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timama [110]
2 years ago
13

ONLY ANSWER IF YOU KNOW FOR SURE PLEASE :)

Physics
1 answer:
sveticcg [70]2 years ago
6 0

Answer:

Question one is b, Question two is b, and question three is b, im pretty positive that what it is

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Harmfful effect of earthquake
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Answer:

Death, Destruction, Loss of home

5 0
3 years ago
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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
The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
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A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
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Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

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3 years ago
What is the relationship between thermal energy and conduction convection and radiation
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The higher the thermal energy the faster the conduction convection and radiation take place as the particles have more kinetic (movement) energy
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