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Ugo [173]
3 years ago
8

Displacement can be zero but distance can't. explain this statement with example.​

Physics
1 answer:
harkovskaia [24]3 years ago
5 0

Answer:

1}can displacement be zero even when distance is not zero? Give example.  

2}how can you get speed of an object from its distance time graph ?

3}how can you get distance of an object from its speed time graph ?

4}can a body has zero velocity and still acceleration? give example .

5}is it possible in straight line motion of particle having zero speed and a non zero velocity? explain .

Explanation:

Hi friend,1).yes,displacement can be zero even when distance is not zeroexample :- when a body travels in circular path , after covering a circle the distance cannot be zero,but its displacement is zero..2).by finding the slope of distance time graph,we can find the speed of the body.3).by finding the integrating of speed time graph ,we can find the distance.4).yes,a body can have zero velocity and still acceleration.example- in vertically upward projected body,at its highest point,the velocity becomes zero but still acceleration due to gravity acts on it 5).no its not possible ..when the speed is zero..its velocity will be zero

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A cat runs at 2 m/s for 3 s and then slows to a stop with an acceleration of 0.80 m/s2. What is
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see that the correct one is B

Explanation:

To solve this exercise let us use the kinematic relations

           v² = v₀² - 2 a x

as they indicate that the car stops, therefore the final speed is yield v = 0

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What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
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Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

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Diameter = 3.00 cm

Exit diameter = 9.00 cm

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We need to calculate the pressure

Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)

\Delta P=1.58\times10^{6}\ N/m^2

(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

8 0
3 years ago
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