Answer:
See the answers below
Explanation:
This problem and its respective questions can be easily solved using Newton's law of universal gravitation. Which can be calculated by means of the following expression.
where:
G = it is the universal gravitation constant. = 6.673 x 10⁻¹¹ [N*m²/kg²]
m1 = mass of the first body [kg]
m2 = mass of the second body [kg]
r = distance among the bodies [m]
a. the mass of one is doubled?
When this happens we see that the force is increased twice as well, since the mass is in the numerator of the expression.
b. The masses of both are doubled?
If both masses are doubled the force is increased to four times its original value since the terms of the masses are in the numerator of the expression.
c. The distance between them is doubled?
In this case the force is decreased to half of its original value, since the distance is in the denominator of the expression of universal gravitation.
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
<h3>
What is vertical motion of a projecile?</h3>
The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;
Vy = Vsinθ
<h3>
What is horizontal motion of a projecile?</h3>
The horizontal velocity of a projectile is given by the following formula;
Vx = Vcosθ
<h3>Direction of the motion</h3>
The direction of the motion is calculated as follows;
tanθ = Vy/Vx
Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
Learn more about vertical motion here: brainly.com/question/24216590
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Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
m1= mass 1 = 1.1 kg
Vi1 = initial velocity 1 = 2.7 m/s
m2= 2.4 kg
V2i = -1.9 m/s
We assume east as positive and west as negative.
Apply the formulas:
Vf1 = ?
Replacing:
Answer: 3.6 m/s west
Answer:
E=-1.51 eV.
Explanation:
The nth level energy of a hydrogen atom is defined by the formula,
Given in the question, the hydrogen atom is in the 3p state.
Then energy of n=3 state is,
Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.
Now, the value of L can be calculated as,
For 3p state, l=1
Therefore, the value of L of a hydrogen atom in 3p state is .
