( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =
(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =
(0.03388) x (psc) x (10³) =
33.88 parsecs
At a distance r from a charge e on a particle of mass m the electric field value is 8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.
<h3>what is magnitude ?</h3>
Magnitude can be defined as the maximum extent of size and the direction of an object.
It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.
There are different ways where magnitude is used Magnitude of earthquake, charge on an electron, force, displacement, Magnitude of gravitational force
For more details regarding magnitude, visit
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Answer:
Vf = 15 m/s
Explanation:
First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)
h = height =?
Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)
Vi = Initial Velocity = 15 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²
h = (-225 m²/s²)/(-19.6 m/s²)
h = 11.47 m
Now, we consider downward motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height = 11.47 m
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²
Vf = √(224.812 m²/s²)
<u>Vf = 15 m/s</u>
Answer:
106.03 meters
Explanation:
The height is given by the formula for motion under the influence of gravity.
h = -4.9t^2 +162.7
Height is 0 when ...
0 = -4.9t^2 +162.7
4.9t^2 = 162.7
t^2 = 162.7/4.9
t = √(162.7/4.9)
The horizontal distance traveled in that time is ...
(18.4 m/s)√(162.7/4.9) s ≈ 106.03 m
The object will strike the ground about 106.03 meters from the base of the cliff.
