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3 years ago

PLS HELP. I WILL MARK U THE BRAINLIEST IF U ANSWER EVERYTHING STEP BY STEP AND CORRECTLY.

Physics

2 answers:

oee [108]3 years ago

5 0

The distance covered would be 15m and the displacement is 5m . Mark me brainliest bro

dsp733 years ago

3 0

distance is the length of his trip so 15( 3+4+8)

displacement is how far he is from his original point ( -5)( -4^2 + 3^2 = 5^2) he was displaced 5 meters from his start position

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The work done by a magnetic feild on a chraged particle moving in it is zero

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An object with a density of 0.85 g/cc is dropped into each of the two beakers shown below. Beaker 1 has a density of 0.5 g/cc. B

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Answer:

The object will sink in the liquid in beaker 1.

The object will float in the liquid in beaker 2

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The density of an object relative to the density of a fluid determines if the object floats or sink in a fluid. The density of a material is the measure of the amount of mass of that material packed into a unit volume of that material.

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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

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Answer:

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

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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.

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