A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
face. Determine the image distance and magnification. Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
1 answer:
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
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Answer:
F = 236063.6N
Explanation:
Please see attachment below.
Answer:
See the explanation below.
Explanation:
Solving the first image question:
C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.
For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.
W = m*g
where:
W = weight [N]
m = mass [kg]
g = gravity acceleration [m/s²]
Now we have
m = 50 [kg]
ge = Earth gravity acceleration = 10 [m/s²]
gp = Distant planet gravity acceleration = 4 [m/s²]
We = ge*m
We = 10*50 = 500 [N]
Wp =gp*m
Wp = 4*50 = 200 [N]
Therefore the answer is D
For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.
m = 70 [kg]
First, we must calculate the acceleration, by means of the following equation of kinematics.
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 0 (because stars from the rest)
a = acceleration [m/s²]
t = time = 4 [s]
20 = 0 + a*4
20 = 4*a
a = 5 [m/s²]
Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.
F = m*a
where:
F = force [N] (units of Newtons)
m = mass = 2 [kg]
a = acceleration = 5 [m/s²]
F = 2*5
F = 10 [N]
The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]
50 + 40 - 35 - 30 = F
F = 25 [N]