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Sever21 [200]
3 years ago
6

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

perfect alloy and saves a sample of it. Which mass and volume measurements correspond to this sample?
7 g and 2.3 cm³
10 g and 7 cm³
15 g and 5 cm³
21 g and 8 cm³
Physics
1 answer:
pochemuha3 years ago
5 0

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 />

<u>For m = 21 [g] and V = 8 [cm³]</u>

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

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10 kg of R-134a fill a 1.115-m^3 rigid container at an initial temperature of -30∘C. The container is then heated until the pres
Sidana [21]

Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

Explanation :

First we have to calculate the initial pressure of gas.

As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT

where,

P = pressure of gas = ?

V = volume of gas = 1.115m^3

T = temperature of gas = -30^oC=273+(-30)=243K

R = gas constant = 8.314m^3Pa/mole.K

w = mass of gas = 10 kg = 10000 g

M = molar mass of R-134a gas = 102.03 g/mole

Now put all the given values in the ideal gas equation, we get:

P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)

P=177587.9687Pa=177.6kPa

Thus, the initial pressure of gas is, 177.6 kPa

Now we have to calculate the final temperature of gas.

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure of gas = 177.6 kPa

P_2 = final pressure of gas = 200 kPa

T_1 = initial temperature of gas = -30^oC=273+(-30)=243K

T_2 = final temperature of gas = ?

Now put all the given values in the above equation, we get:

\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}

T_2=273.6K

Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa

4 0
3 years ago
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Answer:

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Explanation:

6 0
3 years ago
If a 4.5 kg object is dropped from a height of 6.0 m, what will be its velocity when it is halfway toward the ground? (Use g = 9
dangina [55]
When an object falls or is dropped from rest it's initial velocity is zero.
Using the equations for a motion in straight line. I can find the time it takes to reach 3.0 m down (half way).
x = vt - 4.9t²
-3 = 0 - 4.9t²
-3/-4.9 = t²
0.6122 = t²
0.7825 sec = t

v = v - gt
v = 0 - 9.8(0.7825)
v = -7.67 m/s
the negative denotes downward direction.

You  could also solve the problem using potential and kinetic energy.

Since it starts with maximum PE and gets converted to KE when it hits the ground. mgh = mv²/2
mass cancels, use 3 meters for the halfway distance
-9.8(-3) = v²/2
29.4 * 2 = v²
√(58.8) = 7.67 m/s downwards
7 0
3 years ago
Read 2 more answers
A 0.280 m radius, 475 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
KonstantinChe [14]
Hope it helps u mate. Plz follow me .

8 0
3 years ago
Please help<br> Right answers please<br> Will mark brainliest
balandron [24]

Answer:

a. 45 N. / b. 0.08 m/s^2. / c. 102 N

F = ma

F = 15(3)

F = 45 newtons

F/m = a

20/250 = a

0.08 m/s^2 = a

R = ma

R =1.5(68)

102 N

3 0
3 years ago
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