Answer:
The magnitude of rate of change of electric field is 
.
Explanation:
Given that,
Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m
Total displacement current through a cross section of the region, 
We need to find the rate of change of electric field. Its is given by the formula as follows :
So, the magnitude of rate of change of electric field is .
First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
The force tending to lift the load (vertical force) is equal to <u>22.5N.</u>
Why?
Since the boy is pulling a load (150N) with a string inclined at an angle of 30° to the horizontal, the total force will have two components (horizontal and vertical component), but we need to consider the given information about the tension of the string which is equal to 105N.
We can calculate the vertical force using the following formula:
Hence, we can see that <u>the force tending to lift the load</u> off the ground (vertical force) is equal to <u>22.5N.</u>
Have a nice day!
Answer:
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Explanation:
The average speed is defined by the variation of the position between the time spent
v = Δx / Δt
since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is
Δx = 20 - 15 +20
Δx = 25 m
therefore we calculate
v = 25/75
v = 1/3 m / s = 0.333 m / s
in the direction of the truck
Summary:
a= 12.0 m/(s^2)
v= 100m/s
t1= 2.0s => s1=?
t2=5.0s => s2=?
t3=10.0s => s3=?
——————
Solution:
• when t1=2.0 s, I have gone:
S1= v*t1 + 1/2*a*(t1^2)
=100.0 *2 + 1/2*12.0*(2.0^2)
=224 (m)
• when t2=5.0s, I have gone
S2=v*t2+ 1/2*a*(t2^2)
= 100*5.0+ 1/2*12.0*(5.0^2)
=650 (m)
•when t3= 10.0s, I have gone:
S3=v*t3+ 1/2*a*(t3^2)
=100*10.0+ 1/2*12*(10.0^2)
=1600 (m)
