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3 years ago

5. Which of the following conduct electricity the best?

Physics

2 answers:

Stels [109]3 years ago

5 0

Answer:

Silver is the best conductor of electricity because it contains a higher number of movable atoms (free electrons). For a material to be a good conductor, the electricity passed through it must be able to move the electrons; the more free electrons in a metal, the greater its conductivity.

ratelena [41]3 years ago

5 0

The answer Silver.
This is because the electrons in this are more free thus making it able to move more and conduct better.

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Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

Air flows upward in the wick of a lantern because of the liquid property called

vovangra [49]

Adhesive.

Adhesive is the force of attraction between molecules of different kind. Liquid flows upward the wick because the adhesive force between the wick and the liquid is higher than cohesive forces in the liquid.

When the adhesive force between the wick and the liquid is high we have capillarity taking place. This cause the liquid to move up the wick.

7 0

3 years ago

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and

cricket20 [7]

Answer:

The frequency $f$ = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = $\frac{d}{2}$ = $\frac{44}{2}$ = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = $\frac{d}{2}$ = $\frac{3.0}{2}$ = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = $\frac{d}{2}$ = $\frac{2.0}{2}$ = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

$P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2$

$pgy_1=\frac{1}{2}pv^2_2 +pgy_2$

$v_2=\sqrt{2g(y_1-y_2)}$

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = $\sqrt{2*9.8*(0.35-0)}$

v₂ = $\sqrt {6.86}$

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

$v_3 = \frac{v_2r_2^2}{v_3^2}$

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = $(2.62 m/s)*$ $(\frac{1.5mm^2}{1.0mm^2} )$

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

$f=\frac{v_s}{4(h-v_3t)}$

where;

$v_s$ = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

$f=\frac{343}{4(0.40-(0.0589)(0.4)}$

$f= \frac{343}{0.6576}$

$f$ = 521.59 Hz

∴ The frequency $f$ = 521.59 Hz

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

$\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})$

$\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)$

$\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}$

where;

$v_s$ (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

$\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}$

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.

8 0

3 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be<u> zero</u> when it hits the ground)

<u>0 </u>= 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

1 year ago

Please help easy 8th grade question

Tamiku [17]

Answer:

whats the question?

Explanation:

7 0

3 years ago