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4 years ago

The Pacific Ocean has a surface area of about 166 241 700 km2 and an average depth of 3940 m. Estimate the

Physics

1 answer:

Tatiana [17]4 years ago

4 0

Answer:

Volume = 6.5 x 10¹⁸ cm³

Explanation:

The volume of an object is the space enclosed by three dimensional surfaces of the object. The volume of an object can be given in terms of area and height. The formula of the volume of an object in terms of area and height is given as follows:

Volume = Area x Height

for Pacific Ocean we have the following data:

Area = (166241700 km²)(10⁵ cm/1 km) = 1.66 x 10¹³ cm²

Height = Depth = (3940 m)(10² cm/1 m) = 3.94 x 10⁵ cm

Therefore, using these values, we get:

Volume = (1.66 x 10¹³ cm²)(3.94 x 10⁵ cm)

Volume = 6.5 x 10¹⁸ cm³

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Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away

Rasek [7]

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

The pitch does not depend on the distance of the object from the observer.

As per the given data

pitch = frequency

Frequency = $f_{0}$ $\frac{V +- V_{0}}{V +- V_{s}}$

$f^{'}$ = $f_{0}$ $\frac{V }{V - V_{s}}$

Hence, the pitch of the two sirens remains the same for the observer.

5 0

3 years ago

Read 2 more answers

A rock tied to the end of a string moves in a circle at a constant speed of 2.5 m/s and experiences an acceleration of 4.0 m/s.

artcher [175]

The radius of the circular motion at the given speed is 1.56 m.

The given parameters;

speed of the rock, v = 2.5 m/s
acceleration of the rock, a = 4 m/s²

The radius of the circular motion is calculated by using centripetal acceleration formula as follows;

$a_c = \frac{v^2}{r} \\\\r = \frac{v^2}{a_c} \\\\r = \frac{2.5^2}{4} \\\\r = 1.56 \ m$

Thus, the radius of the circular motion at the given speed is 1.56 m.

Learn more about centripetal acceleration here: brainly.com/question/79801

8 0

2 years ago

A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan

Nesterboy [21]

Answer:

Compression distance: $d \approx 0.102\,m$

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy ( $K$ ), in joules, at the expense of elastic potential energy ( $U$ ), in joules, and overcoming work losses due to friction ( $W_{l}$ ), in joules:

$K + W_{l} = U$ (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$ (2)

Where:

$m$ - Mass of the block, in kilograms.

$v$ - Final velocity of the block, in meters per second.

$\mu$ - KInetic coefficient of friction, no unit.

$g$ - Gravitational acceleration, in meters per square second.

$s$ - Width of the rough patch, in meters.

$k$ - Spring constant, in newtons per meter.

$d$ - Compression distance, in meters.

If we know that $m = 1.2\,kg$ , $v = 2.3\,\frac{m}{s}$ , $\mu = 0.44$ , $g = 9.807\,\frac{m}{s^{2}}$ , $s = 0.05\,m$ and $k = 730\,\frac{N}{m}$ , then the compression distance of the spring is:

$\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2} \cdot k \cdot d^{2}$

$m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}$

$d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }$

$d \approx 0.102\,m$

4 0

3 years ago

11) We are past the year 1990: A.) Observation B.) Inference

Sliva [168]

A an observation hehdhhdhdhdhdhhdhdhdhdhdb

7 0

3 years ago

Read 2 more answers

What determines whether or not work is being done?

FromTheMoon [43]

C is the correct answer but the best possible answer is that work is done when a force is imposed on an object and the object moves in the same direction as the force

Hope this helps!

3 0

3 years ago