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3 years ago

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The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are- a

. the positively charged protons b. the positively charged electrons c. the neutrons of heavy metallic nuclei d. the negatively charged electrons e. the negatively charged protons

1 answer:

SVEN [57.7K]3 years ago

3 0

Answer:

D) The negatively charged electrons

Electricity passes through metallic conductors as a flow of negatively charged electrons. The electrons are free to move from one atom to another. We call them a sea of delocalised electrons. Current was originally defined as the flow of charges from positive to negative. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

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A uniform solid disk of mass 5.00 kg and diameter 47.0 cm starts from rest and rolls without slipping down a 40.0 ∘ incline that

saveliy_v [14]

Answer:

The speed it reaches the bottom is

$v=6.51m/s$

Explanation:

Given: $m=5.0kg$ , $r=47cm\frac{1m}{100cm}=0.47m$

Using the conservation of energy theorem

$U_i=K_E+K_{ER}$

$m*g*h=\frac{1}{2}*m*v^2+\frac{1}{2}*I*w^2$

$v=r*w$ , $I=\frac{1}{2}*m*r^2$

$m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2$

$m*g*h=\frac{3}{4}*m*r^2*w^2$

$g*h=\frac{3}{4}*r^2*w^2$

Solve to w'

$w^2=\frac{4*g*h}{3*r^2}$

$h=x*sin(30)=6.5m*sin(30)=3.25m$

$w=\sqrt{\frac{4*9.8m/s^2*3.25m}{3*(0.235m)^2}}$

$w=27.74rad/s$

$v=27.74rad/s*0.235m=6.51m/s$

7 0

3 years ago

5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?

Degger [83]

Answer:

$\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}$

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

$\boxed{ \sf{velocity = \frac{distance \: travelled \: }{time \: taken}}}$

$\dashrightarrow{ \sf{velocity = \frac{200 \: m}{24 \: s} }}$

$\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}$

Hope I helped!

Best regards!

6 0

3 years ago

A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

denis-greek [22]

The radius of the cylinder is equal to half the diameter:

$r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm$

The volume of the cylinder is given by:

$V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3$

where h is the heigth of the cylinder. Converting into meters,

$V=1.28 \cdot 10^{-4} m^3$

And the density of the material will be given by the ratio between the mass and the volume:

$d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3$

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3 years ago

Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the

kifflom [539]

Answer:

Height will be 3.8971 m

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity $\omega =4.46rad/sec$

Moment of inertia is given by $I=\frac{1}{2}mr^2$

Transnational Kinetic energy is given by $KE=\frac{1}{2}mv^2$ as we know that v = $v=\omega r$

So $KE=\frac{1}{2}m(\omega r)^2$

Rotational kinetic energy is given by $KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2$

Potential energy is given by mgh

According to energy conservation

$mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2$

$h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m$

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3 years ago

A) Find the displacement for the given time intervals ,

Olegator [25]

Answer:

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Explanation:

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3 years ago

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