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3 years ago

5. What is the percent composition of sulfur in H2SO4?

Physics

2 answers:

N76 [4]3 years ago

8 0

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 %

AVprozaik [17]3 years ago

7 0

A. I looked it up and it should be somewhere around A

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How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?

aalyn [17]

There is too much information given, it's hard to understand exactly which variables are important in this problem.

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2 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

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2 years ago

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gogolik [260]

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2 years ago

Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert

damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

$F = k \frac{q\cdot q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

$q+\frac{q}{2}=\frac{3}{2}q$

while the other charge will be

$q-\frac{q}{2}=\frac{q}{2}$

So, the new force will be

$F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F$

So, the force will decrease to 3/4 of its original value.

6 0

2 years ago

You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

Cloud [144]

Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

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$s = ut + \frac{1}{2} gt^2$

Here s is the distance of the object from Barney head ,

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Solving the above equation using quadratic formula

The value of t obtained is $t = 8 \ s$

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2 years ago