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Morgarella [4.7K]
2 years ago
7

Q3. Recall the equation that links gravitational field strength, mass and weight

Physics
1 answer:
Ierofanga [76]2 years ago
7 0
Gravitational potential energy = mass x gravitational field strength (g) x height
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Properties seen when one substance changes to another are know as ____ properties
maxonik [38]

Answer: chemical property

Explanation:

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3 years ago
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Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.
nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

  1. The main Ionic bond between the HCOO^- radical and Na^+.
  2. The  sigma covalent bonds between atom of H, atom of C and  both atoms of O.
  3. The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0
3 years ago
What factor affects a solutes solubility rather than its rate of solution
blagie [28]

Answer:

temperature

Explanation:

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3 years ago
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Two equally charged insulating balls each weigh 0.16 g and hang from a common point by identical threads 35 cm long. The balls r
Dafna11 [192]

Answer:

Q₁ = Q₂ = 8.84 x 10⁻⁹ C

Explanation:

given,

mass of ball, m = 0.16 g = 1.6 x 10⁻⁴ Kg

ball each other, r = 6.8 cm

Weight of the ball

F_w = m g

F_w = 1.6 x 10⁻⁴ x 9.8

F_w = 1.56 x 10⁻³ N

The tension in each string is a force directed along the length of the string and is the hypotenuse of a right triangle.

we have to find the horizontal component of the forces.

The length of the string,L is 35 cm so, it will be the hypotenuse.

θ be the angle made with imaginary vertical line and the string.

now,

sin \theta = \dfrac{r\2}{L}

sin \theta = \dfrac{3.4}{35}

   θ = 5.57°

horizontal component of the force = ?

vertical component of force,F_v = 1.56 x 10⁻³ N

tan\theta = \dfrac{F_H}{F_v}

tan(5.57^0) = \dfrac{F_H}{1.56\times 10^{-3}}

 F_h = 1.52 x 10⁻⁴ N

now, each ball will be repelled by

F = 1.52 x 10⁻⁴ N

now calculation of charges

F = \dfrac{kQ_1Q_2}{r^2}

Q₁ = Q₂ because both charge are same

1.52\times 10^{-4} = \dfrac{9\times10^9Q^2}{0.068^2}

    Q² = 7.809 x 10⁻¹⁷

   Q = 8.84 x 10⁻⁹ C

hence the change on the balls were Q₁ = Q₂ = 8.84 x 10⁻⁹ C

5 0
3 years ago
How do the charges compare when two objects are charged through friction?
spayn [35]

Answer:

The objects become oppositely charged and have equal amounts of charge.

Explanation:

There are three methods for charging objects:

- Conduction: a charged object is brought in contact with a neutral object. Electrons are transferred from the charged object to the neutral one, which also becomes charged

- Induction: a charged object is brought close (but not in contact) to a neutral object. The charges inside the neutral object redistribute, such that those of opposite sign to the charge in the charged object migrate on the side closer to the charged object, while the charges of same sign migrate towards the opposite side. If the neutral object is then grounded, the charges on the opposite side flow to the ground, leaving the neutral object charged as well

- Friction: two objects initially neutral are rubbed against each other. Electrons move from one object to the other one: therefore, one object becomes positively charged while the other one becomes negatively charged. Since the charge gained by one object is equal to charge lost by the other object, it follows that the two objects have same magnitude of charge, but with opposite sign.

3 0
3 years ago
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