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3 years ago

Which of the following is the correct name for CCl4

Physics

2 answers:

Ira Lisetskai [31]3 years ago

7 0

Answer: Carbon tetrachloride Or Tetrachloromethane

Explanation: Carbon tetrachloride is an important nonpolar covalent compound. You determine its name based on the atoms present in the compound.

kogti [31]3 years ago

3 0

Answer:

Tetrachloromethane

Explanation:

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A car that is standing still accelerates up a hill with a slope of 6.4% at an average acceleration of 2.93 ft/s^2. The hill is 1

alexgriva [62]

Answer:

213 s

Explanation:

Slope is the ratio of change in vertical distance to change in horizontal distance.

Slope = vertical height / horizontal height

Therefore:

6.4% = vertical height / 12.42

vertical height = 6.4% * 12.42

vertical height = 0.8 miles

The distance travelled by the car (s) is:

s² = 0.8² + 12.42²

s² = 154.9

s = 12.45 miles

Acceleration (a) = 2.93 ft/s^2 = 0.00055 mile/s²

initial velocity (u) = 0, final velocity = 203 mph

Using:

s = ut + 0.5at²

12.45 = 0.5(0.00055)t²

t =213 s

5 0

3 years ago

A yellow ball with a mass of 2 kg is rolling across the floor at 3 m's A red ball with a mass of 3 kg is rolling across the

Wewaii [24]

Answer:
The red ball has a greater kinetic energy, because it has a greater mass

Explanation:
Mass is directly proportional to kinetic energy, as one increases, so does the other

6 0

3 years ago

Read 2 more answers

Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a

butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0

3 years ago

What would you expect to cause a drop in air pressure?

Murrr4er [49]

Answer:

Explanation:

6 0

2 years ago

Read 2 more answers

Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

$KE = \frac{1}{2}mv^2$