one layer of earth's atmosphere is called the stratosphere. At one point above earth's surface the stratosphere extends from an
2 answers:
You might be interested in
Answer:
1) d
2) 5 m/s
3) 100
Explanation:
The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:
(i)
The equation for velocity v and a constant acceleration a is:
(ii)
1) Solve equation (ii) for acceleration a and plug the result in equation (i)
(iii)
(iv)
Simplify equation (iv) and use the given values v = 0, x₀ = 0:
(v)
2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v:
3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):
Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.
Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.
Case 1: Terminal velocity of a piece of tissue paper.
The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.
Downward gravitational force, F = mg
Upward air resistance or friction or drag force will be
So, paper will attain terminal velocity when mg =
Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.
Downward force on rock, F = Mg
Drag force =
Rock will attain terminal velocity when Mg =
Mg > mg
so, >
And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.
Answer:
<em>3.15 N towards the positive x-axis</em>
<em></em>
Explanation:
first charge has charge q1 = 10 μC = 10 x 10^-6 C
second charge has charge q2 = 20 μC = 20 x 10^-6 C
third charge has charge q3 = -30 μC = -30 x 20^-6 C
According to coulomb's law, force between two charged particle is given as
F =
Where
F is the force between the charges
k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.
Q is the magnitude of one charge
q is the magnitude of the other charge
is the distance between these two charges
For the force on q2 due to q1,
distance r between them = 0 - (-1.0) = 1 m
F = = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive x-axis)
For the force on q2 due to q3,
distance between them = 2.0 - 0 = 2 m
F = = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)
Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>