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3 years ago

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A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1587 g. From this information, calculate the d

ensity of lead.

1 answer:

IrinaVladis [17]3 years ago

3 0

Density = mass / volume. Mass = 1587g. Volume = 4.5cm x 5.2cm x 6cm = 140.4 cubic-cm. Mass/volume = 1587/140.4 = 11.3 g/cm^3.

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An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

<u>For m = 7 [g] and V = 2.3 [cm³]</u>

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

<u>For m = 10 [g] and V = 7 [cm³]</u>

<u /> $Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$ <u />

<u>For m = 15 [g] and V = 5 [cm³]</u>

<u /> $Ro=15/5\\Ro=3[g/cm^{3} ]\\$ <u />

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<u /> $Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$ <u />

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3 years ago

7. The S.l. unit for force is<br> A. kg<br> C. m/s<br> B. m/s2<br> D. N

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D.N(Newton) this is the S.I unit for force

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3 years ago

Two objects of the same size are both perfect blackbodies. One has a temperature of 3000 K, so its frequency of maximum emission

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Answer:

a) The colder body (3000k), b) hearter body c) 12000K body

Explanation:

This exercise should know the power emitted by the objects and the distribution of this emission in the energy spectrum, for this we will use Stefan's laws and that of Wien's displacement

Stefan's Law P = σ A e T⁴

Wien displacement law λ T = 2,898 10⁻³ m K

Let's calculate the power emitted for each object.

As they are perfect black bodies e = 1, they also indicate that they have the same area

T = 3000K

P₁ = σ A T₁⁴

T = 12000K

P₂ = σ A T₂⁴

P₂ / P₁ = T₂⁴ / T₁⁴

P₂ / P₁ = (12000/3000)⁴

P₂ / P₁ = 256

This indicates that the hottest body emission is 256 times the coldest body emission.

Let's calculate the maximum emission wavelength

Body 1

T = 3000K

λ T = 2,898 10-3

λ₁ = 2.89810-3 / T

λ₁ = 2,898 10-3 / 3000

λ₁ = 0.966 10-6 m

λ₁ = 966 nm

T = 12000K

λ₂ = 2,898 10-3 / 12000

λ₂ = 0.2415 10-6 m

λ₂ = 214 nm

a) The colder body (3000k) emits more light in the infrared, since the emission of the hot body is at a minimum (emission tail)

b) The two bodies have emission in this region, the body of 3000K in the part of rise of the emission and the body to 12000K in the descent of the emission even when this body emits 256 times more than the other, so this body should have the highest broadcast in this area

c) The emission of the hottest 12000K body is mainly in UV

d) The hottest body emits more energy in UV and visible

e) No body has greater emission in all zones

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3 years ago

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Answer:

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where,

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θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

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3 years ago

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Answer:

The correct option is;

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Explanation:

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A loud sound has a high amplitude and a soft sound has a low amplitude, such that as the amplitude of the sound is increased, due to increased energy input, the sound becomes louder, and as the amplitude of the sound is decreased due to reduced energy input, the sound becomes softer.

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3 years ago