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3 years ago

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Seismometers Choose one: A. measure vertical (up-and-down) ground motion but cannot measure horizontal (back-and-forth) motion.

B. are only sensitive enough to record ground movements down to about 1 mm. C. can "feel" (or record) large earthquakes that happen on the opposite side of the world. D. begin recording when the ground shakes and surface waves arrive.

1 answer:

solniwko [45]3 years ago

3 0

Answer: C) Can "feel" (or record) large earthquakes that happen on the opposite side of the world

Explanation:

If a seismogram has recorded P-waves and surface waves, but not S-waves, the seismograph may have been on the other side of the planet from the earthquake. Therefore seismometers can feel large earthquakes from the other side of the world

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A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o

kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

$= K _{i}$

$K_{i} = \frac{1}{2} mv ^{2}$

The work done by the frictional force is,

$Work \: done \: by \: frictional \: force =W_{f}$

$W _{f} = μ_{k}mgd$

Work done by frictional force= Initial kinetic energy of the sled

$W_{f} = K_{i}$

$μ_{k}mgd= \frac{1}{2} mv ^{2}$

So, the distance traveled by the sled before stopping is

$d= \frac{1mv ^{2} }{2 \:μ_{k}mg}$

$d= \frac{1v ^{2} }{2 \:μ_{k}g}$

$d= \frac{2^{2} }{2 \times \:0.100 \times 9.8}$

$d= 2.04 \: m$

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

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2 years ago

You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in

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$\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}$

<h3>Explanation</h3>

$I = v \cdot A \cdot n \cdot q$ ,

where

$I$ is the current;
$v$ is the drift speed;
$A$ is the cross-section area of the wire,
$n$ is the number of charge carrier per unit volume, and
$q$ is the charge on each charge carrier.

Area of a circular cross-section:

$A = \pi \cdot r^{2}$ ,

where

$r$ is the radius of the wire.

$n$ and $q$ are the same for all four samples, for they are made out of the same material.

As a result, $I$ of each wire is directly proportional to $v \cdot r^{2}$ where the value of $\pi \cdot n \cdot q$ is constant.

For each of the four wires:

$\begin{array}{ccc|c}\\& r & v &I \propto v\cdot r^{2}\\a) & 3 & 1 & 9\\b) & 4 & 0.5 & 8\\c) & 1 & 5 & 5\\d) & 2 & 2.5 & 10\\\end{array}$ .

How do the four wires rank by their current?

d > a > b > c.

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4 years ago

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

Sunny_sXe [5.5K]

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = $1000\times 8 = 8000 bits$

Distance, d = 2500 km = $2.5\times 10^{6}\ m$

Speed of propagation, s = $2.5\times 10^{8}\ m/s$

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

$t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s$

t = 10 ms

In general, propagation time, t is given by:

$t = \frac{link\ distance}{Propagation\ speed}$

No, this delay is independent of the length of the packet.
No, this delay is independent of the rate of transmission.

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4 years ago

How many significant figures are in 0.0069

posledela

Two significant figures, the 6 and the 9

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Which of the following statements best describe destructive forces? A. Forces that build up, create, landmasses. B. Forces that

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<span>An event that breaks objects into smaller objects or pieces is called destructive force
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4 years ago