Explanation:
Equation of motion, mathematical formula that describes the position, velocity, or acceleration of a body relative to a given frame of reference.
Answer:
Surface charge density will be
Explanation:
We have given speed of the belt v = 43 m/sec
Width of the belt w = 61 cm
We know that charge is equal to Q = It, here Is current and t is time
And time is equal to , here L is distance and v is speed
Putting the value of t in charge equation
Surface charge density is equal to
We know that width is equal to
So
Answer:
The magnetic field reverses direction.
Explanation:
Answer:
892 Hz or 842 Hz
Explanation:
Given
The frequency of first guitar is 867 Hz
Beat is produced for every 0·04 s
∴ Beat has a time period of 0·04 s
<h3>Beat frequency is defined as the inverse of the time period of the beat </h3>
Beat frequency = 1 ÷ (Time period) = 1 ÷ 0·04 = 25 Hz
Let the frequency of the second guitar be f Hz
<h3>Beat frequency is defined as the absolute difference between the two frequencies</h3>
∴ Beat frequency = (867 - f) or (f - 867)
First case
25 = 867 - f ⇒ f = 842 Hz
Second case
25 = f - 867 ⇒ f = 892 Hz
∴ Possible frequencies of second guitar could be 892 Hz or 842 Hz
Answer:
Time is 14.8 s and cannot landing
Explanation:
This is a kinematic exercise with constant acceleration, we assume that the acceleration of the jet to take off and landing are the same
Calculate the time to stop, where it has zero speed
Vf² = Vo² + a t
t = - Vo² / a
t = - 110²/(-7.42)
t = 14.8 s
This is the time it takes to stop the jet
Let's analyze the case of the landing at the small airport, let's look for the distance traveled to land, where the speed is zero
Vf² = Vo² + 2 to X
X = -Vo² / 2 a
X = -110² / 2 (-7.42)
X = 815.4 m
Since this distance is greater than the length of the runway, the jet cannot stop
Let's calculate the speed you should have to stop on a track of this size
Vo² = 2 a X
Vo = √ (2 7.42 800)
Vo = 109 m / s
It is conclusion the jet must lose some speed to land on this track