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Pepsi [2]
3 years ago
6

A tsunami, an ocean wave generated by an earthquake, propagates along the open ocean at 700 km/hr and has a wavelength of 750 km

. What is the frequency of the waves in such a tsunami?A. 6.8 HzB. 0.93 HzC. 0.00026 HzD. 1.1 HzE. 0.15 Hz
Physics
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

C) 0.00026 Hz

Explanation:

  • In any wave, there exists a fixed relationship between the speed v, the frequency f and the wavelength λ, as follows:

       v = \lambda * f (1)

  • Replacing by the givens in (1), and solving for f, we get:

       f = \frac{v}{\lambda} = \frac{700km/hr}{750 km} = 0.93 1/hr (2)

  • Converting the units to Hz (1/sec), we get:

       f = 0.93 \frac{1}{hr} *\frac{1 hr}{3600sec} = 2.6e-4 = 0.00026 Hz (2)

  • The answer C. is the right one.
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A proton enters a uniform magnetic field of strength 2 T at 300 m/s. The magnetic field is oriented perpendicular to the proton’
sattari [20]

Answer:

Magnetic force, F=9.6\times 10^{-17}\ N

Explanation:

It is given that,

Magnetic field, B = 2 T

Velocity of the proton, v = 300 m/s

Charge on the proton, q=1.6\times 10^{-19}\ C

The magnetic field is oriented perpendicular to the proton’s velocity. The magnetic force on the charged particle is given by :

F=qvB\ sin\theta

The magnetic field is oriented perpendicular to the proton’s velocity, \theta=90^{\circ}

F=1.6\times 10^{-19}\times 300\times 2

F=9.6\times 10^{-17}\ N

So, the magnitude of the force that the proton experiences while it moves through the magnetic field is 9.6\times 10^{-17}\ N. Hence, this is the required solution.

7 0
3 years ago
a group of scientists claim that a new type of fuel will help reduce greenhouse gas emissions. to evaluate this claim you can. A
Mariana [72]
A group of scientists claim that a new type of fuel will help reduce greenhouse gas emissions. to evaluate this claim you can a. calculate the price of producing the fuel. b. study the effect of a fuel spill on wetland ecosystems. c. determine the efficiency of the fuel in different types of engines. d. test the amount
5 0
3 years ago
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you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.
vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

                                 F = Fₐ - Fₓ

Substituting the given values in the above equation

                                  F = 485 - 232

                                    = 253 N

The acceleration of the crate is given by

                                   a = F/m

                                      = 253 / 51

                                      = 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

3 0
3 years ago
The _______ is the the distance between two crests or two troughs on a transverse wave. It is also the distance between compress
Natasha2012 [34]
The answer is wavelength :)
3 0
4 years ago
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An incline plane has Ф = 40.0° and μ k = 0.15. Starting from rest, how long will it take a 4.0 kg block to reach a speed of 12 m
lesya [120]
     The block on the action of two forces, the Force of Friction and the Tangential Weight. Using the Newton's Secound Law, we have:

P_{t}-Fat=ma \\ mgsen\O-mgucos\O=ma \\ a=g(sen\O-ucos\O)
 
     Using the Velocity Hourly Equation, we get:

V=V_{o}+a\Delta t \\ \Delta t= \frac{V}{a}
 
     Uniting the equations:

\Delta t=  \frac{V}{g(sen\O-ucos\O)}
 
     Entering the unknowns:

\Delta t= \frac{V}{g(sen\O-ucos\O)} \\ \Delta t= \frac{12}{10(sen40^o-0.15cos^o)}  \\ \Delta t= \frac{12}{10(0,64-0.15x0.77)}  \\ \Delta t= \frac{12}{5.27}  \\ \boxed {\Delta t=2.28s}

Obs: Approximate results

If you notice any mistake in my english, please let me know, because i am not native.

5 0
3 years ago
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