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Hoochie [10]
2 years ago
10

8 kg cat is running with a velocity of 4 m/s. What is the cat's momentum?

Physics
1 answer:
Svetllana [295]2 years ago
4 0

Answer:

Kinetic energy = (1/2) (mass) (speed)²

Kinetic energy = (1/2) (8 kg) (4 m/s)² =

 

                                 (4 kg) (16 m²/s²) =

                                  64  (kg-m/s²) (m) = 64 newton-m = 64 joules

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Communication with submerged submarines via radio waves is difficult because seawater is conductive and absorbs electromagnetic
REY [17]

Answer:

\lambda=4000\ km

Explanation:

It is given that,

Frequency for submarine communications, f = 76 Hz

We need to find the wavelength of those extremely low-frequency waves. the relation between the wavelength and the frequency is given by :

c=f\times \lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{76\ Hz}

\lambda=3947368.42\ m

\lambda=3947\ km

or

\lambda=4000\ km

So, the wavelength of those extremely low-frequency waves is 4000 km. Hence, this is the required solution.

4 0
3 years ago
The ocean thermal energy conversion project uses the surface water near tropical islands with a temperature of 20°C as the hot t
choli [55]

Answer:

\eta=0.0512\ or\ 5.12\%

Explanation:

Heat engine is device operating in a continuous cycle between two reservoirs such that it transfers heat from a high temperature reservoir to a low temperature reservoir giving some work output in synchronization to Kelvin-Plank statement of the second law of thermodynamics.

<u>Given that:</u>

  • temperature of higher temperature reservoir, T_H=20+273=293\ K
  • temperature of lower temperature reservoir, T_L=5+273=278\ K

<u>For maximum efficiency:</u>

\eta=1-\frac{T_L}{T_H}

were:

T_H\ \&\ T_L are the temperatures of higher temperature reservoir and lower temperature reservoir respectively.

\eta=1-\frac{278}{293}

\eta=0.0512\ or\ 5.12\%

3 0
3 years ago
You matter .<br>Until you Multiply yerself by the speed of Light Squared. <br>Then you Energy.<br>​
iragen [17]

Answer:

Hellow ❤️

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6 0
3 years ago
Read 2 more answers
What property of a wave determines the pitch? What property of a wave determines the volume? Describe your reasoning and include
anzhelika [568]

Answer:

Frequency determines the pitch we perceive. Amplitude determines the volume, and loudness perceived (see more detail below)

Explanation:

A most fundamental wave shape is the sine (or cosine). A sine wave has three parameters: (1) frequency, (2) phase, and (3) amplitude. Out of the three, two play a role in your question, as follows.

The frequency, namely the number of periods (repetitions) per second, is perceived by our ear of "the pitch." The higher the frequency, the higher the pitch. However, our ears perceive pitch as a logarithm of the physical frequency. Every doubling of the frequency causes us to hear the pitch to increase by one octave.

The volume is determined by the amplitude of the wave. The higher the amplitude (positive or negative) the higher the volume. Our ears perceive volume as "loudness" and, again, not linearly - some frequencies are perceived as louder than others, even though they have the same physical volume. Frequencies between about 200 and 2000 Hz are perceived as louder than the rest of the audible spectrum (between 20 and 20000 Hz).

Interestingly, the phase does not play all too prominent role in perception, at least as far as music and speech are concerned.

6 0
3 years ago
A rocket is launched from Earth (mass ME, radius RE) with velocity v° and reaches radial distance r=6RE with velocity v°/10. Exp
Aliun [14]

The velocity, expressed in terms of  ME, RE is given as V_0 = \sqrt{98.99R_E}.

The maximum height that the rocket could reach if launched vertically is H = (¹/₂v₀²)/g.

<h3>Maximum height of the rocket</h3>

The maximum height reached by the rocket can be modeled using conservation of energy as shown below;

P.Ei + K.Ei = K.Ef + P.Ef

M_EgR_E + \frac{1}{2} M_EV_0^2= \frac{1}{2} M_E(\frac{V_o}{10} )^2+ M_Eg(6R_E)\\\\\frac{1}{2} M_EV_0^2 - \frac{1}{2} M_E(\frac{V_o}{10} )^2 = M_Eg(6R_E) - M_EgR_E\\\\0.495M_EV_0^2= 5gM_ER_E\\\\0.495V_0^2= 5gR_E\\\\V_0 = \sqrt{\frac{5gR_E}{0.495} } \\\\V_0 = \sqrt{98.99R_E}

<h3>Maximum height when it is launched vertically</h3>

P.E = K.E

mgH = ¹/₂mv²

H = (¹/₂v₀²)/g

Learn more about conservation of energy here: brainly.com/question/166559

5 0
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