The calculation for kinetic energy is this
KE = 1/2mv^2
KE = 1/2(50)(7^2)
KE = 1/2(49•50)
KE = 1225 kgm^2/s^2.
Or simply 1225 J.
She possess this much energy when she runs.
Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.
The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
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Answer:
it must be possible to prove it wrong
Explanation:
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
