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Anestetic [448]
3 years ago
11

A man paddles a canoe in a long, straight section of a river. The canoe moves downstream with constant speed 5 m/s relative to t

he water. The river has a steady current of 1 m/s relative to the bank. The man's hat falls into the river. Five minutes later, he notices that his hat is missing and immediately turns the canoe around, paddling upriver with the same constant speed of 5 m/s relative to the water. How long does it take the man to row back upriver to reclaim his hat?
Physics
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

Explanation:

We shall consider all movement with respect to water assuming that river is at rest or motionless .

speed of canoe = 5 m /s

in five minutes , distance between hat and canoe = 5 x 60 x 5

1500 m

This distance will be covered by man in return journey . In this case his speed is 5 m /s again considering river constant .

So this will be covered at 5 m /s

time taken = 1500 / 5 = 300 s

= 5 minutes .

so it will take 5 minutes to row back to reclaim his hat .

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Consider the data collected in science class. Different masses were thrown with varied amounts of
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Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

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3 years ago
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Which values are equivalent to the fraction below ? check all that apply. 2^4/2^7
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2^4/2^7 = 16/128 = 0.125
(1/2)^3= 0.125
1/8= 0.125
a and f are equivalent
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A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
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a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

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4 0
2 years ago
The tendency for an object to remain at rest in continue in motion is called:
nekit [7.7K]

Answer:

A Inertia

Explanation:

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3 years ago
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A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a dist
kvasek [131]

Answer:

a)  x_{cm} = m₂/ (m₁ + m₂)   d , b)   x_{cm} = 52.97 pm

Explanation:

The expression for the center of mass is

                x_{cm} = 1 / M  ∑ x_{i} m_{i}

Where M is the total masses, mI and xi are the mass and position of each element of the system.

Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon

              x_{cm} = 1 / (m₁ + m₂)   (0+ m₂ x₂)

Let's reduce the magnitudes to the SI system

             m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg

             m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg

             d = 128 pm = 128 10⁻¹² m

The equation for the center of mass is

               x_{cm} = m₂/ (m₁ + m₂)   d

b) let's calculate the value

            x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷    128 10-12

            x_{cm} = 52.97 10⁻¹² m

            x_{cm} = 52.97 pm

7 0
3 years ago
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