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3 years ago

The sun has more gravity than planets in our solar system because the sun is

Physics

1 answer:

Makovka662 [10]3 years ago

5 0

Answer:

the sun is larger in mass and therefore gravitational pull.

Explanation:

Fun fact the moon has more gravitational force on the earth than the sun because its closer to the earth.

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An atom is the smallest particle of matter

zheka24 [161]

Answer:

yes

Explanation:

it makes up everything but you cant see it with out a microscope

3 0

3 years ago

(b) The output current of the generator is 2.0A. The output potential difference (p.d.) of the

Luden [163]

Answer:

P = 22 watts

Explanation:

Given that,

The output power of the generator = 2 a

The output potential difference = 11 V

We need to find the output power of the generator. The formula for the output power is given by :

$P=V\times I\\\\P=11\times 2\\\\P=22\ W$

So, the output power of the generator is equal to 22 Watts.

8 0

3 years ago

In a horizontal pipe with a gradually decreasing cross-section (in the direction of flow) there is a clear fluid of unknown dens

qwelly [4]

Answer:

the density of this clear fluid is 1140.35 kg/m^3

Explanation:

Given that :

$P_p = P__{Q}} + 260 \ Pa \\ \\ v_p =0.37 \ m/s \\ \\ v_Q = 0.77 \ m/s$

According to Bernoulli's Equation.

$P_p + \frac{1}{2} pv^2_p+pgh_p= P_Q+ \frac{1}{2} pv^2_Q + pgh_Q$

∴ $260 = \frac{1}{2}p (v^2_Q-v^2_p)$ since ( $h_p = h_Q$ )

$\rho = \frac{2(260)}{v^2_Q-v^2_p}$

$\rho = \frac{2(260)}{0.77^2-0.37^2}$

$\rho = \frac{520}{0.5929-0.1369}$

$\rho =$ $1140.35 \ kg/m^3$

5 0

4 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

i will give the brainliest answer to whoever answers this, explain how Sir William Gilbert used models in his investigation of m

levacccp [35]

Explanation:

Lodestone am iron were the only known magnetic materials in Gilbert's day, and his task was to investigate magnetism. Gilbert was so sure that the earth was a giant lodestone and used the earth as a primary reference, defining the north(magnetic) pole of a needle, or a a nail floating on a piece of cork, to be that which turns towards the Earth's north geographic pole. he wanted to prove this with a model Terella, using short pieces of iron.

7 0

3 years ago