In which case would electrical potential energy be built up and stored in the electric field?

2. A 1.30-m long gas column that is open at one end and closed at the other end has a fundamental resonant frequency 80.0 Hz. Wh

Elina [12.6K]

To solve this problem, it will be necessary to apply the concepts related to the fundamental resonance frequency in a closed organ pipe.

This is mathematically given as

$f_n (2n+1)(\frac{v}{4L})$

For fundamental frequency n is 0, then,

$f_0 = \frac{v}{4L}$

When,

v = Velocity of sound

L = Length,

Rearranging to find the velocity,

$v = f_0 (4L)$

$v = (80Hz)(4)(1.3m)$

$v = 416m/s$

Therefore the speed of sound in this gas is 416m/s

7 0

3 years ago

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

nlexa [21]

Answer:

Explanation:

hi

5 0

2 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

11 months ago

A child is sitting on the outer edge of a merry-go-round that is 1.5 m in diameter. If the merry-go-round makes 3.2 rev/min, wha

Troyanec [42]

Answer:

the velocity of the kid is 5.6 m/s

Explanation:

r is the radius and w is the frequency.

so we should know that the diameter is 18m and the diameter is equal to two times the radius, so r = 18m/2 = 9m

we should also know that the circumference of a circle is equal to c = 2pi*r, so each revolution has this length. if the kid does 5.9 revolutions in one minute then the kid spins at v = 5.9*2pi*9m/min

so we want to write this in meters per second and this means that we need to divide it by 60!

v = (5.9*2pi*9/60)m/s = 5.56 m/s

so your answer will be 5.6 m/s glad i could help!

4 0

2 years ago

Read 2 more answers

Hang two sheets of paper vertically from adjacent corners. The sheets should be parallel and close to each other with a small ga

Mrrafil [7]

Answer:

a. The sheets move toward each other and the gap narrows.

Explanation:

This exercise is related to fluid mechanics, when blowing between the two sheets, we can apply Bernoulli's equation, where the index 2 is the space between the two sheets

P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y²

if the two leaves are at the same height

y₁ = y₂

whereby

P₁ + ½ ρ g v₁² = P₂ + ½ ρ v₂²

for the air velocity between the leaves let us use the continuity equation

A₁ v₁ = A₂ v₂

the area between the leaves is less than the external area, so the air speed must increase. If we use this in Bernoulli's equation, increasing the speed 2 (between the leaves) to maintain equality the pressure must decrease.

If the pressure decreases, the blades should move closer

When resisting the answers, the correct one is a

4 0

3 years ago