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2 years ago

What are action-reaction force pairs?

2 answers:

8 0

Exerted by two objects on each other are often called an action- reaction force pair. Either force can be considered the action force or the reaction force.

Cloud [144]2 years ago

7 0

Two objects on each other

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Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

disa [49]

Answer:

The net magnetic field ta the center of square is $1.36\times10^{-4} T$ .

Explanation:

Current, I = 12 A , side ,a = 10 cm = 0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

$B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 + sin 45 \right )\\\\B = 3.4\times 10^{-5} T$

Net magnetic field at the center of the square is

B' = 4 B

$B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T$

4 0

3 years ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago

A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.

Natali5045456 [20]

Answer:

B) electrons transferred from sphere to rod.

(2) 1.248 x 10¹¹ electrons were transferred

Explanation:

Given;

initial charge on the plastic rod, q₁ = 15nC

final charge on the plastic rod, q₂ = - 5nC

let the charge acquired by the plastic rod = q

q + 15nC = -5nC

q = -5nC - 15nC

q = -20 nC

Thus, the plastic rod acquired excess negative charge from the metal sphere.

Hence, electrons transferred from sphere to rod

B) electrons transferred from sphere to rod.

2) How many charged particles were transferred?

1.602 x 10⁻¹⁹ C = 1 electron

20 x 10⁻⁹ C = ?

= 1.248 x 10¹¹ electrons

Thus,1.248 x 10¹¹ electrons were transferred

7 0

3 years ago

Can an ordinary object, like a motorcycle, be mass-less? Yes or No

Drupady [299]

Answer:

no.

Explanation:

because the mass of an object never changes.

4 0

3 years ago

A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu

yarga [219]

Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

Temperature = 296 K

If the initial volume is V then the final volume will be V/3.

We need to calculate the work done

Using formula of work done

$W=nRT\ ln(\dfrac{V_{f}}{V_{i}})$

Put the value into the formula

$W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})$

$W=1.90\times8.314\times296\ ln(\dfrac{1}{3})$

$W=−5136.88\ J$

The Work done on the system.

Hence, The work done is 5136.88 J.

5 0

3 years ago