Could someone try to help me solve those 3 questions by giving me clues or equations?
With what speed (SI Units) must a ball be thrown directly upward so that it remains in the air for 12.5 second?
An object of mass 197kg is at rest. If a force of 2.5N is applied for 30 seconds, what will its final velocity be?
How much work is done by slowly lifting a 1.75kg sack of flour to a height of 2.0m?
U R dumm
Sure! I can help you solve these three questions by providing you with the necessary clues and equations.
1. For the first question: "With what speed (SI Units) must a ball be thrown directly upward so that it remains in the air for 12.5 seconds?"
To solve this question, we can use the kinematic equation:
Δy = v₀t + 0.5at²
where Δy represents the change in height (which is equal to zero in our case, as the ball returns to its initial position), v₀ is the initial velocity, t is the time in the air, and a is the acceleration due to gravity (-9.8 m/s²).
Since the ball goes up and then comes back down, the total time in the air is 12.5 s. However, we need to find the initial velocity for it to remain in the air for this time.
Plugging in the values, we get:
0 = v₀(12.5) + 0.5(-9.8)(12.5)²
Solving this equation for v₀ will give us the required initial velocity in meters per second (m/s).
2. For the second question: "An object of mass 197 kg is at rest. If a force of 2.5 N is applied for 30 seconds, what will its final velocity be?"
To solve this question, we can use Newton's second law of motion:
F = ma
where F represents the force applied, m is the mass of the object, and a is the acceleration.
In this case, the force applied is 2.5 N, the mass is 197 kg, and we need to find the final velocity. Rearranging the equation, we get:
a = F/m
Plugging in the values, we can calculate the acceleration.
Once we have the acceleration, we can use the kinematic equation:
vf = v₀ + at
where vf is the final velocity, v₀ is the initial velocity (which is zero since the object is initially at rest), a is the acceleration, and t is the time.
Substituting the values, we can calculate the final velocity.
3. For the third question: "How much work is done by slowly lifting a 1.75 kg sack of flour to a height of 2.0 m?"
To solve this question, we can use the work-energy principle:
Work = force × distance × cos(θ)
where work represents the work done, force is the applied force, distance is the height or distance moved, and θ is the angle between the force vector and the direction of movement (which is 0° in this case, as the force is applied vertically upward).
Here, the force required to lift the sack of flour is the gravitational force acting on it, which is given by:
Force = mass × gravity
Plugging in the values, we can calculate the force.
After that, we can substitute the force and the distance into the work equation to find the amount of work done.