Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C
Answer:
137200000 watts or 137200 kilowatts
Explanation:
The formula for power is P= dhrg
Where P = Power in watts
d = density of water (1000 kg/m^3)
h = height in meters
r = flow rate in cubic meters per second,
g = acceleration due to gravity of 9.8 m/s^2,
Plugging in the known values,
we get
P = 1000 kg/m^3 * 80 m * 175 m^3/s * 9.8 m/s^2
P = 80000 kg/m^2 * 175 m^3/s * 9.8 m/s^2
P = 14000000 kg m/s * 9.8 m/s^2
P = 137200000 kg m^2/s^3
P = 137200000 watts or 137200 kilowatts
The above figure assumes 100% efficiency which is impossible. A good efficiency would be 90% so the actual power available would be close to 0.90 * 137200 = 123480 kilowatts
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
Explanation:
since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.
We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.
range can be calculated by the formula :-
u is the velocity during its take off and 
is the angle at which its thrown
Given that
- u = 8m/ s
- = 40°
calculating range using the above formula
value of sin 80 = 0. 985
Hence,
Answer:
Two point charges are separated by 6.4 cm . The attractive force between them is 10 N .
units.
Explanation:
