Can someone please help with this problem?

A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?

Irina-Kira [14]

F = 400 Pa x 2.5 m2
F = 1 kN

4 0

3 years ago

The gas tank is made from A-36 steel and has an inner diameter of 1.50 m. If the tank is designed to withstand a pressure of 5 M

Tanzania [10]

Answer:

a. t = 23mm, b. t = 20mm

Explanation:

Obtain the value of yield strength in tension for A- 36 steel from table ‘Average Mechanical Properties of typical Engineering Materials’, which is σ(y) = 250 MPa.

a.

Assume that the thin wall analysis is valid, Calculate the hoop stress

σ(1) = pd/2t, where p is the pressure in the tank, d is the internal diameter of the tank and t is the thickness.

Substitute 5MPa for p and 1.5m for d

σ(1) = 5 x 10⁶x 1.5/2t

σ(1) = (3.75 x 10⁶)/t

Calculate the longitudinal stress

σ(2) = pd/4t

σ(2) = (5 x 10⁶ x 1.5)/4t

Apply Maximum Shear stress theory which states that failure occurs when the maximum shear stress from a combination of principal stresses <u>equals or exceeds</u> the value obtained for the shear stress at yielding in the uni-axial tensile test. Hence

τ(abs.max) ≤ τ (allowed)

τ (abs.max) ≤ σy /2FS, where FS is the factor of safety

Substitute σ(1)/2 for τ (abs.max) as both the principal stresses have same sign.

σ(1)/2 ≤ σy/2FS

3.75 x 10⁶/2t = 250 x 10⁶/2 x 1.5

T = 0.0225m = 22.5mm = 23mm to the nearest millimeter

Hence the required minimum thickness using the maximum shear stress theory is t = 23mm

b.

Apply maximum distortion energy theorem

σ²(allowed) =σ²(1) – σ(1) x σ(2) + σ²(2)

σ²(y)(allowed)/FS = (3.75 x 10⁶/t)² – (3.75 x 10⁶/t) x (1.875 x 10⁶/t) + (1.875 x 106/t)²

250 x 10⁶/1.5 = (3.2476 x 10⁶)/t

t = 3.2476/166.67

t = 0.0195 m = 19.50 mm = 20mm to the nearest millimeter

Hence, the required minimum thickness using the maximum distortion energy theory is t = 20 mm

7 0

3 years ago

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the

Alexxx [7]

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is (using (1)):
$A=28.7 m$

(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$

5 0

3 years ago

5. A 5.5 x10-6 C charge is located 0.28 m from a -3.5 x 10-6 C charge.

ra1l [238]

(a) The magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

Explanation:

The force acting between two charged particles separated by a distance is termed as Coloumb's force or electrostatic force. It can be termed as electrostatic force of attraction if the the force acting between the charges are oppositely charged. And it can be termed as electrostatic force of repulsion if the charges are similar or like charges.

In the present case, there is a positive and negative charge, so electrostatic force of attraction will be acting between them. As per Coloumb's law, the electrostatic force of attraction is directly proportional to the product of charges and inversely proportional to the square of distance of separation.

$F = \frac{kQq}{d^{2} }$

Here, k is the constant of proportionality which is equal to 9 × $10^{9}$ and Q, q are the two charges, d is the distance of separation.

So here Q = 5.5 × $10^{-6} C$ and q = - 3.5 × $10^{-6} C$ and d = 0.28 m

Then, $F=-\frac{9*10^{9}*5.5*10^{-6} * 3.5 * 10^{-6} }{(0.28)^{2} } = 2209.82*10^{-3}$

So the magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.

(b) If the negative charge is doubled, then the force will also get doubled.

The new force will be F = -4.418 N.

3 0

4 years ago

Displacement is the change in velocity of an object.<br><br> True or false

Elden [556K]

Words less true are seldom if ever spoken.

7 0

3 years ago

Read 2 more answers