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Alinara [238K]
3 years ago
13

If the mass of a material is 45 grams and the volume of the material is 11 cm^3, what would the density of the material be?

Physics
2 answers:
Jobisdone [24]3 years ago
5 0
D=m^v
d=45g^11cm^3
D=4.1g/cm^3
Stella [2.4K]3 years ago
3 0

The density of the material would be 4.1 g/cm³.

Density is calculated by dividing the mass by the volume.

D=m÷v

D=45 g÷11 cm³

D=4.1 g/cm³

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Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
3 years ago
What describes why voters xhoose political canidates?
Alexus [3.1K]
The voters political opinions and what they think what is right and wrong.
4 0
3 years ago
An ordinary flashlight battery has a potential difference of 1.2 V between its positive and negative terminals. How much work mu
Maru [420]

The work done to transport an electron from the positive to the negative terminal is 1.92×10⁻¹⁹ J.

Given:

Potential difference, V = 1.2 V

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

Calculation:

We know that the work done to transport an electron from the positive to the negative terminal is given as:

W.D = (Charge on electron)×(Potential difference)

       = e × V

       = (1.6 × 10⁻¹⁹ C)×(1.2 V)

       = 1.92 × 10⁻¹⁹ J

Therefore, the work done in bringing the charge from the positive terminal to the negative terminal is 1.92 × 10⁻¹⁹ J.

Learn more about work done on a charge here:

<u>brainly.com/question/13946889</u>

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4 0
2 years ago
A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the followi
nikitadnepr [17]

Answer:

a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged

Explanation:

When a ray of light passes from one transparent material to another, it must comply with the law of refraction

     n1 sin θ₁ = n2 sin θ₂

Where index v1 is for the incident ray and index 2 for the refracted ray

With this expression let's examine the questions

a) They indicate that the refractive index increases,

      sin θ₂ = n₁ / n₂ sin θ₁

     θ₂ = sin⁻¹ (n₁ /n₂   sin θ₁)

    As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases

Answer is decrease

b) The wave velocity eta related to the wavelength and frequency

      v = λ f

The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.

The speed decreases with the indicated refraction increases and therefore the wavelength decreases

      λ = λ₀ / n

The answer is decrease

c) from the previous analysis the frequency remains unchanged

d) the refractive index is defined by

       n = c / v

So if n increases, the speed must decrease

The answer is decrease

e) the energy of the photon is given by the Planck equation

      E = h f

Since the frequency does not change, the energy does not change either

Answer remains unchanged

7 0
3 years ago
The tendency for an object to remain at rest in continue in motion is called:
nekit [7.7K]

Answer:

A Inertia

Explanation:

7 0
3 years ago
Read 2 more answers
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