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3 years ago

7

If the moon disappeared, the Earth's tides would a)not happen b)not be as large c)be much larger d)all be extreme just answer pl

ease

2 answers:

murzikaleks [220]3 years ago

7 0

Your answer would be B because although the Moon is gone, the sun still has a gravitational pull towards the oceans waves

vichka [17]3 years ago

3 0

B not as large is the correct answer

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2 equal charges, 27 micro Coulomb each, are separated by 5 cm. Find force between those.

padilas [110]

Answer:

The force between charges is $F= 2.624*10^3N$ .

Explanation:

The Coulomb force $F$ between the two charges $q_1$ and $q_2$ separated by distance $d$ is given by the equation

$F = k\dfrac{q_1q_2}{d^2}$

where $k$ is the coulombs constant, and has the value

$k= 9*10^9N\cdot C\cdot m^2$ .

Now in our case

$q_1=q_2=27\mu C =27*10^{-6}C$

and

$d= 5cm =0.05m$ ,

therefore, the Coulomb force between the charges is

$F =( 9*10^9N\cdot C\cdot m^2)*\dfrac{(27*10^{-6}C)(27*10^{-6}C)}{(0.05m)^2}$

$\boxed{ F= 2.624*10^3N}$

3 0

4 years ago

. An iron cube (ρ = 7860 kg/m3) having sides of 1 cm is submerged in open fresh water (ρ = 1000kg/m3) with its top surface at a

Korolek [52]

Answer:

$\Delta P = 98.07\,Pa$

Explanation:

The difference between top and bottom surfaces is computed by the following hydrostatic equivalence:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.01\,m)$

$\Delta P = 98.07\,Pa$

5 0

3 years ago

Why pressure above the surface is greater than in the air

nexus9112 [7]

Answer:

At higher elevations, there are fewer air molecules above a given surface than a similar surface at lower levels. ... Since most of the atmosphere's molecules are held close to the earth's surface by the force of gravity, air pressure decreases rapidly at first, then more slowly at higher levels.

Explanation:

7 0

4 years ago

a hammer of mass 5 kg travelling at 4 metre per second is a nail directly and does not rebound what is the impulse of the hammer

dmitriy555 [2]

20 kg.m/s
p =m*v = 5*4 = 20

5 0

3 years ago

A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc

hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

$E V = log_2(\dfrac{N^2}{t})$

N is the diameter of the aperture and t is the time of exposure

now,

$log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})$

$\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}$

inserting all the values

$\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}$

N₂² = 4

N₂ = 2 mm

hence , the correct answer is option E

4 0

3 years ago