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Andrei [34K]
3 years ago
9

A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative

to the water), starts to swim directly across a 3.0-km-wide river. However, the current is 0.91 m/s, and it carries the swimmer downstream.
(a) How long does it take the swimmer to cross the river?
(b) How far downstream will the swimmer be upon reaching the other side of the river?
Physics
1 answer:
Vesnalui [34]3 years ago
5 0
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o
dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching  the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

F\propto\frac{d}{dt} (p)

\Rightarrow F=\frac{d}{dt} (m.v)

since mass is constant

F=m\frac{d}{dt}v

when dt=10t

then,

F'=m.\frac{v}{10\times t}

F'=\frac{1}{10} \times \frac{m.v}{t}

F'=\frac{F}{10} the body will experience the tenth part of the maximum force.

where:

\frac{d}{dt} = represents the rate of change in dependent quantity with respect to time

p= momentum

m= mass of the person jumping

v= velocity of the body while hitting the ground.

7 0
3 years ago
A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg
Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s.  Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0
3 years ago
What is a gravitational wave and why was it so hard to detect?
Alika [10]

Answer:

gravitational waves are ripples in spece-time caused primarily when objects are accelerated and the energy for the acceleration is transpoted as gravitational radiation.

they are difficult to detect because they require very sensitive technology or you will have to wait unitl black holes collide.

5 0
3 years ago
Which property of gold allows it to be used this way?
densk [106]

the electric conductivity of gold is very high

3 0
3 years ago
How do you find the capacitance in this?
Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

Ceq = 11C/6

3 0
2 years ago
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