How can x-rays make pictures of the inside of solid objects
1 answer:
You might be interested in
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q =
Therefore, substitute the values into the above formula as follows.
Q =
=
=
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .
Answer:
C. At the bottom of the circle.
Explanation:
Lets take
Radius of the circle = r
Mass = m
Tension = T
Angular speed = ω
The radial acceleration towards = a
a= ω² r
Weight due to gravity = mg
<h3>At the bottom condition</h3>T - m g = m a
T = m ω² r + m g
<h3>At the top condition</h3>T + m g = m a
T= m ω² r -m g
From above equation we can say that tension is grater when ball at bottom of the vertical circle.
Therefore the answer is C.
C. At the bottom of the circle.
Answer:
In the vertical direction the acting forces are the normal force and the weight of the bobsleder plus the sled. In the horizontal direction the acting force is the friciton force.
Explanation:
Hi there!
Please, see the attached figure for a graphic representation of the forces acting on the sled after the bobsleder jumped in.
In the vertical direction, the acting forces are the normal force (N) and the weight of the sled plus the bobsledder (W).
Since the sled is not being accelerated in the vertical direction, the sum of forces in that direction is zero:
∑Fy = W + N = 0 ⇒ W = N
The weight is calculated as follows:
W = (mb + ms) · g
Where:
mb = mass of the bobsleder.
ms = mass of the sled.
g = acceleration due to gravity.
In the horizontal direction the only acting force is the friction force (Fr). The friction force is calculated a follows:
Fr = N · μ
Where:
N = normal force.
μ = kinetic friction coefficient.
Since N = W = (mb + ms) · g
Fr = (mb + ms) · g · μ
If we want to find the acceleration of the sled after the bobsleder jumps in, we can apply Newton's second law:
∑F = m · a
Where "a" is the acceleration and "m" is the mass of the object (in this case, the mass of bobsleder plus the mass of the sled).
∑F = Fr = (mb + ms) · g · μ = (mb + ms) · a
(mb + ms) · g · μ = (mb + ms) · a
Solving for "a":
g · μ = a
