Ask question

Ask question

All categories

3 years ago

12

Two particles, one with charge − 3.77 μC −3.77 μC and one with charge 4.39 μC, 4.39 μC, are 4.34 cm 4.34 cm apart. What is the m

agnitude of the force that one particle exerts on the other?

1 answer:

Fudgin [204]3 years ago

8 0

Answer:

the magnitude of the force that one particle exerts on the other is 79.08 N

Explanation:

given information:

q₁ = 3.77 μC = -3.77 x 10⁻⁶ C

q₂ = 4.39 μC = 4.39 x 10⁻⁶ C

r = 4.34 cm = 4.34 x 10⁻² m

What is the magnitude of the force that one particle exerts on the other?

lFl = kq₁q₂/r²

= (9 x 10⁹) (3.77 x 10⁻⁶) (4.39 x 10⁻⁶)/(4.34 x 10⁻²)²

= 79.08 N

You might be interested in

The baseball diamond is a square with sides 27.4 m long. If the positive x-axis points from home plate to first base and the pos

miss Akunina [59]

Answer:

What is the magnitude of the player 's displacement? A baseball "diamond" is a square, each side of length 27.4 m, with home plate and if each length of the base is 27.4 m, and the baseball player ran to third base,Then, the baseball player is 27.4 feet away from home plate (where he started)below the horizontal.

Explanation:

5 0

3 years ago

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is:

By (1):

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$

$m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}$

$v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}$

$v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)$

$v_{1} = 3.280\,\frac{m}{s}$

By (2):

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$

$v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}$

$v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}$

$v_{2} = 0.024\,\frac{m}{s}$

5 0

3 years ago

An ion of an element has 30 protons 32 neutrons and 29 electrons what is that charge and how did you make that determination

ale4655 [162]

The ion has a charge of 1 +.

A neutral ion would have the same number of protons and neutrons. The neutrons have no charge (they're neutral) so they don't matter when it comes to the charge of an ion. This ion has 30 protons (+) and 29 electrons (-). There is 1 more proton than there are electrons so the ion has a charge of 1 +.

4 0

3 years ago

Read 2 more answers

A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can b

vazorg [7]

Answer:

Explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

$I_{1}=\frac{ML^{2}}{12}$

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

$I_{2}=2\times m\times \frac{L^{2}}{4}$

$I_{2}=\frac{mL^{2}}{2}$

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

$I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}$

$I=\frac{(M +6m)L^{2}}{12}$

8 0

3 years ago

3. What can you infer about an ion that has a positive charge?

IRINA_888 [86]

Answer: This ion is derived from atom who lost one electron.

4 0

3 years ago