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bagirrra123 [75]

3 years ago

11

A positively charged rod is brought close to one end of an uncharged metal rod but does not actually touch it. What type of char

ge does the end of the metal rod closest to the positively charged rod acquire?

1 answer:

svlad2 [7]3 years ago

7 0

Answer:

The end of the neutral rod which is the closest part to the charged rod would acquire a negative charge.

Explanation:

One of the rods is positively charged and one of them is neutral.

And the important part is that <u>they do not touch one another</u>, but get close to each other.

In this case, the end of the neutral rod which is the closest part to the charged rod would acquire a negative charge. This is because of the Coulomb's Law. The opposite charges exert an attractive force to each other. The positive charges attract the negative charges on the neutral rod.

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1. The volume of a given mass of gas is 20cm when its

Pavel [41]

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2 Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

5 0

3 years ago

Which phenomenon occurs when the moon and Earth are aligned with the sun? A. Retrograde motion B. Solstice C. Equinox D. Eclipse

muminat

Answer: Eclipse

Explanation: A lunar eclipse occurs when the full moon moves through the shadow of the Earth. This can only happen when the Earth is between the Moon and the Sun and all three are lined up in the same plane, called the ecliptic. The ecliptic is the plane of Earth's orbit around the Sun.

7 0

3 years ago

An object is dropped from rest from the top of a 400 m cliff on Earth. If air resistance is negligible, what is the distance the

IrinaVladis [17]

Answer:

176.58 m

Explanation:

t = Time taken = 6 seconds

u = Initial velocity = 0

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 6^2\\\Rightarrow s=176.58\ m$

The object travels 176.58 m from the cliff in 6 seconds.

8 0

3 years ago

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$

where

$m_1 = m_2 = m = 10 kg$ is the mass of each puck

$u_1 = 10 m/s$ is the initial velocity of the 1st puck

$u_2 = -20 m/s$ is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

$mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s$

Learn more about momentum:

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8 0

2 years ago

A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be

stepan [7]

Answer:

$p_2 = 664081 N/m^{2}$

Explanation:

from the ideal gas law we have

PV = mRT

$P = \rho RT$

$\rho = \frac{P}{RT}$

HERE R is gas constant for dry air = 287 J K^{-1} kg^{-1}

$\rho = \frac{7.00 10^{5}}{287(18+273)}$

$\rho = 8.38 kg/m^{3}$

We know by ideal gas law

$\rho = \frac{m_1}{V_1}$

$m_1 = \rho V_1 = 8.38 *2*10^{-3}$

$m_1 = 0.0167 kg$

for m_2

$m_2 = \rho V_i - V_removed$

$m_2 = 8.38*(.002 - 10^{-4})$

$m_2 = 0.0159 kg$

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

$\frac{p_2}{p_1}=\frac{m_2}{m_1}$

$p_2 = p_1 * \frac{m_2}{m_1}$

$p_2 =7*10^{5} * \frac {.0159}{0.0167}$

$p_2 = 664081 N/m^{2}$

8 0

3 years ago