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2 years ago

Write one general properties of non contact force

Physics

1 answer:

Gekata [30.6K]2 years ago

3 0

Answer: Gravitational force between the two masses does not depend on the medium separating two masses.

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A motorcyclist drives around a bend with a 20 m radius, with a constant velocity of 3 m/s. The motorcyclist and the motorcycle h

Pavel [41]

Answer:

$a=0.45\ m/s^2$

Explanation:

Given that,

The radius of a bend, r = 20 m

Velocity of motorcyclist, v = 3 m/s

The combined mass of motorcyclist and the motorcycle is 50 kg

We need to find the motorcyclist’s centripetal acceleration. The formula used to find the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(3)^2}{20}\\\\a=0.45\ m/s^2$

So, the acceleration of the motorcyclist is $0.45\ m/s^2$ .

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3 years ago

What force is required to accelerate a body with a mass of 15 kilograms at a rate of 8 m/s²?

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3 years ago

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kap26 [50]

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2 years ago

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Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

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6/10

g100num [7]

Answer:pounds

Explanation:

7 0

2 years ago

Write one general properties of non contact force​

Write one general properties of non contact force