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If a reaction starts with 30 grams how many should it end with ?

zysi [14]

30grams

Explanation:

If a reaction starts with 30grams then the reaction should end with 30grams.

This in conformity with the law of conservation of mass.

The law states that "in an isolated system, mass is neither created nor destroyed during chemical transformation".
Mass is the quantity of matter contained in a substance.
In chemical reactions, the mass of reactants must always be the same with the mass of the product baring any loss.
In an isolated system, there is no exchange of energy and mass.
Chemical systems are usually treated as isolated systems in which mass is conserved.

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4 0

3 years ago

The period of the wave is

Karo-lina-s [1.5K]

Answer:

2.5 s

Explanation:

3 0

3 years ago

Robin is standing terrified at the end of a diving board, which is high above the water. If Robin has a mass of 76 kg and is sta

masya89 [10]

Answer:

Torque = 1191.68 N-m

Explanation:

Given data

mass m = 76 kg

standingdistance r = 1.6 m

Solution

we get here torque that si express as

torque = force × distance ................1

torque = r × F sin(theta)

and we know that

F = mg .........2

and g = 9.8 m/s²

put here value in equation 1 we get

Torque = 76 × 1.6 × 9.8 × sin(90)

Torque = 1191.68 N-m

5 0

3 years ago

What capacitance is needed to store 3μC of charge at a voltage of 120V?

katrin [286]

<u>Answer:</u>

C = 0.025F

<u>Explanation:</u>

Charge =q=3× $10^{-6}$ C

Voltage=V=120V

Q=CV

C=Q/V

=3× $10^{-6}$ /120

=1/40× $10^{-6}$

C = 0.025F

C=2.5 $10^{-4}$ C 0r 25 × $10^{-3}$ F

3 0

2 years ago

A long, straight wire carries a current of 5.20 A. An electron is traveling in the vicinity of the wire. Part A) At the instant

Gre4nikov [31]

A) $2.2\cdot 10^{-19} N$

The strength of the magnetic field produced by a current-carrying wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I is the current in the wire

r is the distance from the wire

In this situation,

I = 5.20 A

r = 4.60 cm = 0.046 m is the distance of the electron from the wire

Therefore the magnetic field strength at the electron's location is

$B=\frac{(4\pi \cdot 10^7)(5.20)}{2\pi (0.046}=2.26\cdot 10^{-5} T$

The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by

$F=qvB$

where

q is the magnitude of the charge of the particle

v is its velocity

B is the magnetic flux density

For the electron of this problem,

$q=1.6\cdot 10^{-19} C$ is the charge

$v=6.10\cdot 10^4 m/s$ is the speed

$B=2.26\cdot 10^{-5} T$ is the magnetic field

Substituting,

$F=(1.6\cdot 10^{-19})(6.10\cdot 10^4)(2.26\cdot 10^{-5})=2.2\cdot 10^{-19} N$

B) Same direction as the current in the wire

First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:

- The thumb gives the direction of the current -> upward

- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)

Now the direction of the force can be found by using the right-hand rule. We have:

- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)

- middle finger = direction of the magnetic field (to the north)

- Thumb = direction of the force --> downward

However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.

8 0

3 years ago