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Vikentia [17]
3 years ago
8

O2 is a molecule.

Physics
1 answer:
cestrela7 [59]3 years ago
5 0
1. true is the answer
2. The correct option is this: FIGURE II AND FIGURE IV ARE BOTH COMPOUNDS.Figure two and figure four have components that are of different colors, this means that different atoms had combined together to form a compound.3. dont know yet
4. The figure is a molecule and an element is the correct answer
5. Elements
6. Co2
7. Both are made of two or more atoms.
8. the answer is atom
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Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

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3 years ago
If you double the pressure on the surface of a can of water, the buoyant force on a stone placed in that water will
Bingel [31]
The buoyant force won't change.
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3 years ago
What shape is the Milky way galaxy
gizmo_the_mogwai [7]
It appears to be a <span>spiral shape. </span>
5 0
3 years ago
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Use examples to explain how the geosphere interacts with two other of Earth's spheres. Explain the interaction for each using co
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The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.

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3 years ago
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A rectangular barge, 5.2 m long and 2.4 m wide, floats in fresh water. Suppose that a 410-kg crate of auto parts is loaded onto
sesenic [268]
<h2>Answer:</h2><h2>The depth of barge float=3 cm</h2><h2>Explanation:</h2>

Length of rectangular barge=5.2 m

Width of rectangular barge=2.4m

Mass of crate=410 kg

Let h be the height of barge float

Volume of barge float=l\times b\times h=5.2\times 2.4\times h=12.48h

Density of water=10^3kg/m^3

Weight of water displaced by barge=Buoyant force=-Weight of horse

Volume\;of\;water\times density\;of\;water\times g=410\times g

12.48h\times 1000=410

h=\frac{410}{12.48\times 1000}=0.03 m

1 m=100 cm

0.03 m=0.03\times 100=3cm

Hence, the depth of barge float=3 cm

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3 years ago
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