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2 years ago

A bicyclist is initially traveling at 3 m/s. The bicyclist accelerates at 1 m/s2 for 5 seconds.

1 answer:

7 0

The change in velocity is 5m/s which added to the initial 3m/s makes the final velocity 8m/s

Distance = (3*5) + (1/2*1*5^2)= 15+12.5= 27.5m

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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

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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, $\rho = 1030 kg/m^{3}$

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

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

$v_{b} = 28.63 m/s$

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3 years ago

Q1. In a 100m track event, the time taken by four runners is 10 sec., 10.2 sec., 10.4 sec and 10.6 sec. Find the ratio of their

Anna35 [415]

Distance (s)=100m
time
T₁=10
T₂=10.2
T₃=10.4
T₄=10.6
by v= $\frac{s}{t}$
we get
V₁= $\frac{100}{10}$
V₁=10m/s

V₂= $\frac{100}{10.2}$
V₂=9.8m/s

V₃= $\frac{100}{10.4}$
V₃=9.61 m/s

V₄= $\frac{100}{10.6}$
V₄=9.43 m/s

V₁:V₂:V₃:V₄ = 10 : 9.8 : 9.61 : 9.43

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2 years ago