Answer:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,
when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
Therefore, the magnitude of the electric field is 0.1108 N/C
Answer:
You drop a rock from rest out of a window on the top floor of a building, 30.0 m above the ground. When the rock has fallen 3.00 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the ...... rest out of a window on the top floor of a building, 30.0m above the ground. ... You Notice That Both Rocks Reach The Ground At The Exact Same Time. ... You drop a rock from rest out of a window on the top floor of a building, 30.0m ... When the rock has fallen 3.20 m, your friend throws a second rock straight down from ...
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
a = g = 9.81[m/s^2]
Explanation:
This problem can be solve using the second law of Newton.
We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.
m*g = m*a
where:
g = gravity = 9.81[m/s^2]
a = acceleration [m/s^2]
Note: If the skydiver will be under air resistance forces his acceleration will be different.
