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3 years ago

II. IDENTIFY THE TYPE OF FORCE

Physics

1 answer:

uysha [10]3 years ago

4 0

Answer:

Explanation:

1. Frictional force is responsible for running of car and buses on roads.

2.Gravitational force exists between astronauts in space.

3. Magnetic for is responsible to attract the iron objects using a magnet

4.Electrostatic force is responsible for fiber to stick on the skin.This force occurs due to the presence of charge.

5.When a person is pushing a trolley then object experience a normal reaction from ground.

6.Gravitational force makes the planet to move in their orbits.

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A switch can

oee [108]

Answer: a switch can do A, B and E

Explanation:

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2 years ago

A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla

aivan3 [116]

Answer:

4 x 10^-9F

Explanation:

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3 years ago

A violin string vibrates at 260 Hz when unfingered. At what frequency will it vibrate if it is fingered one fourth of the way do

Advocard [28]

Answer:

346.66 Hz

Explanation:

$l_1$ = Length of string which is unfingered = l

$l_2$ = Length of string which is vibrate when fingered = $l-\dfrac{1}{4}l=\dfrac{3}{4}l$

$f_1$ = Unfingered frequency = 260 Hz

$f_2$ = Fingered frequency

Frequency is inversely proportional to length

$f=\dfrac{1}{l}$

So,

$\dfrac{f_1}{f_2}=\dfrac{l_2}{l_1}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{\dfrac{3}{4}l}{l}\\\Rightarrow \dfrac{f_1}{f_2}=\dfrac{3}{4}\\\Rightarrow f_2=\dfrac{4}{3}f_1\\\Rightarrow f_2=\dfrac{4}{3}260\\\Rightarrow f_2=346.66\ Hz$

The frequency of the fingered string is 346.66 Hz

3 0

3 years ago

Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.

mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, $\rho=10^{-6}\ \Omega -m$

Density, $d=8.31\times 10^3\ kg/m^3$

(A) The length of the wire is given by using the definition of resistance as :

Volume,

$V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3$

Area,

$A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}$ ....(1)

$R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m$

(b) Equation (1) becomes :

$A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m$