All categories

3 years ago

A book that weighs 19 Newtons sits on a table. With what force

Physics

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.

Explanation:

There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

You might be interested in

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago

Why do hurricanes tend to form in warmer climates

aniked [119]

Warm air rises causing an area of lower air pressure below.

4 0

3 years ago

Read 2 more answers

When people use plastic combs on their hair, the combs become negatively charged. Which statements about this situation are true

Vadim26 [7]

The answer would be:
The comb gains electrons.
The hair loses electrons.

In chemistry, the charge depends on the electrons and protons. The electron will give negative charge and proton will give the positive charge. Proton is located in the nucleus of the atoms so it won't easily move like electron which located in the orbit in the atoms perimeter. So, ignore the option with the proton.
If the combs become negatively charged, that means it gain some electron. Since something gain electron, that means another thing is losing an electron. That electron comes from the hair.

3 0

4 years ago

Read 2 more answers

When a sound source is moving away from an observer, the observer will hear a lower pitch. This is due to

iren2701 [21]

Lower Pitch Due to the change to a lower frequency

7 0

3 years ago

Read 2 more answers

How long does it take long a light from the sun to reach the earth if it travels a distance of 1.5 × 10^11m? (velocity of light

mojhsa [17]

Answer:

Therefore, light travelling at 3.0x10^8 meters per second takes 500 seconds (8 minutes, 20 seconds) to reach the Earth, which is 1.5x10^11 meters away from the sun

Explanation:

7 0

3 years ago