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Step2247 [10]
3 years ago
9

Three swimmers can swim equally fast relative to the water. They have a race to see who can swim across a flowing river in the l

east time. Swimmer A swims perpendicular to the current and lands on the opposite shore downstream, because the current has swept him in that direction. Swimmer B swims upstream at angle to the current, so that when she reaches the other side, she is straight across from where she started. Swimmer C swims downstream at an angle theta in an attempt to take advantage of the current. Who crosses the river in the least time?
Physics
1 answer:
katrin [286]3 years ago
8 0
I think its swimmer B because a and c are swimming downstream when they are trying to go across
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Answer:

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Explanation:

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3 years ago
!!!!BRAINLIEST!!!!!
kykrilka [37]

Answer:

The jp2003parker guy is extremely wrong

So he says that the size wont matter and a physical change should occur, but how would the size change without having a physical change occur.

Explanation:

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2 years ago
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2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos
Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh

there, no vertical movement

So, on neglect of height term

P_{1}=P+\dfrac{1}{2}\rho v_{1}^2

Where, P= atmospheric pressure

\rho = air density

v = speed

Put the value in the equation

P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)

P_{1}=2126.0\ lb/ft^2

P_{1}=\dfrac{2126.0}{144}

P_{1}= 14.76\ psi

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{2}=P+\dfrac{1}{2}\rho v_{2}^2

Put the value in the equation

P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)

P_{2}=2246.17\ lb/ft^2

P_{2}=\dfrac{2246.17}{144}

P_{2}= 15.59\ psi

(c).  Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

P_{3}=P+\dfrac{1}{2}\rho v_{3}^2

Put the value in the equation

P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)

P_{3}=2257.93\ lb/ft^2

P_{3}=\dfrac{2257.93}{144}

P_{3}= 15.68\ psi

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0
3 years ago
g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses
statuscvo [17]

Answer:

v₁f = 0.5714 m/s   (→)

v₂f = 2.5714 m/s   (→)

e = 1  

It was a perfectly elastic collision.

Explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i +  ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) +  ((2*6m) / (m + 6m)) * (2)  

v₁f = 0.5714 m/s   (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i +  ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s   (→)

e = - (v₁f - v₂f) / (v₁i - v₂i)   ⇒   e = - (0.5714 - 2.5714) / (4 - 2) = 1  

It was a perfectly elastic collision.

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podryga [215]
I believe it is -1.11 m/s^2. I will let you know if its correct
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