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3 years ago

What would most likely happen if one of the runners bagan jogging in knee-deep water

Physics

1 answer:

Nesterboy [21]3 years ago

7 0

Then they will most likely slow down unless they do high knees or something like that

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At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta

Lubov Fominskaja [6]

Answer:

a=2330

b= 0.223secs

Explanation:

pb=2330m

t=0.223secs

6 0

3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago

One positve and one negative charhe jave a force of -5. 00 N between them. If the force changed to -170 N , by what factor did t

stepladder [879]

Answer:

The distance is shortenend by factor .1715

Explanation:

5 n = 1/r^2

sqrt (1/5) = r

170 n = 1 / ( x sqrt(1/5))^2

(xsqrt 1/5)^2 = 1/170

x sqrt 1/5 = .076696

x = .1715

5 0

2 years ago

In this diagram, which force is represted by L?

Dmitry [639]

Answer is a) the force caused by the wend nag encountering wind and air.

8 0

3 years ago

Three resistors having resistances of LaTeX: 4.0\;\Omega4.0 Ω, LaTeX: 6.0\;\Omega6.0 Ω, and LaTeX: 10.0\;\Omega10.0 Ω are connec

aalyn [17]

Answer:

Explanation:

Given that three resistor of resistance

R1 = 4 Ω

R2 = 6 Ω

R3 = 10 Ω

The resistor are connected in parallel, the equivalent resistance is

1 / Req = 1 / R1 + 1 / R2 + 1 / R3

Rearranging this we have

Req = R1•R2•R3 / (R2•R3 + R1•R3 + R1•R2)

Req = 4 × 6 × 10 / (6×10 + 4 × 10 + 4 × 6)

Req = 240 / (60 + 40 + 24)

Req = 240 / 124

Req = 1.94 Ω

The parallel connection is connected in series with a battery and a resistor.

Resistor resistance is

r = 2 Ω

Supply voltage of the battery is 12V

V = 12V

Let find the current flowing in the circuit.

V = I(Req + r)

I = V / (Req + r)

I = 12 / (1.94 + 2)

I = 12 / 3.94

I = 3.05A

So, this is the current flowing in the circuit and It is the same current that will be shared by the parallel resistance.

We can calculated the voltage across the parallel resistance

From ohms las

V = iR

V = I × Req

V = 3.05 × 1.94

V = 5.92 V.

Then, this the voltage across each parallel resistor.

Then, to know the current in the 10ohms resistance

V = iR

I = V / R3.

I = 5.92 / 10

I = 0.592 A.

The current in the 10 ohms resistor is 0.59A

5 0

3 years ago