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Arisa [49]
3 years ago
8

An aluminium bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 30mm in diameter and i

s 100mm long. If the modulus of elasticity for the aluminium is 85GN/m2, calculate the total contraction on the bar due to a compressive load of 180kN.

Physics
1 answer:
larisa86 [58]3 years ago
8 0

Answer: 1.228mm

Explanation:

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What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.
Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0
3 years ago
What do you notice about where the independent and dependent variables are located on each hypothesis above?!
nata0808 [166]

Answer:

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4 0
3 years ago
Two very large parallel metal plates, separated by 0.20 m, are connected across a 12-V source of potential. An electron is relea
Semmy [17]

Answer:

{\rm K} = 2.4\times 10^{-19}~J

Explanation:

The electric field inside a parallel plate capacitor is

E = \frac{Q}{2\epsilon_0 A}

where A is the area of one of the plates, and Q is the charge on the capacitor.

The electric force on the electron is

F = qE = \frac{qQ}{2\epsilon_0 A}

where q is the charge of the electron.

By definition the capacitance of the capacitor is given by

C = \epsilon_0\frac{A}{d} = \frac{Q}{V}\\\frac{Q}{\epsilon_0 A} = \frac{V}{d} = \frac{12}{0.20} = 60

Plugging this identity into the force equation above gives

F = \frac{qQ}{2\epsilon_0 A} = \frac{q}{2}(\frac{Q}{\epsilon_0 A}) = \frac{q}{2}60 = 30q

The work done by this force is equal to change in kinetic energy.

W = Fx = (30q)(0.05) = 1.5q = K

The charge of the electron is 1.6 \times 10^{-19}

Therefore, the kinetic energy is 2.4\times 10^{-19}

8 0
4 years ago
WILL GIVE BRAINLIEST
lisabon 2012 [21]
B 200N; 200N/10kg=20m/s2
4 0
4 years ago
String linear mass density is defined as mass/unit length. Calculate the linear mass density in kg/m of a string with mass 0.3g
SIZIF [17.4K]

Answer:

Linear mass density,\lambda=2\times 10^{-4}\ kg/m

Explanation:

Given that,

Mass of the string, m = 0.3 g = 0.0003 kg

Length of the string, l = 1.5 m

The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :

\lambda=\dfrac{m}{l}

\lambda=\dfrac{0.0003\ kg}{1.5\ m}

\lambda=0.0002\ kg/m

or

\lambda=2\times 10^{-4}\ kg/m

So, the linear mass density of a string is 2\times 10^{-4}\ kg/m. Hence, this is the required solution.

5 0
4 years ago
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