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3 years ago

What does deep current mean

Physics

1 answer:

irakobra [83]3 years ago

4 0

Deep current is a big wave or a deep wave

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Where is the epicenter (it's about earthquakes)

sweet [91]

Answer:

The epicenter is the point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins.

<h2><u>PLEASE MARK ME BRAINLIEST, PLEASE</u></h2>

4 0

3 years ago

Read 2 more answers

Which of these objects must have the greatest force acting on it? pls help fast

8_murik_8 [283]

Let's look at Newton's second law

F=ma

Force is directly proportional towards mass

If mass is more force will be more.

Between baseball and bowling ball Bowling ball has higher mass

So it would expert most force

Option D

8 0

2 years ago

The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply

kykrilka [37]

Answer: Got It!

<em>Explanation: </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0 0

4 0

3 years ago

A hollow spherical planet is inhabited by people who live inside it, where the gravitational is zero. When a very massive space

Ulleksa [173]

Answer:

c) nonzero, directed toward the spaceship

Explanation:

As we know that net gravitational force due to spherical shell inside all its points will always be zero

So if planet is a spherical shell then inside the planet net gravitational force is zero on the people living in it

So when massive spaceship land on the surface of planet then the gravitational force of the spaceship is experienced by the people inside the shell

So here the gravitational force on the people is nonzero and it is towards the spaceship which landed on the surface of planet

5 0

3 years ago

It is interesting to speculate on the properties of a universe with different values for the fundamental constants.

qwelly [4]

Answer:

Part a)

$\lambda = 0.345 m$

Part b)

$\Delta x = 0.274 m$

Part c)

$r = 2.8 \times 10^{11} m$

Explanation:

Part a)

De broglie wavelength is given as

$\lambda = \frac{h}{mv}$

$\lambda = \frac{1}{(0.145)(20)}$

$\lambda = 0.345 m$

Part b)

By principle of uncertainty we know that

$\Delta x \times \Delta P = \frac{h}{4\pi}$

$\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}$

$\Delta x = 0.274 m$

Part c)

As we know that

$\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}$

also we know

$mvr = \frac{nh}{2\pi}$

$v = \frac{h}{2\pi mr}$

now we have

$\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}$

$r = \frac{h^2}{4\pi^2mke^2}$

$r = 2.8 \times 10^{11} m$

5 0

3 years ago