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3 years ago

Weight of a man is less at coal mine.Why?

Physics

2 answers:

gavmur [86]3 years ago

6 0

There is less gravity underground

dmitriy555 [2]3 years ago

5 0

If someone is underground, then therefore there is less planet/ground underneath them, so there would be less gravity. Gravity directly affects weight.

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A capacitor is charged to a potential difference of 3 volt it delivers 30% store energy to lamp what is the final potential diff

nexus9112 [7]

Answer:

3.98V

Explanation:

Given

Pontential difference V as 3v

Energy delivered is 30%,

Recall that Enery E=1/2cv^2 from this E=V^2(since Current C is not provided we can assume a value 2)

So E=V^2

E=3^2=9

At full charge E=9,30%of 9,0.3*9=2.7 energy in capacitor is 9-2.7=6.3

But E=V^2

✓E=V

✓6.3=3.98V

4 0

2 years ago

Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction

Norma-Jean [14]

Answer:

The ball would hit the floor approximately $0.55\; \rm s$ after leaving the table.

The ball would travel approximately $5.5\; \rm m$ horizontally after leaving the table.

(Assumption: $g = 9.81\; \rm m \cdot s^{-2}$ .)

Explanation:

Let $\Delta h$ denote the change to the height of the ball. Let $t$ denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let $v_0(\text{vertical})$ denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible: $\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t$ .

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was $v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}$ .

The height of the table was $1.5\; \rm m$ . Therefore, after hitting the floor, the ball would be $1.5\; \rm m \!$ below where it was before leaving the table. Hence, $\Delta h = -1.5\;\rm m$ .

The equation becomes:

$\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}$ .

Solve for $t$ :

$\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55$ .

In other words, it would take approximately $0.55\; \rm s$ for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at $v(\text{horizontal}) =10\; \rm m \cdot s^{-1}$ ) until the ball hits the floor.

The ball was in the air for approximately $t = 0.55\; \rm s$ and would have travelled approximately $v(\text{horizontal})\cdot t \approx 5.5\;\rm m$ horizontally during the flight.

4 0

2 years ago

Why is copper NOT used as a filament?

chubhunter [2.5K]

Answer:

Short answer, because copper wire does not have high resistance.

Explanation:

6 0

2 years ago

Which is a correct reason to physically connect objects with a bond wire when transferring flammable liquids? A. To create a pat

RideAnS [48]

<h2>When Transferring Flammable Liquids - Option A</h2>

To create a pathway for the liquid to flow easily is a correct reason to physically connect objects with a bond wire when transferring flammable liquids. Physically connect the two conductive objects together with a bond wire to eliminate a difference in static charge potential among them. The bond wire is attached between the containers in the process of flammable filling liquid actions except for a metallic path is found.

5 0

3 years ago

Read 2 more answers

An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 13 V

MrMuchimi

Answer:

9.2 V

Explanation:

The RMS value of an AC is the effective value of a varying voltage or current in DC, that is the equivalent value of the AC which produces the same effect as an DC. For example if a motor is supplied by a 9V RMS voltage, it will rotate as if the voltage applied was 9V DC.

The RMS value is given by:

RMS voltage = Peak voltage * 1/√2

Given that the maximum voltage should not exceed 13 V, this means that the peak voltage is 13 V. The maximum RMS voltage is:

RMS voltage = Peak voltage * 1/√2 = 13 * 1/√2 = 9.2 V

4 0

3 years ago