What would happen if you changed the position of the screen, but kept the other factors the same?

Explain the relationship between the earth's crust and the earth's ocean sizes.

valina [46]

the outermost layer of Earth’s lithosphere that is found under the oceans and molded at scattering centres ono ceanic ridges, which occur at deviating plate boundaries

Oceanic crust is about 6 km (4 miles) thick.

hope it helps

3 0

3 years ago

Consider a boat heading due east at 15 miles/hour. The water's current is moving at 7.1 miles/hour at 45º south of east. Drag ve

givi [52]

If a boat is going East at 15mph and there is a water current going southeast at 45° then the boat is being drifted southward. So since the current is going at an angle then it has a x and y component. So Rx refers to the x-component force of the current and Ry refers to the y-component of the current, and |R| refers to the magnitude of these forces.

7 0

3 years ago

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A heavy bank-vault door is opened by the application of a force of 300 N directed perpendicular to the plane of the door at a di

kondaur [170]

The answer is 240 Nm.

5 0

3 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

8 0

3 years ago

An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq

wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / $\sqrt{(R)^{2} + (Xl -Xc)^{2} }$

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0

3 years ago