Her acceleration is 0.33 meters per second squared.
The work done in lifting a 60kg crate a vertical distance of 10 meters is given by the formula: work = force x distance x cos(theta), where force = weight x gravitational acceleration = 60kg x 9.8 m/s^2, distance = 10m, theta is the angle between the force and direction of displacement (which is 0 in this case, as it's lifting vertically). Plugging in the values, the work done is approximately 5,880 Joules.
Okay. You have to use the formula F=M*A (force = mass (times) acceleration). You are given the mass and the force while he is sitting on his chair on the launch pad. Set up the problem. 588 N (Newtons) = 60 kg * Acceleration. 588N / 60 kg = Acceleration Acceleration = 9.8 m/s^2 (meters per second squared) That is the acceleration due to gravity, looks like we are on Earth. To get to your question though, you have to add both accelerations in the formula F=M*A. You are given 8 m/s^2 and we just found the other to be 9.8 m/s^2. So... F = 60kg * (9.8m/s^2 + 8 m/s^2) That will give you your answer, which is 1068 N. Hope this helps.
The GPE (gravitational potential energy) of the diver can be calculated using the formula GPE = mgh, where m is the mass (60kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height (10m). Plugging in the values, GPE = 60kg * 9.81 m/s^2 * 10m = 5886 Joules.
60kg is equivalent to 132.28 pounds.
A bag of rice would typically weigh around 3kg, not 60kg.
F=mass * acceleration 60kg m/s^2=10kg * acceleration 6m/s^2 = acceleration
The work done in lifting a 60kg crate a vertical distance of 10 meters is given by the formula: work = force x distance x cos(theta), where force = weight x gravitational acceleration = 60kg x 9.8 m/s^2, distance = 10m, theta is the angle between the force and direction of displacement (which is 0 in this case, as it's lifting vertically). Plugging in the values, the work done is approximately 5,880 Joules.
Okay. You have to use the formula F=M*A (force = mass (times) acceleration). You are given the mass and the force while he is sitting on his chair on the launch pad. Set up the problem. 588 N (Newtons) = 60 kg * Acceleration. 588N / 60 kg = Acceleration Acceleration = 9.8 m/s^2 (meters per second squared) That is the acceleration due to gravity, looks like we are on Earth. To get to your question though, you have to add both accelerations in the formula F=M*A. You are given 8 m/s^2 and we just found the other to be 9.8 m/s^2. So... F = 60kg * (9.8m/s^2 + 8 m/s^2) That will give you your answer, which is 1068 N. Hope this helps.
12% of 60kg is 7.2kg
The GPE (gravitational potential energy) of the diver can be calculated using the formula GPE = mgh, where m is the mass (60kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the height (10m). Plugging in the values, GPE = 60kg * 9.81 m/s^2 * 10m = 5886 Joules.
60kg is equivalent to 132.28 pounds.
2/3 of 60kg is 40kg
A bag of rice would typically weigh around 3kg, not 60kg.
your weight would be 120 kg but your mass would be 60 kg
To find the mass, you can use Newton's second law, which states that force equals mass times acceleration (F = ma). Rearranging the formula to solve for mass, the mass of you and the skateboard together is given by mass = force/acceleration. Therefore, the mass would be 60 kg (mass = 30 N / 0.5 m/s^2).
To find out how many packets weighing 480g can be made from a quantity of 60kg, first convert 60kg to grams: 60kg = 60,000g. Then, divide the total weight by the weight of each packet: 60,000g ÷ 480g = 125. Therefore, 125 packets weighing 480g can be made from 60kg.
To find the force on an object, you multiple the mass of the object by its accelerationFor example, let's say a ball is moving at 7 m/s squared and has a mass of12 kg.The formula for force is: F = maF = 12 kg x 7 m/s squaredF = 84 N ( the unit for force is N, which is Newtons)*Be careful, an object does not "have a force". A force is an action that can modify the shape of an object and/or modify its velocity. Therefore, you do not calculate the force of an object, but the force required to give it a certain acceleration.