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3 years ago

How do scientist know that seismic waves can be either compressional or transverse

1 answer:

8 0

Hey there!

<span>Scientists know that seismic waves can be either compressional or transverse because there are horizontal and vertical movements during an earthquake. Horizontal movements are compressional while vertical movements are transverse.

Thank you!</span>

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If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

A certain runner averages 4.1 m/s over a 10

Ghella [55]

Answer: 40.650406504065 or 40 minutes and 39 seconds.

Explanation:

1 k = 1000m

race = 10000m

runner time = 10000 / 4.1

runner time = 2439.0243902439024 seconds

runner time = 2439.0243902439024/60 = 40.650406504065 or 40 minutes and 39 seconds.

6 0

2 years ago

Which of the following substance are not formed by chemical bonds

Sliva [168]

<span>Which of the following substance are not formed by chemical bonds? </span>A MIXTURE

3 0

2 years ago

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

3 0

2 years ago

d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?

Anna71 [15]

<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules

5 0

3 years ago

Read 2 more answers