Answer:
Also 3s.
Explanation:
Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.
According to the law of conservation of energy,
A. an object loses most of its energy as friction
<u>B. the total amount of energy for a system stays the same</u>
Energy is never lost due to the law of conservation
C. the potential energy of an object is always greater than its kinetic energy
D. the kinetic energy of an object is always greater than its potential energy
Answer:
Mass of the vehicle and small bug.
Explanation:
- By Newton's third law, force on bug and vehicle will be same when they collide with each other irrespective of their masses.
- But according to Newton's second law, force is mass times acceleration. Since the force on each mass is same, the smaller mass will accelerate more and the heavier mass will accelerate less for the same force.
- Therefore the acceleration of bug will be very greater than vehicle as the mass of the bug is very small as compared to vehicle.
Learn more about Newton's law.
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Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
Average speed = (distance covered) / (time to cover the distance)
Average speed = (4 meters) / (5 seconds)
Average speed = (4/5) (meters/seconds)
Average speed = 0.8 m/s
