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3 years ago

How is electrostatic force impacted by charge and distance?

1 answer:

7 0

Answer: In electrostatics, the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. ... Electrostatic force and distance are inversely related.

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Bob is pulling a 30kg filing cabinet with a force of 200n , but the filing cabinet refuses to move. the coefficient of static fr

puteri [66]

The cabinet is being pulled with 200N and is being rested by a force equal to 200N. That is why it is not being moved.

<span>Although the force of static friction can equal Fk=µs*F=m*g*µs=(30kg)*(9.8m/s^2)*(0.80)=235 N. It is not resisting the 200N force with 235N. Imagine if you pushed something with 200N and it pushed you back with 235N, especially a cabinet. You would think that the cabinet was alive.</span>

8 0

3 years ago

Water at 20oC flows through a long elliptical duct 30 cm wide and 22 cm high. What average velocity, in m/s, would cause the wei

mars1129 [50]

Explanation:

The given data is as follows.

Fluid is water so, density $\rho = 1000 kg/m^{3}$

Weight flow rate = 500 lbf/s = 2224.11 N/sec

Cross-sectional area (A) = $\pi \times \frac{30}{2} \times \frac{22}{2}$

= 0.05184 $m^{2}$

Hence, weight flow rate will be given as follows.

w = $\rho \times g \times A \times V$

2224.11 N/sec = $\rho \times g \times A \times V$

V = $\frac{2224.11}{1000 \times 9.81 \times 0.05184}$ m/s

= 4.373 m/s

Thus, we can conclude that average velocity in the given case is 4.373 m/s.

8 0

3 years ago

Read 2 more answers

Page 423 in the 8th grade science fusion textbook what at the answers

balu736 [363]

Go to slader.com and type in the full name of the text book and page number. It should have the andwers

3 0

2 years ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

If a projectile fired beneath the water, straight up, breaks through the surface at a speed of 13m/s, to what height above the w

OlgaM077 [116]

Well if you had either the velocity or distance traveled i could tell you. But since you haven't all i can say for sure is that the water slowed the bullet down to 13m/s so lets say you knew the distance you would calculate how many meters it traveled and you would have your answer because in this situation, meters (height) =how many seconds spent going into the air.

3 0

3 years ago