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alekssr [168]
3 years ago
13

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

Physics
1 answer:
olasank [31]3 years ago
8 0

Answer: b) A positive charge is moved toward a positive charge.

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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
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