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vfiekz [6]
3 years ago
15

What is the velocity of a bicycle in meters per second if it travels 1 kilometer west in 4.1 minutes

Physics
1 answer:
valentinak56 [21]3 years ago
6 0
1 kilometre is equal to 1000m
and 4.1 minutes is equal to 246 seconds
thus 1000/246 = 4.065 m/s
and the direction is towards the west
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While studying bacterial cells, scientists measure the lengths of the cells in one colony. The chart shows their data. The scien
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Answer:

Following are the solution to the given question:

Explanation:

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2 years ago
A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th
ankoles [38]
<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 3 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 3 + 0.5 x 9.81 x 3²

                      s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0
3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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