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3 years ago

How long is a pendulum with a period of 1.0 S on the moon which has 1/6 of the earths gravity

1 answer:

7 0

For small deflections, T = 2*pi*sqrt(L / g) where T is period, L is length and g is gravity. Setting the equations to the same period, 2*pi*sqrt(3.85 / g) = 2*pi*(L / (1/6 * g)) The equation reduces to 3.85 m = 6 * L so L = 0.642 m chrsclrk · 7 years ago

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what is the mass of a pure platinum disk with a volume of 113 cm3? the density of platinum is 21.4 g/cm3

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A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging

Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

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3 years ago

A lawyer drives from her home, located 88 milesmiles east and 1818 milesmiles north of the town courthouse, to her office, lo

myrzilka [38]

Answer:

the distance between the lawyer's home and her office is 124 miles

Explanation:

given information:

first lets assume that

x-axis (west = positive, east = negative)

y-axis (north = positive, south = negative)

thus,

distance of the house = (-88,18)

distance of the office = (13, -54)

thus, the distance between the lawyer's home and her office

R = √(x₂ - x₁)² + (y₂ - y₁)²

= √(13 - (-88))² + (-54 -18)²

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3 years ago

A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

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

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

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3 years ago