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2 years ago

Costal residents must do many things to prepare for hurricanes

Physics

1 answer:

tangare [24]2 years ago

3 0

The answer is A it’s more safer that way

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 tesla?

Nonamiya [84]

Complete Question:

When specially prepared Hydrogen atoms with their electrons in the 6f state are placed into a strong uniform magnetic field, the degenerate energy levels split into several levels. This is the so called normal Zeeman effect.

Ignoring the electron spin what is the largest possible energy difference, if the magnetic field is 2.02 Tesla?

Answer:

ΔE = 1.224 * 10⁻²² J

Explanation:

In the 6f state, the orbital quantum number, L = 3

The magnetic quantum number, $m_{L} = -3, -2, -1, 0, 1, 2, 3$

The change in energy due to Zeeman effect is given by:

$\triangle E = m_{L} \mu_{B} B$

Magnetic field B = 2.02 T

Bohr magnetron, $\mu_{B} = 9.274 * 10^{-24} J/T$

$\triangle E = 6 * 9.274 * 10^{-24} * 2.2\\$

ΔE = 1.224 * 10⁻²² J

5 0

3 years ago

Maddie wanted to make a sugar-water solution to put in her hummingbird feeder. She found some sugar cubes and was about to drop

NNADVOKAT [17]

Answer:

this is a good suggestion

Explanation:

when the sugar cubes are crushed and they become a powder so its surface area increases. And as surface area is directly proportional to rate of reaction so the desired solution will be formed rapidly

3 0

3 years ago

Read 2 more answers

IP The x and y components of a vector r⃗ are rx = 16 m and ry = -8.5 m , respectively

fiasKO [112]

as it is given that

$r_x = 16 m$