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labwork [276]
3 years ago
9

A shark is swimming at 10 m/s west when it spots some floating bait caught in some seaweed straight ahead. After 10 s, and trave

ling at a velocity of 15 m/s, the shark reaches the bait. What was the shark's average acceleration? A. 0.5 m/s² B. 0.5 m/s² east C. 0.5 m/s² west D. 10 m/s²
Physics
2 answers:
Volgvan3 years ago
8 0
<h2>Answer:</h2>

<u>The average acceleration is </u><u>0.5 m/s² west</u>

<h2>Explanation:</h2>

As we know that

Acceleration =  

Putting the values

Acceleration =

Acceleration = 0.5

Since the shark was moving in west so the Acceleration will be in west too

Andreyy893 years ago
3 0

Answer:

C. 0.5 m/s² west

Explanation:

The shark average acceleration is given by the following equation:

a=\frac{v-u}{t}

where:

u = 10 m/s west is the initial velocity of the shark

v = 15 m/s west is the final velocity

t = 10 s is the time taken

Plugging numbers into the formula, we find

a=\frac{15 m/s-10 m/s}{10 s}=0.5 m/s^2

and the direction is west, because the shark speeds up in the same direction of the initial velocity, so the acceleration is positive and therefore in the same direction.

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nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

v = r\omega \rightarrow \omega = \frac{v}{r}

Here,

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\omega= Angular velocity

r = Radius

Our values are

v = 6/ms

r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m

Replacing to find the angular velocity we have,

\omega = \frac{6m/s}{0.06m}

\omega = 100rad/s

Convert the units to RPM we have that

\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})

\omega = 955.41rpm

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A glass tube 1 mm in diameter is dipped into glycerin. The density of the glycerin is 1260 kg/m3, surface tension is 6.3x10-2 N/
Sunny_sXe [5.5K]

Answer:

The capillary rise of the glycerin is most nearly  y  =  0.0204 \ m

Explanation:

From the question we are told that

  The diameter of the glass tube is  d =  1 \ mm =  0.001 \ m

   The density of glycerin is  \rho =  1260 \ kg /m^3

   The surface tension of the glycerin is \sigma   =  6.3 *10^{-2} \ N /m

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       y  =  \frac{4 * \sigma  *  cos (\theta )}{ \rho * g *  d}

substituting value  

       y  =  \frac{4 * 6.3 *10^{-2}  *  cos (0 )}{ 1260 * 9.8 *  0.001}

      y  =  0.0204 \ m

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