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2 years ago

Electrical insulating gloves marked with a beige label have a maximum use voltage level of _____ v.

Physics

1 answer:

zlopas [31]2 years ago

3 0

The maximum use voltage level for electrical insulating gloves with a beige label is 17000 v AC.

<h3 /><h3>By electrical insulating gloves, what do you mean?</h3>

To prevent burns and electrical shocks, users should wear electrically insulated rubber gloves. To shield workers from electrical shock, these are constructed of dielectric materials. Electricians working on electrical lines don rubber gloves, sometimes known as lineman's gloves. The gloves are tested at voltages higher than those for which they are rated. Rubber insulating gloves are the first line of defense, offering protection against shock and electrocution when coming into touch with electrified electrical parts and equipment, in contrast to arc flash protective suits, which are intended to lessen or eliminate injuries after an incident occurs. Rubber is a reliable form of shock protection because it doesn't conduct electricity. Rubber is a natural insulator, thus it will prevent electric currents from passing through your skin, unlike many other materials that behave as conductors.

To learn more about electrically insulated rubber gloves, click here:

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A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h

zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

$m_{1}$ $u_{1}$ + $m_{2}$ $u_{2}$ = $m_{1}$ $v_{1}$ - $m_{2}$ $v_{2}$

$m_{1}$ = 3 kg, $u_{1}$ = 8 m/s, $m_{2}$ = 2 kg, $u_{2}$ = 0 m/s ( since it is at rest), $v_{1}$ = 2 m/s, $v_{2}$ = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x $v_{2}$ )

24 + 0 = 16 - 2 $v_{2}$

2 $v_{2}$ = 16 - 24

2 $v_{2}$ = -8

$v_{2}$ = $\frac{-8}{2}$

= -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0

3 years ago

Usted arrastra los pies sobre un piso alfombrado en un día seco y la alfombra adquiere una carga positiva neta. a) Usted a defic

gregori [183]

Answer:

I would need more electrons 1.3125 10¹⁰ electrons

Explanation:

Static electricity is formed by induced charges in each part of the system, so the net charge does not change.

Let us analyze the situation presented, if the carpet acquires a positive charge, so that it does not alter the total charge in man, it should acquire a negative charge, to have a net negative charge I would need more electrons that have a negative charge.

We can know the number of electrons by a rule of proportions, if an electron has a charge of 1.6 10-19 C for a charge of 2.15 10-9 C how many electrons are needed

# = 2.15 10⁻⁹ /1.6 10⁻¹⁹

# = 1.3125 10¹⁰ electrons

The number is very high for which we will not worry about rounding since the number of electrons must be an integer

6 0

3 years ago

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

DerKrebs [107]

The ball's horizontal and vertical velocities at time $t$ are

$v_x=v_{xi}$

$v_y=v_{yi}-gt$

but the ball is thrown horizontally, so $v_{yi}=0$ . Its horizontal and vertical positions at time $t$ are

$x=v_{xi}t$

$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

3 years ago

A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to

White raven [17]

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^{-1}$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^{-1}$

speed of the swimmer with respect to water, $v=4\ ft.s^{-1}$

<u>Now the resultant of the two velocities perpendicular to each other:</u>

$v_r=\sqrt{v^2+v_s^2}$

$v_r=\sqrt{4^2+8^2}$

$v_r=8.9442\ ft.s^{-1}$

<u>Now the angle of the resultant velocity form the vertical:</u>

$\tan\beta=\frac{v_s}{v}$

$\tan\beta=\frac{8}{4}$

$\beta=63.43^{\circ}$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d\times \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=\sqrt{d^2-w^2}$

$\Delta s=\sqrt{1118.034^2-500^2}$

$\Delta s=1000\ ft$

On swimming 37° upstream:

<u>The velocity component of stream cancelled by the swimmer:</u>

$v'=v.\cos37$

$v'=4\times \cos37$

$v'=3.1945\ ft.s^{-1}$

<u>Now the net effective speed of stream sweeping the swimmer:</u>

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^{-1}$

<u>The component of swimmer's velocity heading directly towards the opposite bank:</u>

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^{-1}$

<u>Now the angle of the resultant velocity of the swimmer from the normal to the stream</u>:

$\tan\phi=\frac{v_n}{v'_r}$

$\tan\phi=\frac{4.8055}{2.4073}$

$\phi=63.39^{\circ}$

Now let the distance swam in this direction be d'.

$d'\times \cos\phi=w$

$d'=\frac{500}{\cos63.39}$

$d'=1116.344\ ft$

<u>Now the distance swept downstream:</u>

$\Delta s'=\sqrt{d'^2-w^2}$

$\Delta s'=\sqrt{1116.344^2-500^2}$

$\Delta s'=998.11\ ft.s^{-1}$

Time taken in crossing the rive in case 1:

$t=\frac{d}{v_r}$

$t=\frac{1118.034}{8.9442}$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=\frac{d'}{v'_r}$

$t'=\frac{1116.344}{2.4073}$

$t'\approx463.733\ s$

7 0

4 years ago

Three capacitors are connected as follows: 2.8 F capacitor and 5.57 F capacitor are connected in series, then that combination i

DedPeter [7]

ANSWER

8.99 F

EXPLANATION

We know that two capacitors of capacitances 2.8 F and 5.57 F are connected in series, while a third capacitor of capacitance 7.13 F is connected in parallel to that combination,

The capacitance works similarly to the resistance, except that when capacitors are connected in parallel, their capacitances add up, while when they are connected in series, the equivalent capacitance is like we were finding the equivalent resistance of resistors connected in parallel,

$C_{eq}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}+C_3=\frac{1}{\frac{1}{2.8F}+\frac{1}{5.57F}}+7.13F\approx8.99F$

Hence, the total capacitance is 8.99 F, rounded to the nearest hundredth.

6 0

2 years ago