Answer:
W = (F1 - mg sin θ) L, W = -μ mg cos θ L
Explanation:
Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane
Y Axis
N -
=
N = W_{y}
X axis
F1 - fr - Wₓ = 0
fr = F1 - Wₓ
Let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
We substitute
fr = F1 - W sin θ
Work is defined by
W = F .dx
W = F dx cos θ
The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1
W = -fr x
W = (F1 - mg sin θ) L
Another way to calculate is
fr = μ N
fr = μ W cos θ
the work is
W = -μ mg cos θ L
TLDR: It will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.
This is an example that requires you to investigate the properties that occur in electric generators; for example, hydroelectric dams produce electricity by forcing a coil to rotate in the presence of a magnetic field, generating a current.
To solve this, we need to understand the principles of electromotive forces and Lenz’ Law; changing the magnetic field conditions around anything with this potential causes an induced current in the wire that resists this change. This principle is known as Lenz’ Law, and can be described using equations that are specific to certain situations. For this, we need the two that are useful here:
e = -N•dI/dt; dI = ABcos(theta)
where “e” describes the electromotive force, “N” describes the number of loops in the coil, “dI” describes the change in magnetic flux, “dt” describes the change in time, “A” describes the area vector of the coil (this points perpendicular to the loops, intersecting it in open space), “B” describes the magnetic field vector, and theta describes the angle between the area and mag vectors.
Because the number of loops remains constant and the speed of the coils rotation isn’t up for us to decide, the only thing that can increase or decrease the emf is the change in magnetic flux, represented by ABcos(theta). The magnetic field and the size of the loop are also constant, so all we can control is the angle between the two. To generate the largest emf, we need cos(theta) to be as large as possible. To do this, we can search a graph of cos(theta) for the highest point. This occurs when theta equals 90 degrees, or a right angle. Therefore, the electromotive potential will reach a maximum when the angle between the area vector and the magnetic field vector are perpendicular to one another.
Hope this helps!
Answer:
An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?
A.
We’ll see the bell move, but we won’t hear it ring.
B.
We won’t see the bell move, but we’ll hear it ring.
C.
We’ll see the bell move and hear it ring.
D.
We won’t see the bell move or hear it ring.
E.
We’ll see the sound waves exit the vacuum pump.
Explanation:
so, the answer to the question is
A.
We'll see the bell move, but we won’t hear it ring.
Given data:
- It is a graphical display where the data is grouped in to ranges
- A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.
- It is an accurate representation of the distribution of numerical data.
<em>From Figure:</em>
Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).
<em>So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.</em>
Answer:

Explanation:
The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

where
is the horizontal velocity
t is the time of flight
The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.
In the first situation, the horizontal distance covered is

in the second case, the horizontal velocity is increased to

And so the new distance travelled will be

So, the distance increases linearly with the horizontal velocity.