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3 years ago

Matter can be classified as elements, compounds, and mixtures. Choose all of the elements from the following examples of matter.

Physics

1 answer:

trapecia [35]3 years ago

7 0

Answer: gold ,oxygen magnesium

Explanation:

we cannot conclude air because it is mixture of gases , blood is also mixture of plasma , thrombocyte ; co2 is a compound ,

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Please help! 100 points + 50 + brainlist!!!

irakobra [83]

Explanation:

First find the displacement in the x direction:

dₓ = 449 cos 66° + 1112 cos 169° + 1571 cos 26°

dₓ = 182.6 − 1091.6 + 1412

dₓ = 503 km

Next, find the displacement in the y direction:

dᵧ = 449 sin 66° + 1112 sin 169° + 1571 sin 26°

dᵧ = 410.2 + 212.2 + 688.7

dᵧ = 1311 km

The magnitude is:

d² = dₓ² + dᵧ²

d² = (503)² + (1311)²

d = 1404 km

The angle is:

tan θ = dᵧ / dₓ

tan θ = 1311 / 503

tan θ = 2.61

θ = 69°

1404 km and 69° north of east from New Orleans is approximately Toledo.

4 0

4 years ago

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To discover the laws of reflection it is necessary to use a ...<br> 1 point

tamaranim1 [39]

Isn’t it a light box , mirror and / or angle measurer?

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4 years ago

Three point charges are on the x axis: −1 µC

Nina [5.8K]

Answer:

0.0078 N

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

$q_1 = -1 \mu C = -1 \cdot 10^{-6}C$ located at $x_1=-3 m$

$q_2=+9 \mu C = +9\cdot 10^{-6}C$ located at $x_2=0 m$

$q_3 = -5 \mu C = -5\cdot 10^{-6} C$ located at $x_3=+3 m$

The force between charge 1 and charge 2 is:

$F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

$F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net force on charge 1 is:

$F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N$

towards the right.

5 0

3 years ago

URGENT PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST !!!

kykrilka [37]

Answer:

C would be the answer

Explanation:

Hope this helps! :)

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3 years ago

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Which quantity or quantities is/are increasing for the object represented by line B?

torisob [31]

Answer:

C. Velocity and Position

Explanation:

7 0

3 years ago