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3 years ago

This image below shows a ship using sonar to map the seafloor. pls help i need this :(

Physics

1 answer:

Umnica [9.8K]3 years ago

5 0

Answer:

d and i think a?

Explanation:

im so sorry if its wrong

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Why is random sampling and random assignment in experimental research important?

Hunter-Best [27]

So the result is not biased or affected in some way

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3 years ago

Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .

vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

$Ke=\frac{1}{2}Mass*Velocity^2$

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

$Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules$

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0

3 years ago

Read 2 more answers

What is the energy in joules of a mole of photons associated with visible light of wavelength 486 nm?

ivann1987 [24]

Answer:

$2.46\cdot 10^5 J$

Explanation:

The energy of a single photon is given by:

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength

For the photon in this problem,

$\lambda=486 nm=4.86\cdot 10^{-7}m$

So, its energy is

$E_1=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.86\cdot 10^{-7}m}=4.09\cdot 10^{-19} J$

One mole of photons contains a number of photons equal to Avogadro number:

$N_A = 6.022\cdot 10^{23}$

So, the total energy of one mole of photons is

$E=N_A E_1 = (6.022\cdot 10^{23})(4.09\cdot 10^{-19} J)=2.46\cdot 10^5 J$

7 0

3 years ago

Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

____________ is a medium-sized flat-topped hill with cliff-face sides that is taller than it is wide.

Luda [366]

Answer:

mesa

Explanation:

A mesa is a flat-topped mountain or hill. It is a wide, flat, elevated landform with steep sides. ... Spanish explorers of the American southwest, where many mesas are found, used the word because the tops of mesas look like the tops of tables.

5 0

2 years ago