Answer: 2.4×10^-3 v/m
Explanation: distance between plates of capacitor (d) =5.0×10^-3m
Potential difference between plates (v) = 12v
Force on electronic charge (f) = 3.8×10^-16 N
Strength of electric field (E) =?
The formulae that relates potential difference, eoectiic field strength and distance between plates is given as
v = Ed
By substituting the parameters, we have that
12 = E × 5.0×10^-3
E = 12/ 5.0 × 10^-3
E = 2.4×10^-3 v/m
It could be radiation because radiation means <span>the emission of energy as electromagnetic wave, but potential fits it best</span>
Answer:
7 m/s^2
Explanation:
Given that the jet is traveling 37.6 m/s when the pilot receives the message.
And it takes the pilot 5.37 s to bring the plane to a halt.
Acceleration of the plane can be calculated by using first equation of motion
V = U - at
Since the plane is going to stop, the final velocity V = zero.
And the acceleration will be negative
Substitute all the parameters into the formula
0 = 37.6 - 5.37a
5.37a = 37.6
Make a the subject of formula
a = 37.6 / 5.37
a = 7.0 m/s^2
Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2
Power = Iω (constant) as they are connected together, since effort axle has large radius than resistance axle, so moment of inertia of effort axle is also more as compared to resistance axle, so angular speed of effort axle is less than the resistance axle. So answer is B. resistance axle will have more angular speed as its moment of inertia is less for the same power.
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
