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olchik [2.2K]
3 years ago
7

I need help answering these questions! Please!

Physics
1 answer:
miskamm [114]3 years ago
6 0

Answer:c

Explanation:

Also are u taking a benchmark or cfa

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How did foucaults pendulum and coriolis effect are used to demonstrate earths tilt?
HACTEHA [7]
They do not demonstrate Earth's tilt. In fact, they're not "used" to demonstrate anything. It works the other way:. When you observe the Coriolis effect and the behavior of the Foucault pendulum, and you try to explain why the behave the way they do, one possible simple explanation for both of them is the Earth's ROTATION. Then, when you also observe the rising and setting of the sun and moon, and you also notice how the NUMBERS all go together, the case for the rotating, spherical Earth gets stronger and stronger.
7 0
3 years ago
What is an example of an organism that reproduces
RideAnS [48]
Sea anemones; Coral; Starfish; Some non-flowering plants. Strawberry; Onion; Potato. Mushrooms are an example<span> of asexually </span>reproducing organisms<span>.

hoped i helped my fellow brainily user</span>
4 0
3 years ago
Halley's comet orbits the sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach
Licemer1 [7]

Answer:

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

3 0
3 years ago
An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of
Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, E=390\ keV=390\times 10^3\ eV

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

\lambda=\dfrac{h}{\sqrt{2meE}}

\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}

\lambda=1.96\times 10^{-12}\ m

(b) According to Bragg's law,

n\lambda=2d\ sin\theta

n = 1

For nickel, d=0.092\times 10^{-9}\ m

\theta=sin^{-1}(\dfrac{\lambda}{2d})

\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})

\theta=0.010^{\circ}

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0
3 years ago
A stationary boat in the ocean is experiencing waves from a storm. The waves move at 59 km/h and have a wavelength of 145 m . Th
krek1111 [17]

Answer:

The time elapses until the boat is first at the trough of a wave is 4.46 seconds.

Explanation:

Speed of the wave, v = 59 km/h = 16.38 m/s

Wavelength of the wave, \lambda=145\ m

If f is the frequency of the wave. The frequency of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{16.38\ m/s}{145\ m}\\\\f=0.112\ Hz

The time period of the wave is given by :

T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.112\ Hz}\\\\T=8.92\ s

We need to find the time elapses until the boat is first at the trough of a wave. So, the time will be half of the time period of the wave.

T=\dfrac{8.92}{2}\\\\T=4.46\ s

Hence, this is the required solution.

5 0
3 years ago
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