i know this is alot to ask for but these 10 i can nott get out of the 30
A man pulls a 10-kg box across a smooth floor with a force of 71 newtons at an angle of 23 degrees and for a distance of 97 meters. How much work, to the nearest joule, does he do?
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2
If the floor in the previous question is angled upward at 11.7 degrees and the man pulls the box up the floor at constant speed, what is the work he does to the nearest joule?
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3
In Problem 2, to the nearest joule, what is the work done by gravity?
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4
In Problem 1, if the box starts from rest, what is its final kinetic energy (to the nearest joule)?
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5
If the mass of the box in the previous problem is 17 kilograms, what is its final speed to the nearest tenth of a meter/sec?
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6
You are traveling in your 1542-kg car at 6 m/s and wish to accelerate to 19.2 m/s in 4 seconds. How much work, to the nearest joule, is required?
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7
In the previous problem what is the average power to the nearest watt?
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8
What is the equivalent (nearest) horsepower in Problem 7?
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9
Assuming you have a bow that behaves like a spring with a spring constant of 70 N/m and you pull it to a draw of 59 cm, to the nearest joule how much work do you perform?
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10
In the previous problem, to the nearest tenth of a m/s, what is the speed of the 107-gram arrow when it is released
All of these problems involve using the law of conservation of energy. In the first five, you have to calculate how far the box rises to compute the change in potential energy. Use gelometry for that. In #6, compute the change in kinetic energy of the car of the car. #7 and #8 require dividing work done by the time it takes to do the work
If you are unable to do these, I am skeptical about whether you did 20 other problems correctly.
Show your work if you need further help
work=forcexdistancexsin(angle)
To solve these physics problems, we will need to use several formulas and concepts. Let's go through each problem step by step:
Problem 1:
To find the work done by the man, we can use the formula: work = force * distance * cos(theta), where theta is the angle between the force and the direction of motion. In this case, the force is 71 newtons, the distance is 97 meters, and theta is 23 degrees. Plugging these values into the formula, we get work = 71 * 97 * cos(23). Calculate this expression to find the work done.
Problem 2:
When the man pulls the box up an inclined floor at constant speed, the total work done will be equal to the work done against gravity. Since the force required to lift the box up the inclined floor is equal to the weight of the box (mass * gravity), and the distance is the same as in Problem 1, you can use the same formula as in Problem 1 to calculate the work.
Problem 3:
To find the work done by gravity, we can use the equation: work = force * distance * cos(theta), where the force is the weight of the box (mass * gravity), the distance is the height the box is lifted, and theta is the angle between the weight force and the direction of motion. In this case, the force is the weight of the box (mass * gravity), the distance is the same as in Problem 2, and theta is the angle of the inclined floor. Calculate this expression to find the work done by gravity.
Problem 4:
To find the final kinetic energy of the box, we can use the equation: kinetic energy = 0.5 * mass * velocity^2. In this case, the mass is 10 kg, and the velocity can be calculated using the work-energy theorem. The work done on the box in Problem 1 is equal to the change in kinetic energy. Use the work done and the mass to find the final velocity, and then use this velocity to calculate the kinetic energy.
Problem 5:
To find the final speed of the box, you can use the equation from Problem 4: kinetic energy = 0.5 * mass * velocity^2. In this case, the mass is 17 kg, and you can use the final kinetic energy calculated in Problem 4 to find the final velocity.
Problem 6:
To find the work required to accelerate the car, we can use the equation: work = change in kinetic energy. The change in kinetic energy is given by: 0.5 * mass * (final velocity^2 - initial velocity^2). In this case, the mass is 1542 kg, the initial velocity is 6 m/s, and the final velocity is 19.2 m/s. Use these values to calculate the work required.
Problem 7:
Average power can be calculated using the formula: power = work / time. In this case, the work is the one calculated in Problem 6, and the time is given as 4 seconds.
Problem 8:
To convert power from watts to horsepower, we can use the conversion factor: 1 horsepower = 746 watts. Divide the power obtained in Problem 7 by 746 to get the equivalent horsepower.
Problem 9:
To find the work done when pulling the bowstring, we can use the formula: work = 0.5 * k * (displacement)^2, where k is the spring constant, and the displacement is the distance the bowstring is pulled. In this case, the k value is 70 N/m, and the displacement is 59 cm. Convert the displacement to meters before calculating the work.
Problem 10:
To find the speed of the arrow when released, we can use conservation of mechanical energy. Since there is no work done by external forces (assuming no energy losses), we can equate the initial potential energy (stored in the bowstring) to the final kinetic energy of the arrow. Use the work done and the mass of the arrow to calculate the speed. Remember to convert the mass from grams to kilograms before calculating the speed.
By following these steps and applying the appropriate formulas, you should be able to find the answers to these physics problems.