Help me with these simple nuclear energy reactions:

A lens is designed to work in the visible, near-infrared, and near-ultraviolet. The best resolution of this lens from a diffract

Anestetic [448]

In the near ultraviolet

8 0

3 years ago

JOSEPH JOGS FROM END A TO OTHER END B OF A straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back

zzz [600]

(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

3 0

3 years ago

A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2

Leno4ka [110]

Answer:

A. T=15.54 °C

B. Q/A= 0.119 W/m2

Explanation:

To solve this problem we need to use the Fourier's law for thermal conduction:

$Q= kA\frac{dT}{dx}$

Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:

$\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C$

Then, to find the rate of heat flow per square meter, we have:

$\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}$

$T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness$

7 0

3 years ago

What happens to a visible light wave when you increase the frequency

Anastasy [175]

When frequency increases, the wavelength halves.

6 0

3 years ago

A 1.0-in.-diameter hole is drilled on the centerline of a long, flat steel bar that is 1 2 thick and 4 in. wide. The bar is subj

Dominik [7]

Answer:

The answers are

The average stress = 20000 lb/in²

The maximum tensile stress immediately adjacent to the hole

= 31076.92 lb/in²

Explanation:

To solve the question we have

Weight of tensile load = 30,000 lb

Width of steel bar = 4 in

Thickness of steel bar = 1/2 in

Average Stress = Force/Area

Size of hole drilled = 1.0 in diameter

Available width at cross section where the 1.00 in diameter hole is drilled =

(4 - 1) in = 3 inches

Cross sectional area at the point of reduced cross section due to the drilled hole = Width × Thickness (Since the item is a flat bar)

= 3 in × 1/2 in = 1.5 in²

Therefore Stress = (30000 lb)/(1.5 in²) = 20000 lb/in²

the maximum tensile stress immediately adjacent to the hole.

Bending stress = $\sigma_B= \frac{M_y}{I}$ where $I = \frac{(0.5^2 + 4^2)}{12}$

0.5*30000/I = 11076.92 lb/in²

Max stress = 31076.92 lb/in²

8 0

3 years ago

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