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3 years ago

A table with mass 23.5 kg sits on the ground what is the normal force on the table?

Physics

2 answers:

marin [14]3 years ago

6 0

Answer:

Explanation:

There are two forces on the table: weight and normal force. Newton's second law:

∑F = ma

N - mg = 0

N = mg

N = (23.5 kg) (9.81 m/s²)

N = 230 N

gladu [14]3 years ago

3 0

Answer:

Like the other person said, the answer is 230 N

Explanation:

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Which do you think would stronger the gravitational interaction between an apple and earth or the gravitational interaction betw

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apple, earth. earth more massive than moon. astronauts showed this - armstrong etc

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4 years ago

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Mila [183]

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9. Epithelial tissue

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3 years ago

Arrange the phases of the Moon in order of increasing rising time, from the phase with the earliest rising time at 12:00 a.m. to

ZanzabumX [31]

Answer:

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM : Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

Explanation:

The Moon is the only celestial object which shows visible changes in its shape and rise and set time over a very short period of time i.e. just one day. One can observe it by observing the Moon daily. One will notice the change easily. This happens because of the geometry of the Sun, Earth and Moon. The Moon doesn't have its own light and shines because of the light of Sun.

At any given time half of the Moon would be illuminated by the Sun but how much of this illuminated portion is facing the Earth decides the phase of the Moon visible from the Earth. Due to this the Moon shows us various phases namely: New, Waxing Crescent, Waxing Half, Waxing Gibbous, Full, Waning Gibbous, Waning Half, Waning Crescent.

Also, the Moon revolves around the Earth completing the orbit in 29.5 Days. Everyday the Moon will change its position in the orbit. Due to this the rising time of Moon shifts by approximately 52 minutes daily. So, the New Moon rises with the Sun and Full Moon rises just after the sunset.

Lets assume the Sun rise time to be 06:00 AM. The rise time of different phases of Moon will be as follows:

12:00 AM : Waning Half

01:00 AM - 05:00 AM : Waning Crescent

06:00 AM : New Moon

07:00 AM - 11:00 AM : Waxing Crescent

12:00 PM : Waxing Half

01:00 PM - 06:00 PM : Waxing Gibbous

06:00 PM : Full Moon

07:00 PM - 09:00 PM : Waning Gibbous

5 0

3 years ago

A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea

fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

$Xc = \frac{1}{2\pi fC}$

where,

$Xc\\$ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

$Xc$ =?

$Xc' = \frac{1}{2\pi f'C}$

$= \frac{1}{2\pi 2f C}$

$= \frac{1}{2} (\frac{1}{2\pi fC} )\\$

$Xc' = \frac{1}{2} Xc$

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

The capacitive reactance is doubled.
The capacitive reactance is traduced by a factor of 4.
The capacitive reactance remains constant.
The capacitive reactance is quadrupled.
The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

#SPJ4

3 0

2 years ago

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$