All categories

4 years ago

A book weighing 12 n is placed on a table. how much support force does a table exert on the book?

Physics

1 answer:

Alenkasestr [34]4 years ago

7 0

It should be 12 N. the force of the book on the table should be the same as the force of the table on the book.

You might be interested in

Hi please please help me you’ll get points and I’ll give you brainliest

leonid [27]

Answer:

see below

Explanation:

Question 1)

57kg to ug

$126\:Kg(\frac{1*10^{9}ug }{1kg} )\\\\=126(10^{9} )\\\\=126,000,000,000 ug$ (there are 9 zeros)

Question 2)

126lbs to N

$126lbs(\frac{4.45N}{1lb} )\\\\=126(4.45N)\\\\=506.7N$

7 0

3 years ago

Astrology, that unlikely and vague pseudoscience, makes much of the position of the planets at the moment of oneâs birth. The on

Anton [14]

Answer:

Explanation:

Gravitational force between two objects having mass m₁ and m₂ at a distance R

F = G m₁ m₂ / R²

Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²

= 1.01 x 10⁻⁶ N

b )

Force between baby and Jupiter

F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²

= 1.31 x 10⁻⁶ N

c )

Ratio = 1.01 / 1.31

= .77

4 0

3 years ago

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

3 years ago

Cotton balls could be used to represent clouds because of their white color and appearance of texture. Which of these other thin

iris [78.8K]

I just did it the answer is..... B.cobwebs

8 0

3 years ago

Read 2 more answers

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

3 years ago