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3 years ago

12

Which one of the following statements about sulfuric acid is correct?

1 answer:

kherson [118]3 years ago

6 0

The answer is C. Sulfuric Acid is dangerous to living organism

Direct contact with your skin and it will cause Chemical Burns , Direct contact with your eyes and it could cause a permanent Blindness.

This chemical also suck the water out of everything, so you can guess what will happen if it get into your body.

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A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particl

Sati [7]

Given Information:

KEa = 9520 eV

KEb = 7060 eV

Electric potential = Va = -55 V

Electric potential = Vb = +27 V

Required Information:

Charge of the particle = q = ?

Answer:

Charge of the particle = +4.8x10⁻¹⁸ C

Explanation:

From the law of conservation of energy, we have

ΔKE = -qΔV

KEb - KEa = -q(Vb - Va)

-q = KEb - KEa/Vb - Va

-q = 7060 - 9520/27 - (-55)

-q = 7060 - 9520/27 + 55

-q = -2460/82

minus sign cancels out

q = 2460/82

Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

q = 2460(1.60x10⁻¹⁹)/82

q = +4.8x10⁻¹⁸ C

6 0

3 years ago

Objects in our solar system, including planets and their moons, stay in orbit because of gravity and inertia. Draw a model to sh

coldgirl [10]

Answer:

You cant draw on brainly

Explanation:

6 0

2 years ago

Read 2 more answers

The PE of the box that is on a 2.0 m high self is 1600 J. What is the power expelled to lift the box to this height in 10.0 seco

Pani-rosa [81]

Answer:

160 W

Explanation:

Power is the ratio of work to time:

(1600 J)/(10 s) = 160 J/s = 160 W

7 0

2 years ago

How much equal charge should be placed on the earth and the moon so that the electrical repulsion balances the gravitational for

kumpel [21]

As we know that electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that electrostatic repulsion force is balanced by the gravitational force between them

so here force of attraction due to gravitation is given as

$F_g = 1.98 \times 10^{20} N$

here we can assume that both will have equal charge of magnitude "q"

now we have

$1.98 \times 10^{20} = \frac{kq^2}{r^2}$

$1.98 \times 10^{20} = \frac{(9\times 10^9)(q^2)}{(3.84 \times 10^8)^2}$

$1.98 \times 10^{20} = (6.10 \times 10^{-8}) q^2$

now we have

$q = 5.7 \times 10^{13} C$

6 0

3 years ago

1. I’m in the 2nd column, 4th row, and I’m a metal. Who am I? ________________ 2. I’m a very lonely nonmetal. Who am I? ________

Ghella [55]

Answer:

Explanation:

I'm in 17th column , a nometal, and a solid at room temperature. What am i

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3 years ago