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kirza4 [7]
3 years ago
15

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.3 m/s. The drag force is of the form bv2 What is the value of b?

Physics
1 answer:
Fed [463]3 years ago
6 0

Answer:

The value is  b  =  0.00026 \ kg / m

Explanation:

From the question we are told that

   The mass of the  Ping-Pong is  m  =  2.3 \  g = 0.0023 \  kg

    The terminal  speed is  v  =  9.3 \  m/s

    The drag force is  bv^2

Generally the resultant force on the Ping- Pong is mathematically represented as

      F =  mg  -  bv^2

when  terminal velocity is  attained  , the resultant force is zero  so

      0 =  mg  -  bv^2

=>   b  =  \frac{m *  g}{v^2}

=>    b  =  \frac{0.0023 *  9.8}{ 9.3 ^2}

=>    b  =  0.00026 \ kg / m

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1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].

2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].

3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].

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3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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