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2 years ago

13

The rotational speeds of four generators are listed in RPM (revolutions per minute). Arrange the generators in order based on th

e electric fields they create. Start with the weakest electric field and end with the strongest. Assume that the generators are identical except for their speed.
Tiles
3,600 RPM
3,200 RPM
3,400 RPM
3,000 RPM
Sequence

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2 answers:

artcher [175]2 years ago

4 0

with the same generator, so the only factor for producing the slectric field is only the speed. The faster the rotational speed of the generator the greater it produce electric field. So the sequence is 3000 rpm < 3200 rpm < 3400 rpm < 3600 rpm

ivanzaharov [21]2 years ago

3 0

Answer: 3000

3200

3400

3600

Explanation: just took the test 5 out of 5

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The total energy of a 0.050 kg object travelling at 0.70 c is

quester [9]

Would presume the energy as kinetic energy.

E = (1/2)*mv²

But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s

Ke = (1/2)*mv² = 0.5*0.05*(0.7*<span>3* 10⁸)</span>² = 1.1025 * 10¹⁵ Joules

There is no exact match from the options.

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1 year ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current

sladkih [1.3K]

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero $U = 0.1355 \ m$

Explanation:

From the question we are told that

The distance of wire one from two along the y-axis is y = 0.340 m

The current on the first wire is $I_1 = (27.5i) A$

The force per unit length on each wire is $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

$Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=> $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where $\mu_o$ is the permeability of free space with a constant value of $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

$\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=> $I_2 = 18.23 \ A$

Let U denote the line in the xy-plane where the total magnetic field is zero

So

So the force per unit length of wire 2 from line U is equal to the force per unit length of wire 1 from line (y - U)

So

$\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

$\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

$6.198 -18.23U = 27.5U$

$6.198=45.73U$

$U = 0.1355 \ m$

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1 year ago

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come

Yuliya22 [10]

Answer:

(a) vo = 24.98m/s

(b) t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

$y=y_o+v_ot-\frac{1}{2}gt^2$ (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

$v_o=\frac{y-y_o+1/2gt^2}{t}$

$v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}$

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

$0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s$

The time that the ball takes to arrive to the ground is 5.09s

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1 year ago

Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d

nydimaria [60]

<u>Answer:</u>

<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>

<u>Sources:</u>

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

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1 year ago

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Viktor [21]

To solve this problem we will apply the concepts related to gravity according to the Newtonian definitions. From finding this value we will use the linear motion kinematic equations to find the time. Our values are

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$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

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$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

The time taken for the rock to reach the surface is t = 87.58s

8 0

1 year ago