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3 years ago

1. is the full moon a spring tide or neap tide

2 answers:

7 0

The greatest tidal extremes occur when the Sun, Moon, and Earth
are all lined up along the same straight line. Those occasions are
at the time of the Full Moon ... when the Earth is the middle one ...
and the time of the New Moon ... when the Moon is the middle one.
Those are the occasions of the highest high tides and the lowest
low tides, known collectively as "Spring tides".

The smallest tidal extremes occur when the Sun, Moon, and Earth
form a right angle, with the Earth at the vertex. Those occasions are
at the time of the First Quarter and Third Quarter Moon. Those are
the occasions of the highest low tides and the lowest high tides,
known collectively as "Neap tides".

#1 & #3: Spring tides
#2: False

kkurt [141]3 years ago

5 0

Question 1: the full moon is a spring tide

Question 2 is true. Neap Tides do have higher and lower tides than normal.

Question 3: (the same as Question 1)-- The new moon is a spring tide

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Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

Fudgin [204]

Answer:

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$

Explanation:

We know that the gravity on the surface of the moon is,

$g'=\frac{g}{6}$

$g'=1.63\ m.s^{-2}$

<u>Gravity at a height h above the surface of the moon will be given as:</u>

$g'_h=\frac{G.m}{(r+h)^2}$ ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

$G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}$

$m=7.35\times 10^{22}\ kg$

$r=1.74\times 10^6\ m$

$h=384.4\times 10^6\ m$ is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

$g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}$

$g'_h=1.096\times 10^{-5}\ m.s^{-2}$ is the gravity on the moon the earth-surface.

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3 years ago

The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some elec

PilotLPTM [1.2K]

Answer:

The magnetic force points in the positive z-direction, which corresponds to the upward direction.

Option 2 is correct, the force points in the upwards direction.

Explanation:

The magnetic force on any charge is given as the cross product of qv and B

F = qv × B

where q = charge on the ball thrown = +q (Since it is positively charged)

v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)

B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)

F = qv × B = (+qvî) × (Bj)

F =

| î j k |

| qv 0 0|

| 0 B 0

F = i(0 - 0) - j(0 - 0) + k(qvB - 0)

F = (qvB)k N

The force is in the z-direction.

We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.

Hope this Helps!!!

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3 years ago

A mars prototype carrying scientific instruments has a mass of 1060kg. a net force of 52000 n is applied to this roverat a test

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F = 52000 N

m = 1060 kg

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3 years ago

In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat

olya-2409 [2.1K]

Answer:

Friction force on the bullet is 58.7 N opposite to its velocity

Explanation:

As we know that initial speed of the bullet is 55 m/s

after travelling into the sand bag by distance d = 1.34 m it comes to rest

so final speed

$v_f = 0$