Answer:
Explanation:
given,
side of square loop = a = 2.10 cm
Resistance of the wire = 1.30×10⁻² Ω
Length of the loop = c = 1.10 cm
rate of increasing current = 130 A/s
Answer:
33.0 m
Explanation:
We are given that
Wavelength,
Diameter of pupil=d=5.1 mm=
By Rayleigh criteria
Substitute the values
The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
<h3>Average velocity of the car</h3>
The average velocity of the car is calculated as follows;
x(t) = a + bt + ct2
v = dx/dt
v(t) = b + 2ct
v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s
v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s
<h3>Average velocity</h3>
V = ¹/₂[v(0) + v(10)]
V = ¹/₂ (-10.1 + 11.9 )
V = 0.9 m/s
Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.
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Answer:
Electric field,
Explanation:
It is given that,
Radius of the circular loop, r = 13 cm = 0.13 m
Electric flux in the positive x direction,
Electric flux in the positive y direction,
The formula of the electric flux is given by :
In x- direction,
Where, is the electric field in x direction.
So, the x-component of the electric field is 470.87 N/C. Hence, this is the required solution.
Given :
A farmer is pushing a 75 kg plow across a field with 700 N at an angle of 35 degrees above the ground.
The kinetic frictional force between the plow and the field is 250 N.
To Find :
The Normal Force from the ground onto the plow.
The acceleration of the plow.
Solution :
Normal force = mg + F sin Ф
= 75×10 + 700 × sin 35°
= 1151.50 N
Now,
ma = 700 cos Ф - 250
75×a = 323.4
a = 4.31 m/s²
Hence, this is the required solution.