Janet was pulling a box that weighed 20N across 5m. How much work did she exert?

A force Fof 40000 lbf is applied to rod AC the negative Y-direction. The rod is 1000 inches tall. A Force P of 25 lbf is applied

erastova [34]

Answer:

The answer is "effective stress at point B is 7382 ksi "

Explanation:

Calculating the value of Compressive Axial Stress:

$\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\$

Calculating Shear Transverse:

$\to \frac{4v}{ 3 A} = \frac{4 (75 \ lbf + 25 \ lbf)}{ \frac{3 ( lni)^2}{4}}$

$= \frac{4 (100 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\ = \frac{400 \ lbf)}{ \frac{3 ( lni)^2}{4}} \\\\= 0.17 \ ksi$

$= R \times 200 \ in - P \times 100 \ in = 12500 \ lbf \times\ in$

$\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\$

$= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi$

8 0

2 years ago

Suppose you have two meter sticks, one made of steel and one made of invar (an alloy of iron and nickel), which are the same len

Mekhanik [1.2K]

Answer:

The difference in length for steel is 2.46 x 10⁻⁴ m

The difference in length for invar is 1.845 x 10⁻⁵ m

Explanation:

Given;

original length of steel, L₁ = 1.00 m

original length of invar, L₁ = 1.00 m

coefficients of volume expansion for steel, $\gamma_{st.}$ = 3.6 × 10⁻⁵ /°C

coefficients of volume expansion for invar, $\gamma_{in.}$ = 2.7 × 10⁻⁶ /°C

temperature rise in both meter stick, θ = 20.5°C

Difference in length, can be calculated as:

L₂ = L₁ (1 + αθ)

L₂ = L₁ + L₁αθ

L₂ - L₁ = L₁αθ

ΔL = L₁αθ

Where;

ΔL is difference in length

α is linear expansivity = $\frac{\gamma}{3}$

Difference in length, for steel at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{3.6*10^{-5}}{3} = 1.2*10^{-5} /^oC$

ΔL = 1 x 1.2 x 10⁻⁵ x 20.5 = 2.46 x 10⁻⁴ m

Difference in length, for invar at 20.5°C:

ΔL = L₁αθ

Given;

L₁ = 1.00 m

θ = 20.5°C

$\alpha = \frac{\gamma}{3} = \frac{2.7*10^{-6}}{3} = 0.9*10^{-6}/^oC$

ΔL = 1 x 0.9 x 10⁻⁶ x 20.5 = 1.845 x 10⁻⁵ m

8 0

2 years ago

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ????0 . Find the minimu

Reptile [31]

Answer:

Explanation:

Threshold frequency = 4.17 x 10¹⁴ Hz .

minimum energy required = hν where h is plank's constant and ν is frequency .

E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴

= 27.52 x 10⁻²⁰ J .

wavelength of radiation falling = 245 x 10⁻⁹ m

Energy of this radiation = hc / λ

c is velocity of light and λ is wavelength of radiation .

= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹

= .08081 x 10⁻¹⁷ J

= 80.81 x 10⁻²⁰ J

kinetic energy of electrons ejected = energy of falling radiation - threshold energy

= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰

= 53.29 x 10⁻²⁰ J .

4 0

2 years ago

An emf source of 6.0V is connected to a purely resistive lamp and a current of 2.0 amperes flows. All the wires are resistance-f

IgorC [24]

Resistance = (voltage) / (current)

Resistance = (6.0 v) / (2.0 A)

Resistance = 3.0 ohms

7 0

2 years ago

A car starts from rest and after 7 seconds it is moving at 42 m/s. What is the car’s average acceleration? A. 0.17 m/s2 B. 1.67

aliya0001 [1]

Acceleration =velocity /time
=42/7
=6

5 0

3 years ago

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