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Leni [432]
3 years ago
6

A swimmer is capable of swimming at 1.4m/s in still water. a. How far downstream will he land if he swims directly across a 180m

wide river? b. How long will it take him to reach the other side.
Physics
2 answers:
ozzi3 years ago
8 0

Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

vredina [299]3 years ago
5 0

Answer:

Explanation:

From the given information:

a) the distance(D) showing how dar downstream he will land can be computed as follows:

Assuming the current of the river = 0.2 m/s

D = \dfrac{180 \ m \times 0.2 \ m/s}{1.4 \ m/s}

D = 36 m ÷ 1.4

D = 25.71 m

The required time (t) to reach the other side is:

time (t) = 180 m/ 1.4 m/s

time (t) = 128.57 seconds

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At an instant a traffic light turns green an automobile that has been waiting at an intersection of the road accelerates with 5m
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1) The car overtakes the truck at a distance of 160 m far from the intersection.

2) The velocity of the car is 40 m/s

Explanation:

1)

The car is travelling with a constant acceleration starting from rest, so its position at time t (measured taking the intersection as the origin) is given by

x_c(t) = \frac{1}{2}at^2

where

a=5 m/s^2 is the acceleration

t is the time

On the other hand, the truck is travelling at a constant velocity, therefore its position at time t is given by

x_t(t) = vt

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v = 20 m/s is the velocity of the truck

t is the time

The car overtakes the truck when the two positions are the  same, so when

x_c(t) = x_t(t)\\\frac{1}{2}at^2 = vt\\t=\frac{2v}{a}=\frac{2(20)}{5}=8 s

So, after a time of 8 seconds. Therefore, the distance covered by the car during this time is

x_c(8) = \frac{1}{2}(5)(8)^2=160 m

So, the car overtakes the truck 160 m far from the intersection.

2)

The motion of the car is a uniformly accelerated motion, so the velocity of the car at time t is given by the suvat equation

v=u+at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

For the car in this problem, we have:

u = 0 (it starts  from rest)

a=5 m/s^2

And we know that the car overtakes the truck when

t = 8 s

Substituting into the equation,

v=0+(5)(8)=40 m/s

Learn more about accelerated motion:

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Answer:

The charge resides on the outer surface = 1.245 \times 10^{-12} C

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Dielectric constant (k) = 4.2

Potential difference (\Delta V) = 85.7 \times 10^{-3} V

The capacitance of parallel plate capacitor in free space is given by,

           C = \frac{\epsilon_{o} A }{d}

Where \epsilon_{o}  = permittivity of free space = 8.85 \times 10^{-12}

The Capacitance of capacitor is increase by k times when it placed in dielectric medium.

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So charge on the outer surface is given by,

      Q = \frac{k \epsilon A \Delta V }{d}

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Answer:

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