Answer:
5.9 x 10⁻⁷m
Explanation:
Given parameters:
Frequency = 5.085 x 10¹⁴Hz
Speed of light = 3.0 x 10⁸m/s
Unknown:
Wavelength of the orange light = ?
Solution:
The wavelength can be derived using the expression below;
wavelength =
v is the speed of light
f is the frequency
wavelength = = 5.9 x 10⁻⁷m
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
Gravitational force between the second object and the 42 kg object
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Kinetic Energy of air molecules. for molecules of air, the temperature of the air is directly proportional to the mean kinetic energy of the air molecules.
(a). By the inertia, it's difficult to stop heavier mass than lighter mass.
As, moment of inertia is directly proportional to the mass of the body.
Thus, more force is required to stop the boy with heavier mass than lighter mass.
The problem statement for the given case is,
The time taken by the heavier body to stop swinging is more than the lighter body, then what is the impact of weight on the time period of the swing?
To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:
Where,
x= Displacement
= Initial velocity
a = Acceleration
t = time
Since there is no initial velocity, the same equation can be transformed in terms of length and time as:
For the second cart
When the tenth car is aligned the length will be 9 times the initial therefore:
When the tenth car has passed the length will be 10 times the initial therefore:
The difference in time taken from the second car to pass it is 5 seconds, therefore:
From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:
From the relationship when the car has passed and the time difference we will have to:
Replacing the value found in the equation given for the second car equation we have to:
Finally we will have the time when the cars are aligned is
The time when you have passed it would be:
The difference between the two times would be:
Therefore the correct answer is C.