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3 years ago

The majority of earthquakes worldwide occur at all but one location. That is

1 answer:

5 0

Earthquakes occur in the crust or upper mantle, from the earth’s surface to about 400 miles below the surface. But the very deepest earthquakes only occur at subduction zones where cold crustal rock is being pushed deep into the earth. In California, earthquakes are almost all in the top 15 miles of the crust, except in northern California along the Cascadia Subduction Zone, which extends into Oregon, Washington, and British Columbia.(tectonic plate boundaries)

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago

look at the circuit in the figure. find the current, voltage, and power in each resistor. please list answer

sashaice [31]

Answer:

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6 0

2 years ago

A 7.0 kg bowling ball has a moment of inertia of 2.8x10-2 kg m2, and a radius of 0.10 m. If it rolls down the lane at an angular

slega [8]

Hi there!

Angular momentum is equivalent to:

$\large\boxed{L = I\omega}$

L = angular momentum (kgm²/s)

I = moment of inertia (kgm²)

ω = angular velocity (rad/sec)

Plug in the given values for moment of inertia and angular speed:

$L = (0.028)(40) = \boxed{1.12 kgm^2/s}$

8 0

2 years ago

Two air track carts move along an air track towards each other. Cart A has a mass of 450 g and moves toward the right with a spe

ra1l [238]

Answer:

0.465 kgm/s

Explanation:

Given that

Mass of the cart A, m1 = 450 g

Speed of the cart A, v1 = 0.85 m/s

Mass of the cart B, m2 = 300 g

Speed of the cart B, v2 = 1.12 m/s

Now, using the law of conservation of momentum.

It is worthy of note that our cart B is moving in opposite directions to A

m1v1 + m2v2 =

(450 * 0.85) - (300 * 1.12) =

382.5 - 336 =

46.5 gm/s

If we convert to kg, we have

46.5 / 100 = 0.465 kgm/s

Thus, the total momentum of the system is 0.465 kgm/s

6 0

3 years ago