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natita [175]
3 years ago
6

A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×

10−19 C.
Answer in units of electrons.
Physics
1 answer:
bagirrra123 [75]3 years ago
7 0
Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron 1,6 * 10 ^{-19}C<span>. To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons! 

Q = 4 </span>μC → Q = 4*10^{-6}
e = 1,6 * 10 ^{-19}C
n = ?<span>

We have:
</span>Q = n*e
n =  \frac{Q}{e}
n =  \frac{4*10^{-6}}{1,6 * 10 ^{-19}}
n = 2,5*10^{-6-(-19)}
n = 2,5*10^{-6+19}
\boxed{n = 2,5*10^{13}electrons}
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40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
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Consider a book that weighs 15 N at rest on a flat table. How many newtons of support does the table provide?
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driving down the highway , you find yourself behind a heavily loaded tomato truck. you follow close behind the truck keeping the
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The tomato won't hit the car

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Have a nice day!

4 0
3 years ago
PLEASE HELP ME 45 POINTS
sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a (1)

Mass B

\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a (2)

Where:

T - Tension force in the cord, measured in newtons.

m_{A}, m_{B} - Masses of blocks A and B, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

a - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g

(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g

a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g

If we know that m_{A} = 5\,kg, m_{B} = 2\,kg and g = 9.807\,\frac{m}{s^{2}}, then the acceleration of the masses is:

a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)

a = 4.203\,\frac{m}{s^{2}}

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

T = m_{B}\cdot (a+g)

If we know that m_{B} = 2\,kg, g = 9.807\,\frac{m}{s^{2}} and a = 4.203\,\frac{m}{s^{2}}, then the tension force in the cord:

T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}  \right)

T = 28.02\,N

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

\Delta y  = \frac{1}{2}\cdot a\cdot t^{2} (3)

Where:

a - Net acceleration, measured in meters per square second.

t - Time, measured in seconds.

\Delta y - Covered distance, measured in meters.

If we know that a = 4.203\,\frac{m}{s^{2}} and \Delta y = 1.5\,m, then the time taken by the system is:

t = \sqrt{\frac{2\cdot \Delta y}{a} }

t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }

t \approx 0.845\,s

The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is calculated by the following formula:

v = a\cdot t (4)

Where v is the final speed of the system, measured in meters per second.

If we know that a = 4.203\,\frac{m}{s^{2}} and t \approx 0.845\,s, then the final speed of the system is:

v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

v = 3.551\,\frac{m}{s}

The final speed of the system is 3.551 meters per second.

8 0
3 years ago
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