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4 years ago

8

CAN SOMEONE HELP ME????????

1 answer:

Ymorist [56]4 years ago

8 0

Rising of magma from the seafloor.

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A circular hole in an aluminum plate is 2.739 cm in diameter at 0.000°C. What is the change in its diameter when the temperature

prisoha [69]

Answer:

$L = 2.746 cm$

Explanation:

As we know that thermal expansion coefficient of aluminium is given as

$\alpha = 24 \times 10^{-6} per ^oC$

now we also know that after thermal expansion the final length is given as

$L = L_o(1 + \alpha \Delta T)$

here we know that

$L_o = 2.739 cm$

$\alpha = 24 \times 10^{-6}$

$\Delta T = 108 - 0= 108^oC$

now we will have

$L = 2.739(1 + 24 \times 10^{-6} (108))$

$L = 2.746 cm$

3 0

4 years ago

A cylinder containing an ideal gas has a volume of 2.6 m3 and a pressure of 1.5 × 105 Pa at a temperature of 300 K. The cylinder

frozen [14]

<h2>Answer: $13.5\times 10^{8}$ joules</h2>

Explanation:

From the first law of thermodynamics,

Δ $Q$ =Δ $U$ + $W$

Where $Q$ is the heat given to the gas,

$U$ is the internal energy of the gas,

$W$ is the workdone by the gas.

When pressure is constant,

$\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$

$V_{2}=\frac{2.6\times 900}{300}=7.8m^{3}$

When pressure is constant, $W=P$ Δ $V$

Where $P$ is pressure and $V$ is the volume of the gas.

Given $P=1.5\times 10^{5}Pa$

Δ $V=$ $7.8-2.6=5.2m^{3}$

So, $W=1.5\times 10^{5}\times 5.2=7.8\times 10^{5}J$

Given that Δ $U=6\times 10^{5}$

So,Δ $Q=$ $6\times 10^{5}+7.8\times 10^{5}=13.8\times 10^{5}J$

5 0

4 years ago

What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d

Butoxors [25]

Calculate the pressure due to sea water as density*depth.
That is,
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is 101.3 kPa
Total pressure is 94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx

3 0

4 years ago

What is the maximum acceleration of a platform that oscillates at amplitude 4.70 cm and frequency 6.50 Hz?

mylen [45]

Answer:

78.315 m/s²

Explanation:

Amplitude, A = 4.70 cm = 0.047 m

Frequency, f = 6.50 Hz

Angular frequency, ω = 2 π f = 2 x 3.14 x 6.50 = 40.82 rad/s

Maximum acceleration, a = ω² A

a = 40.82 x 40.82 x 0.047

a = 78.315 m/s²

8 0

4 years ago

You need to design an experiment to measure the free-fall acceleration. You have an inclined plane made of steel and several obj

Nataliya [291]

Attached a diagram of the scheme described in the problem ..

To solve the problem we need to know the relationship between kinetic and normal force, so

$F_k=\mu_k F_N (1)$

Where

$\mu_k =$ coefficient of kinetic

$F_N=$ Normal force

We perform the sum of forces as well,

$\sum F_x = Ma$ (2)

$a= Acceletarion$

For Normal Force in Y,

$F_N= Mgcos \theta$ (3)

The force in X,

$\sum F_x = Mgsin\theta - F_x$ (4)

Replacing in (4)

$Ma = Mgsin\theta - \mu_k F_N$

$Ma = Mgsin\theta - \mu_k F_N$

$Ma = Mgsin\theta - \mu_k Mg cos\theta$

$a= g(sin\theta-\mu_k cos\theta)$

In this way, it does not matter which object is chosen.

6 0

3 years ago