56 J of work was done to increase the kinetic energy of a 1.90 kg mass. If the object’s initial velocity was zero, what is its s
1 answer:
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Answer:
(A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is .
Explanation:
Given that,
Magnetic field = 2.0 T
Diameter = 5.0 cm
Rate of decreasing in magnetic field = 5.00 T/s
(A). We need to calculate the electric field strength inside the solenoid at a point on the axis
Using formula of electric field inside the solenoid
Electric field on the axis of the solenoid
Here, r = 0
The electric field strength inside the solenoid at a point on the axis is zero.
(B). We need to calculate the electric field strength inside the solenoid at a point 1.50 cm from the axis
Using formula of electric field inside the solenoid
Hence, (A). The electric field strength inside the solenoid at a point on the axis is zero.
(B). The electric field strength inside the solenoid at a point 1.50 cm from the axis is .
Answer:
T= 4.24sec
Explanation:
We are going to use the formula below to calculate.
Where T is period
L is length of rod
g is acceleration due to gravity =
From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this
= 4.4625m
thus
T= 4.24sec
There are two rules of reflection for the concave mirror:
1)<span>Any incident ray traveling parallel to the principal axis on the way to the mirror will pass through the focal point upon reflection.
2)</span><span>Any incident ray passing through the focal point on the way to the mirror will travel parallel to the principal axis upon reflection.
Also, keep in that for concave mirror center of curvature is 2f.
Using these two rules we can construct the image.
You should always use these two "special" rays, I will mark then 1 and 2 on the picture.
From the picture, we can see that if you place the object between the focus and vertex you get the virtual image.
The answer is: object should be between the focal point and the vertex</span>
1)<span>Any incident ray traveling parallel to the principal axis on the way to the mirror will pass through the focal point upon reflection.
2)</span><span>Any incident ray passing through the focal point on the way to the mirror will travel parallel to the principal axis upon reflection.
Also, keep in that for concave mirror center of curvature is 2f.
Using these two rules we can construct the image.
You should always use these two "special" rays, I will mark then 1 and 2 on the picture.
From the picture, we can see that if you place the object between the focus and vertex you get the virtual image.
The answer is: object should be between the focal point and the vertex</span>