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2 years ago

Why are different colors used to represent the floor of the ocean?

1 answer:

5 0

The red, yellow, and green wavelengths of sunlight are absorbed by water molecules in the ocean. ... In coastal areas, runoff from rivers, resuspension of sand and silt from the bottom by tides, waves and storms and a number of other substances can change the color of the near-shore waters.

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I threw a plastic ball in the pool for my dog to fetch. The mass of the ball was 125g. What must the volume be to have a density

Stells [14]

Answer:

250 mL

Explanation:

Density equation:

$Density = \frac{mass}{volume}$

Variables:

D = density

m = mass

V = volume

Solve:

M = 125 g

D = 0.500 g/mL

V = ?

$0.500 g/mL = \frac{125 g}{V}$

V = 250 mL

3 0

2 years ago

A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having tha

levacccp [35]

Answer:

The temperature of hot reservoir is 669.24 K.

Explanation:

It is given that,

Efficiency of gasoline engine, $\eta=35.3\%=0.353$

Temperature of cold reservoir, $T_C=160^{\circ}C=433\ K$

We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :

$\eta=1-\dfrac{T_C}{T_H}$

$T_H=\dfrac{T_C}{1-\eta}$

$T_H=\dfrac{433}{1-0.353}$

$T_H=669.24\ K$

So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.

4 0

2 years ago

A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin

padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame S,

Spatial interval of the event, $\rm \Delta x=1000\ m-0\ m=1000\ m.$

Temporal interval of the event, $\rm \Delta t = 1\ \mu s=10^{-6}\ s.$

In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is $\rm \Delta t'=0\ s.$

The speed of the frame S' with respect to frame S = v.

According to the Lorentz transformation,

$\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.$

And positive v means the velocity of the second frame S' is along the positive x-axis direction, i.e., to the right direction relative to the original frame S.

7 0

2 years ago

A stone whirled at the end of the a rope 30cm long, makes 10 complete resolution in 2 seconds Find: A. The angular velocity in r

lord [1]

Answers:

A: Angular velocity $\omega=31.40 \frac{r a d}{s}$

B: Linear velocity $v=9.42 \frac{m}{s}$

C: Linear Distance $d=47.1 \mathrm{m}$

Given:

Radius of the rope r=30cm=0.3m

Angular distance $\Delta \theta$ =10 revolutions

Time taken t=2seconds

To find:

A: Angular velocity in radians

B: Linear speed

C: Distance covered in 5 seconds

Step by Step Explanations:

Solution:

A: Angular velocity in radians;

According to the formula, Angular velocity can be calculated as

Angular Velocity = angular distance/ time

$\omega=\Delta \theta / \Delta t$

Where $\omega$ =Angular velocity

$\Delta \theta$ =Angular distance=10 revolutions

Changing revolutions to radians multiply with $2 \pi$ , so that we get

$=10 \times 2 \pi$

$=10 \times 2(3.14)$

=62.80 rad/rev

$\Delta t$ =Change in time

Substitute the known values in the above equation we get

$\omega$ =62.80 / 2

$\omega=31.40 \frac{r a d}{s}$

B. Linear speed of the rope;

As per the formula

Linear speed = angular speed × radius

$v=\omega \times r$

Where $\omega$ =Angular velocity

v=Linear speed of the rope

r=Radius of the rope

Substitute the known values in the above equation we get

$v=31.40 \times 0.30$

$v=9.42 \frac{m}{s}$

C. Dsitance covered in 5 seconds;

Linear distance = linear speed × time

$d=v \times t$

Where d= Linear distance of the rope

v=Linear speed of the rope

t=Time taken

Substitute the known values in the above equation we get

$d=9.42 \times 5$

$d=47.1 \mathrm{m}$

Result:

Thus A: Angular velocity of the rope $\omega=31.40 \frac{r a d}{s}$

B Linear speed of the rope $v=9.42 \frac{m}{s}$

C: Distance covered in 5 seconds $d=47.1 \mathrm{m}$

6 0

3 years ago

A dolphin in seawater at a temperature of 25°C emits a sound wave directed toward the ocean floor 154 m below. How much time pas

Cerrena [4.2K]

Answer:

0.2s

Explanation:

SO for the dolphin to hear its echo, the sound wave must travel a distance twice as much as the displacement between the dolphin and the ocean floor. So d = 154 * 2 = 308 m

Since the speed of sound in ocean floor is v = 1533m/s we can find out the time by dividing the distance d by the speed of sound

t = d / v = 308 / 1533 = 0.2s

7 0

2 years ago