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Vlad1618 [11]
3 years ago
15

calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​

Physics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

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Explain Resistor in parallel and series. ​
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\sf\huge\underline\blue{Resistor:-}

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\rightarrowResistors reduce the current flow and lower voltage levels within circuits.

\sf\large\underline\purple{Resistors \:in\: Series:-}

\rightarrowA <u>circuit is said to be connected in series</u> when the same amount of <u>current flows through the resistors</u>. In such circuits, the voltage across each resistor is different.

\rightarrowIn a series connection, if any resistor is broken or a fault occurs, then the entire circuit is turned off. The construction of a series circuit is simpler compared to a parallel circuit.

\rightarrowFor the above circuit(attached image-1), the total resistance is given as:

\sf{R_{total}\: = \:R1 + R2 + ….. + Rn}

The total resistance of the system is just the total of individual resistances.

\sf\large\underline\purple{Resistors \:in\: Parallel:-}

\rightarrowA <u>circuit is said to be connected in parallel</u> when the <u>voltage is the same across the resistors</u>. In such circuits, the current is branched out and recombines when branches meet at a common point.

\rightarrowA resistor or any other component can be connected or disconnected easily without affecting other elements in a parallel circuit.

\rightarrowThe figure(attached image -2) above shows ‘n’ number of resistors connected in parallel. The following relation gives the total resistance here

\sf{\frac{1}{R_{total}}\: = \:\frac{1}{R1} + \frac{1}{R2} + ….. + \frac{1}{Rn}}

\rightarrowThe sum of reciprocals of resistance of an individual resistor is the total reciprocal resistance of the system.

_______________________________

Hope it helps you:)

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2 years ago
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This type of bike tire is thinner, lighter, more expensive, and punctures easily.
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There are 3, Tubular, Clincher, and tubeless.
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Being overweight or obese increases the risk of:
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A rope is wrapped around the rim of a large uniform solid disk of mass 325 kg and radius 3.00 m. The horizontal disk is made to
Naily [24]

Answer:

The angular speed is 0.13 rev/s

Explanation:

From the formula

\tau = I\alpha

Where \tau is the torque

I is the moment of inertia

\alpha is the angular acceleration

But, the angular acceleration is given by

\alpha = \frac{\omega}{t}

Where \omega is the angular speed

and t is time

Then, we can write that

\tau = \frac{I\omega}{t}

Hence,

\omega = \frac{\tau t}{I}

Now, to determine the angular speed, we first determine the Torque \tau and the moment of inertia I.

Here, The torque is given by,

\tau = rF

Where r is the radius

and F is the force

From the question

r = 3.00 m

F = 195 N

∴ \tau = 3.00 \times 195

\tau = 585 Nm

For the moment of inertia,

The moment of inertia of the solid disk is given by

I = \frac{1}{2}MR^{2}

Where M is the mass and

R is the radius

∴I = \frac{1}{2} \times 325 \times (3.00)^{2}

I = 1462.5 kgm²

From the question, time t = 2.05 s.

Putting the values into the equation,

\omega = \frac{\tau t}{I}

\omega = \frac{585 \times 2.05}{1462.5}

\omega = 0.82 rad/s

Now, we will convert from rad/s to rev/s. To do that, we will divide our answer by 2π

0.82 rad/s = 0.82/2π rev/s

= 0.13 rev/s

Hence, the angular speed is 0.13 rev/s,

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3 years ago
Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp
Alex787 [66]

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

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