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2 years ago

A bird flies 2km to the south, then 0.2km to the north.What is the displacement of the bird ?

Physics

2 answers:

Lunna [17]2 years ago

6 0

Explanation:

It is given that,

A bird flies 2 km to the south, then 0.2 km to the north. The displacement of the bird is equal to the shortest path covered by it.

Initially, the bird is moving in south direction and then it move to north. Displacement can be calculated using difference between initial position and final position.

d = 2.2 km - 0.2 km = 1.8 km

Hence, the displacement of the bird is 1.8 km

vivado [14]2 years ago

3 0

It would be D because it is flying back to where it started 2-.2=1.8

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A coin of mass m rests on a turntable a distance r from the axis of rotation. The turntable rotates with a frequency of f. What

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Answer:

$\mu = \frac{r (2\pi f)^{2}}{g}$

Explanation:

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Normal force in the upward direction balances the weight of the coin, hence

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Angular velocity of turntable is hence given as

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$\mu$ = minimum coefficient of static friction

static frictional force is given as

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The static frictional force provides the necessary centripetal force , hence

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Answer: A

Explanation:

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2 years ago

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(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

son4ous [18]

Answer:

12.5752053801 m/s

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No.

Explanation:

Q = Volume flow rate = $80\ L/s=80\times 10^{-3}\ m^3/s$

d = Diameter of pipe = 9 cm

A = Area = $\dfrac{\pi}{4}d^2$

Volume flow rate is given by

$Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s$

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

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$a=\dfrac{kq^2}{mr^2}$

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So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

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