All categories

3 years ago

What is the Ring of Fire?

Physics

2 answers:

Artyom0805 [142]3 years ago

5 0

A chain of volcanoes around the Atlantic Ocean...I think

n200080 [17]3 years ago

4 0

Volcanoes around the Pacific basin

You might be interested in

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

If ball 4 has a mass of 2 kg and it is 5m high, what will be its gravitational potential energy? (g=10 N/kg) *

Nastasia [14]

Explanation:

Gravitational potential energy

= mgh

= (2kg)(10N/kg)(5m)

= 100J.

3 0

3 years ago

Read 2 more answers

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t

ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0

2 years ago

PLEASE HELP ME (stop putting links ) Two objects m1 and m2, each with a mass of 5 kg and 6 kg separated by a distance. A third o

BabaBlast [244]

Answer:

Explanation:

Newton's Gravitation Law

$\displaystyle \frac{GmM}{d^2}$

where G is a constant, M and M the masses e d the distance betwen masses.

$\displaystyle G\frac{5\cdot2}{x^2}=G\frac{6\cdot 2}{(2x+1)^2} \quad \sqrt{6}x=(2x+1)\sqrt {5} \quad x=\frac{\sqrt{5}}{\sqrt{6}-2\sqrt{5}}$

7 0

2 years ago

In which material does the light wave have the larger wavelength?

pochemuha

Answer: in the smaller/ thinner pieces.

8 0

2 years ago