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White raven [17]
3 years ago
7

Runner A had the highest velocity. Runner C had the lowest velocity. Runners A and B started at the same position, but runner C

started ahead of runners A and B. What did you include in your response? Check all that apply.
Runner A had the highest velocity.
Runner C had the lowest velocity.
Runners A and B started at the same position.
Runner C started ahead of runners A and B.
Physics
2 answers:
galina1969 [7]3 years ago
6 0

Answer:

A. C. and D.

Explanation:

Elan Coil [88]3 years ago
5 0

Answer:

A, C, and D

Just confirming

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Which substance is uniquely a product of incomplete combustion of a fossil fuel?
Serga [27]

I can think of two of them:

-- carbon monoxide

-- black soot

3 0
3 years ago
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

The  charge on the dust particle is  q_d  = 6.94 *10^{-13} \  C

Explanation:

From the question we are told that

    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

   The charge is  q =  -10\mu C= -10*10^{-6} \ C

    The mass suspended in mid-air is m_a =  5.0 \mu g =  5.0 *10^{-6} \ g =  5.0 *10^{-9} \  kg

   

Generally the electric field on the carpet is mathematically represented as

           E =  \frac{q}{ 2 *  A  *  \epsilon _o}

Where \epsilon _o is the permittivity of free space with value \epsilon_o = 8.85*10^{-12}  \ \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

substituting values

           E =  \frac{-10*10^{-6}}{ 2 *  (2 * 4 )  *  8.85*10^{-12}}

           E = -70621.5 \  N/C

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        F__{E}} =  F__{G}}

=>     q_d *  E  =  m * g

=>      q_d  =  \frac{m * g}{E}

=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

=>     q_d  = 6.94 *10^{-13} \  C

4 0
3 years ago
A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

7 0
3 years ago
What is the main force that must be overcome in order to push an object
My name is Ann [436]

Answer: Friction

Explanation:

Friction and the normal force would be the two initial forces to overcome.

4 0
2 years ago
According to ohm's law what is the resistance of a light bulb if the applied voltage is 21 volts and the current is 3 amps?
Nady [450]
Resistance = voltage/current = 21/3 = 7Ω
6 0
3 years ago
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