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3 years ago

What is the answer be fast

Physics

1 answer:

svetlana [45]3 years ago

5 0

Answer:

Explanation:

Same numbers of protons but different number of neutron so i would go for A same atomic number different number of neutrons

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What is 20kg x 9.8m/s2

trasher [3.6K]

Use newton's second law: net force = m*a

in this case, m is 20kg and a is g. The net Force would be 196 Newtons

4 0

3 years ago

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

5 0

4 years ago

the velocity of sound on a particular day outside is 331 meter/sound. what is the frequency of a tone if it has a wavelength 0.6

nataly862011 [7]

The frequency (f) in hertz (1/s) of the sound waves is the quotient when its speed (S) is divided by the wavelength (n). This is mathematically expressed,

f = S / n ; f = (331 m/s) / (0.6 m) = 551.67 / s

Thus, the frequency of the wave is 551.67 Hz.

6 0

4 years ago

Calculate the diffraction limit of the human eye, assuming a wide-open pupil so that your eye acts like a lens with diameter 0.8

guapka [62]

Hi, thank you for posting your question here at Brainly.

This problem could be solved using this equation:

Diffraction limit = 1.22*wavelength/diameter

diameter = 0.8 cm = 0.008 m
wavelength = 500E-9 m

Diffraction limit = 1.22(500E-9)/0.008
Diffraction limit = 0.00007625

6 0

3 years ago

Brainliest if correct

Bad White [126]

Answer:

D: Increase the distance between the objects.

E: Decrease the mass of one of the objects.

6 0

2 years ago