Aerobie. Frisbee. Discus. Javelin. I suppose an American football to some extent.
<span>Pull! Clay pigeons. Arrows. Wingsuit. Kites. Hang gliders. Sails. sailboat keels/dagger boards. Water skis. Ski jumping skis. Boomerang. </span>
<span>I'm excluding spheres and parachutes as bluff bodies even though aerodynamics often plays a big part in their motion.</span>
Answer:
C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.
Explanation:
When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.
If it were a series connection, then the current would have been the same, but the voltage drop were another story.
The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be
C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
![f=\frac{v_T}{4L}](https://tex.z-dn.net/?f=f%3D%5Cfrac%7Bv_T%7D%7B4L%7D)
where, ![v_T=v\sqrt{\frac{T}{273} }](https://tex.z-dn.net/?f=v_T%3Dv%5Csqrt%7B%5Cfrac%7BT%7D%7B273%7D%20%7D)
Since, T=37+273 K = 310 K
v = 331 m/s
![\therefore v_T=331\sqrt{\frac{310}{273} } = 352.72 \ m/s](https://tex.z-dn.net/?f=%5Ctherefore%20v_T%3D331%5Csqrt%7B%5Cfrac%7B310%7D%7B273%7D%20%7D%20%3D%20352.72%20%5C%20m%2Fs)
Using this, we get:
![f=\frac{352.72}{4(0.240)} \\f=367.42 \ Hz](https://tex.z-dn.net/?f=f%3D%5Cfrac%7B352.72%7D%7B4%280.240%29%7D%20%5C%5Cf%3D367.42%20%5C%20Hz)
Hence, the fundamental frequency is 367.42 Hz.
To learn more about Attention here:
brainly.com/question/14673613
#SPJ4
Answer:
9.9 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
![v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 5+0^2}\\\Rightarrow v=9.9\ m/s](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2as%2Bu%5E2%7D%5C%5C%5CRightarrow%20v%3D%5Csqrt%7B2%5Ctimes%209.81%5Ctimes%205%2B0%5E2%7D%5C%5C%5CRightarrow%20v%3D9.9%5C%20m%2Fs)
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s