Answer:
Microlensing.
Explanation:
This techniques is called Microlensing.
Microlensing is a method of gravitational lensing where light from a backdrop point of origin is curved to develop distorted, numerous and/or lightened images by the gravity field of a foreground lens.
This method is very effective in discovering planets that are far-far from earth.It is actually an astronomical effect that was predicted by Albert Einstein's general theory of relativity.
Answer:
Spring's displacement, x = -0.04 meters.
Explanation:
Let the spring's displacement be x.
Given the following data;
Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg
Number of shrews, n = 49
Spring constant, k = 24 N/m
We know that acceleration due to gravity, g is equal to 9.8 m/s².
To find the spring's displacement;
At equilibrium position:
Fnet = Felastic + Fg = 0
But, Felastic = -kx
Total mass, Mt = nm
Fg = -Mt = -nmg
-kx -nmg = 0
Rearranging, we have;
kx = -nmg
Making x the subject of formula, we have;
Substituting into the formula, we have;
x = -0.04 m
Therefore, the spring's displacement is -0.04 meters.
Answer:
D
Explanation:
Let’s calculate the kinetic energy for all of the choices.
a. (1/2)(100)(100)^2 = 50(10000)=500,000
b. (1/2)(100)(1)^2 = 50
c. (1/2)(10)(100)^2 = 5(10000) = 50,000
d. (1/2)(1)(1)^2 = 0.5
We can see that (d) has the least kinetic energy.
Answer:
F₃ = 122.88 N
θ₃ = 20.63°
Explanation:
First we find the components of F₁:
For x-component:
F₁ₓ = F₁ Cos θ₁
F₁ₓ = (50 N) Cos 60°
F₁ₓ = 25 N
For y-component:
F₁y = F₁ Sin θ₁
F₁y = (50 N) Sin 60°
F₁y = 43.3 N
Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:
F₂ₓ = F₂ = 90 N
F₂y = 0 N
Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:
F₁ₓ + F₂ₓ + F₃ₓ = 0 N
25 N + 90 N + F₃ₓ = 0 N
F₃ₓ = - 115 N
for y-components:
F₁y + F₂y + F₃y = 0 N
43.3 N + 0 N + F₃y = 0 N
F₃y = - 43.3 N
Now, the magnitude of F₃ can be found as:
F₃ = √F₃ₓ² + F₃y²
F₃ = √[(- 115 N)² + (- 43.3 N)²]
<u>F₃ = 122.88 N</u>
and the direction is given as:
θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)
<u>θ₃ = 20.63°</u>
Answer:
Explanation:
given.
magnification(m) = 400 x
focal length (f_0)= 0.6 cm
distance between eyepiece and lens (L)= 16 cm
Near point (N) = 25 cm
focal length of the eyepiece (f_e)= ?
using equation
