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Otrada [13]
2 years ago
8

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

50 kilograms. What can Paul do to increase the bike’s kinetic energy?
A. He can let Ivan off at the next stop.
B. He can pedal harder to increase the rate to 10 meters/second.
C. He can reduce the speed to 3 meters/second.
D. He can pick up a third rider.
Physics
2 answers:
Sonbull [250]2 years ago
5 0
The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.

A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?

The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.

I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.

B <<<<====== answer. 
alisha [4.7K]2 years ago
5 0

Answer: The correct answer is option C.

Explanation:

Kinetic energy is the energy possessed by the an object due to its motion.An its calculated by:

K.E.=\frac{1}{2}mass\times (velocity)^2

Kinetic energy depends upon the mass and velocity of the an object.

So, Paul can increase the bike's kinetic energy by increasing the velocity of its bike. Hence, the correct answer is option C.

Increasing the mass will also increase the kinetic energy . But according to option (D) he has to stop the bike first by applying brakes which will reduce the kinetic energy of the bike.And then again have to perform the work to bring the bike in motion

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When a pendulum is pulled back from its equilibrium position by 10∘, the restoring force is 1.0 N. When it is pulled back to 30∘
jarptica [38.1K]

Answer: B

Explanation: I said B because if you pull something back what is going to be more of a force pulling back or letting it go for a rubier band yes it will have more force if you let it go

5 0
2 years ago
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The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.
Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

W=0.24\times 1\times 10^3=240\ W/m^2

Energy required by the United States is 1\times 10^{19}\ J/yr (assumed)

Power

P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W

Area

A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

1\ m^2=\frac{1}{1000\times 1000}\ km^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2

Converting to mi²

1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2

1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0
2 years ago
A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of kinetic friction between the two boxes and betwe
natta225 [31]
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
4 0
2 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
2 years ago
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Alika [10]

Answer:

d = 2021.6 km

Explanation:

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Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

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2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
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