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Serggg [28]
3 years ago
13

What is another name for a business name or manufacture's name

Physics
2 answers:
stich3 [128]3 years ago
8 0

Answer:

brand name, or company

Explanation:

a brand name is  the kind of product they are selling. a company is the group of peaple that wok together.

Paha777 [63]3 years ago
5 0

Answer: maker or producer

Explanations:

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Oil <span>must be mined from underground 

Oil is NOT a renewable resource
Oil does NOT release more toxins than coal
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4 years ago
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Earth that are large enough to be detected by radar
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A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

\rho = \frac{m}{V} \rightarrow m = V\rho

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

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3 years ago
What is the electric potential v due to the nucleus of hydrogen at a distance of 5.292×10−11 m ?
viktelen [127]

Answer: 27.21 V

Explanation:

The <u>electric potential</u> V_{E} due to a point charge is expressed as:

V_{E}=k\frac{q}{r}

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q=1.6(10)^{-19}C is the <u>electric charge of the hydrogen nucleus</u>, which is positive

r=5.292(10)^{-11}m is the <u>distance</u>

Rewritting the equation with the known values:

V_{E}=9(10)^{9}\frac{Nm^{2}}{C^{2}}\frac{1.6(10)^{-19}C}{5.292(10)^{-11}m}

Finally:

V_{E}=27.21 V

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