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3 years ago

How does the junkyard magnet let go of metal?

Physics

1 answer:

klemol [59]3 years ago

7 0

Answer:

Wrecking yards employ extremely powerful electromagnets to move heavy pieces of scrap metal or even entire cars from one place to another. ... This creates a magnetic field around the coiled wire, magnetizing the metal as if it were a permanent magnet

Explanation:

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Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

$Ac =V^2/R$

where Ac = centripetal acceleration = $98 m/s^2$

R = radius of rotation = 15 m

V = speed of astronaut

Hence, $\frac{V^2}{15} =98$

solving this we get, V = 38.34 m/s

3 0

3 years ago

Read 2 more answers

6. If a drag racer wins the final round of herrace by going an average speed of 320 m/sin 4.5 seconds, what distance did he cove

Ivan

We want to calculate the distance covered by the drag racer. Recall, the formula for calculating distance is expressed as

Distance = speed x time

From the information given,

speed = 320 m/s

time = 4.5 s

By substituting these values into the formula, we have

Distance = 320 m/s x 4.5s

s cancels out. We are left with m. Thus,

Distance = 1440m

4 0

1 year ago

Why does gravity hold objects on earth?

LUCKY_DIMON [66]

Gravity is an attractive force that works to pull objects together. If 2 objects are close the gravitational pull will be stronger
Mass and distance determine gravity. The farther two things are away from each other, the weaker the gravitational forces are, the less mass an object has the less gravitational force it exerts

6 0

3 years ago

Read 2 more answers

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>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

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

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<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>

$P_2V_2 = P_3V_3$

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

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

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<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

A 6.89-nC charge is located 1.76 m from a 4.10-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

$F=\frac{k*q1*q2}{d^{2}}$

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0

3 years ago