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balu736 [363]
3 years ago
15

What happens when one side of the tug-of-war rope has a large net force than the other?

Physics
2 answers:
Jet001 [13]3 years ago
8 0

Answer: Physical Science

Explanation:

When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force. As the sum of forces increases, the speed of the cart goes faster throughout the tug of war.

When both sides of the tug-of-war rope are made equal in force than the rope does not move at all. There is a mass of 350 N on both sides of the cart. The net force in this scenario is 0 because there is no difference in mass.

With the first example, the speed odometer gets up to right next to the “S” in speed. With one small person verse all four on the other side, the speed odometer moves all the way past halfway to the d in Speed.

The objects selected will vary.

The chart will look different depending on the ratio of weights that you put on the skateboard. The answers provide what a chart might look like.

Trial number Mass on Skateboard Applied Force by Person Speed When Let Go

1 330 kg 200 N 3.3 m/s

2 100 kg 200 N 9.2 m/s

3 340 kg 200 N 1.1 m/s

The relationship between speed and force is direct because regardless of the mass on the skateboard and the force involved, they continue to increase. The greater the force applied to the skateboard when it is released, the higher the speed of the skateboard.

The relationship between mass and speed (with constant force) is indirect. The higher the mass that needs to be pushed, the lower the starting speed is of the skateboard.

The longer the force is applied to the skateboard, the higher the speed of the skateboard will be. Even with the heaviest possible skateboard, the speed will max out after a certain period of time. The time needed to do this is shorter when the mass is lighter, indicating a direct relationship between the variables.

Your chart may look different based on the results from your first trial completed in question 5.

Trial number Mass on Skateboard Applied Force by Person Friction Speed When Let Go

1 Same as Trial #1, Question 5 Same as Trial #1, Question 5 None 2.7 m/s

2 Same as Trial #1, Question 5 Same as Trial #1, Question 5 Medium 0.0 m/s

3 Same as Trial #1, Question 5 Same as Trial #1, Question 5 Lots 0.0 m/s

The relationship between speed and friction is indirect because as the friction increases the speed decreases.

The object stays still when there is too much friction applied to the equation with the mass and applied force.

The relationship between force and acceleration is directly proportional. Whatever the force is that is applied to the object, the acceleration of that object increased by the same factor.

beks73 [17]3 years ago
5 0

Answer:

1.When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force. As the sum of forces increases, the speed of the cart goes faster throughout the tug of war.

2.When both sides of the tug-of-war rope are made equal in force than the rope does not move at all. There is a mass of 350 N on both sides of the cart. The net force in this scenario is 0 because there is no difference in mass.

3.With the first example, the speed odometer gets up to right next to the “S” in speed. With one small person verse all four on the other side, the speed odometer moves all the way past halfway to the d in Speed.

4.The objects selected will vary.

5.The chart will look different depending on the ratio of weights that you put on the skateboard. The answers provide what a chart might look like.

Trial number Mass on Skateboard Applied Force by Person Speed When Let Go

1  330 kg  200 N  3.3 m/s

2  100 kg   200 N  9.2 m/s

3  340 kg  200 N  1.1 m/s

6.The relationship between speed and force is direct because regardless of the mass on the skateboard and the force involved, they continue to increase. The greater the force applied to the skateboard when it is released, the higher the speed of the skateboard.

7.The relationship between mass and speed (with constant force) is indirect. The higher the mass that needs to be pushed, the lower the starting speed is of the skateboard.

8.The longer the force is applied to the skateboard, the higher the speed of the skateboard will be. Even with the heaviest possible skateboard, the speed will max out after a certain period of time. The time needed to do this is shorter when the mass is lighter, indicating a direct relationship between the variables.

9.Your chart may look different based on the results from your first trial completed in question 5.

Trial number Mass on Skateboard Applied Force by Person Friction Speed When Let Go

1 Same as Trial #1, Question 5 Same as Trial #1, Question 5  None  2.7 m/s

2 Same as Trial #1, Question 5  Same as Trial #1, Question 5 Medium  0.0 m/s  

3  Same as Trial #1, Question 5 Same as Trial #1, Question 5 Lots  0.0 m/s  

10.The relationship between speed and friction is indirect because as the friction increases the speed decreases.

11.The object stays still when there is too much friction applied to the equation with the mass and applied force.

12.The relationship between force and acceleration is directly proportional. Whatever the force is that is applied to the object, the acceleration of that object increased by the same factor.

Explanation:

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Calculate the rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.​
vredina [299]

Answer: The rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.= 1.25 watt.

Explanation:

  • The rate of energy transfer from one point to another is known as power.
  • Its SI. unit is <em>Watt.</em>

Formula to compute power:

Power = I^2 R, I = cureent, R = resistance.

Given R = 5 ohm , Current = 0.5 A.

Then, power = (0.5)^2\times 5=0.25\times5=1.25\ watt

Hence, the rate of energy transfer by a 5 ohms resistor when there is a current of 0.2 A in it.= 1.25 watt.

5 0
3 years ago
Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants.
Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

\dot{P} = \bar{P}

Where,

\dot{P} = Production of Power

\bar{P} = Consumption of power

Where the production of power would be,

\dot{P} = m \dot{E}\eta

Where,

m = Total mass required

\dot{E} = Energy per Kilogram

\eta =Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

\dot{P} = \bar{P}

m \dot{E}\eta = \bar{P}

Replacing the values we have,

m(8*10^13)(0.45) = 2*10^{12}

Solving for m,

m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}

m = 0.0556 \frac{kg}{s}

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})

m = 1752.14tonnes per year.

8 0
3 years ago
In a power plant, pipes transporting superheated vapor are very common. Superheated vapor flows at a rate of 0.3 kg/s inside a p
grigory [225]

Answer:h=160.84 W/m^2-K

Explanation:

Given

mass flow rate=0.3 kg/s

diameter of pipe=5 cm

length of pipe=10 m

Inside temperature=22

Pipe surface =100

Temperature drop=30

specific heat of vapor(c)=2190 J/kg.k

heat supplied Q=mc\Delta T=0.3\times 2190\times (30)

Heat due to convection =hA(100-30)

A=\pi d\cdot L

A=\pi 0.05\times 10=1.571 m^2

Q_{convection}=h\times 1.571\times (100-22)=122.538 h

Q=Q_{convection}

19,710=122.538 h

h=160.84 W/m^2-K

5 0
3 years ago
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