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Nikitich [7]
3 years ago
9

Earthquakes and volcanic eruptions are examples of rapid events that are caused by the movement of Earth's _______. Mountain bui

lding is a much more slow process, which is caused by the movement of Earth's _______. A. Ocean tides; crustal plates B. Ocean tides; winds C. Crustal plates; crustal plates D. Crustal plates; ocean tides
Physics
2 answers:
andrew11 [14]3 years ago
5 0

The correct answer is - C. Crustal plates, crustal plates.

The movement of the crustal plates of the Earth is the reason why the geologic processes are occurring on our planet. With their movement, the crustal plates manage to create lot of pressure, open up gaps for the magma from the mantle, as well as make adjustments deeper into the ground.

Because of the adjustments deep into the ground, or rather the crust, lot of force is released, manifested as strong and very quick vibrations, better known as earthquakes.

The gaps that are opening between the plates let the magma reach the surface, thus enabling the volcanic activities on the planet.

The pressure from the plates' collision makes the land lift up slowly, and over time a mountain range starts to form because of the continuous lifting up of the area.

nydimaria [60]3 years ago
4 0

the answer is c just had this one

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3 years ago
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A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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3 years ago
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

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{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

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Q1.
Vesnalui [34]

Answer:

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Explanation:

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Mass of a child, m = 18 kg

The vertical distance from the top to the bottom of the slide is 2.5 metres.

The Gravitational field strength = 10 N/kg

We need to find the decrease in gravitational potential energy of the child sliding from the  top to the bottom of the slide.

The formula for the gravitational potential energy is given by :

P = mgh

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P = 18 kg × 10 m/s² × 2.5 m

P = 450 J

Hence, the decrease in gravitational potential energy is 450 J.

4 0
3 years ago
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