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3 years ago

10

Cross section of a solid circular rod is subject to a torque of t = 2.7 kn⋅m . if the diameter of the rod is d = 8 cm , what is

the maximum shear stress?

1 answer:

polet [3.4K]3 years ago

3 0

Shear Stress= Force along the tangent/ Area
= 2700/π*r^2
=2700/3.14*16
=54

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A 2.51 kg ball is attached to a ceiling by a 1.19 m long string. The height of the room is 3.45 m. The acceleration of gravity i

ANTONII [103]

Answer:

55,42 J

Explanation:

Since the height of the room is 3.45 m (distance between the floor and the ceiling) the difference between this value and the length of the rope 1.19 m; it will be equal to (3.45-1.19) =2.26 m. If we take as a reference point (Ep=0) the floor of the room, then the potential energy will be equal to Ep = M * g * h, replacing values in this equation (2.5 kg * 9.81 m/s2 * 2.26 m) will be 55,42 (N * m) or Jules.

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2 years ago

Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the

astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

$B = \frac{\mu_oI}{2\pi R}$

Where $\mu_o$ is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

$B_1 = \frac{\mu_oI_1}{2\pi R}$

$B_2 = \frac{\mu_oI_2}{2\pi R}$

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T = $\frac{\mu_oI_1}{2\pi R}$ - $\frac{\mu_oI_2}{2\pi R}$

or

4.0×10⁻⁶T = $\frac{\mu_o}{2\pi R}\times (I_1-I_2 )$

also

$\frac{I_1}{I_2} = \frac{3}{1}$

⇒ $I_1 = 3\times I_2$

substituting the values in the above equation we get

4.0×10⁻⁶T = $\frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)$

⇒ $I_2 = 1A$

also

$I_1 = 3\times I_2$

⇒ $I_1 = 3\times 1A$

⇒ $I_1 = 3A$

Hence, the larger of the two currents is 3A

3 0

3 years ago

The recoil of a shotgun can be significant. Suppose a 3.6-kg shotgun is held tightly by an arm and shoulder with a combined mass

devlian [24]

Answer:

The velocity of the recoil is $v=1.001 \frac{m}{s}$

Explanation:

Kinetic Energy

$m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\m_{gun}= 15.0kg+3.6kg$

The mass of th gun is the both mass the shotgun and the arm shoulder combination

$m_{bullet}=0.049kg\\v_{bullet}=380\frac{m}{s} \\m_{bullet}*v_{bullet}=m_{gun}*v_{recoil}\\0.049kg*380\frac{m}{s}=(15.kg+3.6kg)* v_{recoil}\\v_{recoil}=-\frac{18.62 kg \frac{m}{s} }{18.6 kg}\\ v_{recoil}=-1.0010 \frac{m}{s}$

The velocity is negative because is in opposite direction of the bullet

6 0

2 years ago

Read 2 more answers

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

Energy produced by nuclear fusion is ?

pentagon [3]

<h3><u>Answer;</u></h3>

<em>D. Produces no wastes</em>

<h3><u>Explanation;</u></h3>

<em><u>Nuclear fusion are nuclear reactions in which two or more atomic nuclei fuse or join to generate different atomic nuclei together with subatomic particles, such as protons or neutrons. </u></em>
Additionally,<em><u> nuclear fusion reactions yields lots energy.</u></em> The energy produced is usually more than the energy consumed by the reaction.
<em><u>Fusion power is a form of power generation in which energy generated using fusion reactions is used to produce heat that is used in generation of electricity.</u></em> Energy from nuclear fusion reactions is not coupled by production of wastes.

4 0

3 years ago

Read 2 more answers