Answer: 24.97 kg
Explanation:
The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:
F = G*(M1*M2)/R^2
Where G is the gravitational constant:
G = 6.67*10^-11 m^3/(kg*s^2)
In this case, we know that
R = 0.002m
F = 0.0104 N
and that M1 = M2 = M
And we want to find the value of M, then we can replace those values in the equation to get
0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2
(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2
623.69 kg^2 = M^2
√(623.69 kg^2) = M = 24.97 kg
This means that the mass of each object is 24.97 kg
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
Answer:
θ = 28.9
Explanation:
For this exercise let's use the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where we use index 1 for air and index 2 for water where the fish is
sin θ₂ = n₁ / n₂ sin θ₁
in this case the air repair index is 1 and the water 1.33
we substitute
sin θ₂ = 1 / 1.33 sin t 40
sin θ = 0.4833
θ = sin⁻¹ 0.4833
θ = 28.9
