Answer:
Air resistance have some significant role in projectile motion if the motion lasts for some time.
Explanation:
- Air resistance or air drag seems to be important in daily actvities and games like baseball.
- The trajectory of the projectile with or without air resistance or air drag is totally different.
- When we neglect air drag, the only acting force is gravity against the motion so the maximum height and range are suppose Hmax and R.
- Now, when we consider air drag, it is important to notice that there are two forces against the motion of the ball and along the direction of gravity. It seems that both maximum height and range are lesser Hmax'< Hmax and R'<R.
Answer:
B
Explanation:
Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.
A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.
C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.
D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.
Answer:
F_Balance = 46.6 N ,m' = 4,755 kg
Explanation:
In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in
∑ F = 0
Fe –W + F_Balance = 0
F_Balance = - Fe + W
The electric force is given by Coulomb's law
Fe = k q₁ q₂ / r₂
The weight is
W = mg
Let's replace
F_Balance = mg - k q₁q₂ / r₂
Let's reduce the magnitudes to the SI system
q₁ = + 8 μC = +8 10⁻⁶ C
q₂ = - 3 μC = - 3 10⁻⁶ C
r = 0.3 m = 0.3 m
Let's calculate
F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²
F_Balance = 49 - 2,397
F_Balance = 46.6 N
This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.
Mass reading is
m' = F_Balance / g
m' = 46.6 /9.8
m' = 4,755 kg
We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is
From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as
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