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3 years ago

You have a 16-kg suitcase. what work did you do when you have slowly lifted it 0.80 m upward?

1 answer:

8 0

The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s² = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.

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Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi

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Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ $V=20\ \Omega,\ V_1=30\ \Omega$

⇒ $G=1680\ \Omega,\ G_1=2930\ \Omega$

Concept to be used:

Conversion of galvanometer into voltmeter.

Let $G$ be the resistance of the galvanometer and $I_g$ the maximum deflection in the galvanometer.

To measure maximum voltage resistance $R$ is connected in series .

So,

⇒ $V=I_g(R+G)$

We have to find the value of $R$ we know that in series circuit current are same.

For $G=1680$ For $G_1=2930$

⇒ $I_g=\frac{V}{R+G}$ equation (i) ⇒ $I_g=\frac{V_1}{R+G_1}$ equation (ii)

Equating both the above equations:

⇒ $\frac{V}{R+G} = \frac{V_1}{R+G_1}$

⇒ $V(R+ G_1) = V_1 (R+G)$

⇒ $VR+VG_1 = V_1R+V_1G$

⇒ $VR-V_1R = V_1G-VG_1$

⇒ $R(V-V_1) = V_1G-VG_1$

⇒ $R =\frac{V_1G-VG_1}{(V-V_1)}$

⇒ Plugging the values.

⇒ $R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}$

⇒ $R =\frac{(50400 - 58600)}{(-10)}$

⇒ $R=\frac{-8200}{-10}$

⇒ $R=820\ \Omega$

The coil resistance of the circuit is 820 Ω .

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3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

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3 years ago