Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
yes
Explanation:
objects with constant velocity also have zero net external force. this means the forces on the object are balanced. this mean they are in equilibrium
It is as a result of gravity. (D)
This is as stated by Newton's law of universal gravitation. That two objects in the universe attract one another with a force that is proportional to the product of their masses and inversely proportional to the square of the distance apart.
The constant of proportionality is the Universal Gravitational Constant.
G = 6.673 × 10⁻¹¹ Nm²kg⁻²
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Explanation:
Given that,
Initial speed of a car, u = 12 m/s
Aceleration, a = -1.6 m/s²
(a) The displacement of car after 6 seconds is :

(b) The displacement of the car after 9 seconds is :

Hence, this is the required solution.