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:
angular acceleration is -0.2063 rad/s²
Explanation:
Given data
mass m = 95.2 kg
radius r = 0.399 m
turning ω = 93 rpm
radial force N = 19.6 N
kinetic coefficient of friction μ = 0.2
to find out
angular acceleration
solution
we know frictional force that is = radial force × kinetic coefficient of friction
frictional force = 19.6 × 0.2
frictional force = 3.92 N
and
we know moment of inertia that is
γ = I ×α = frictional force × r
so
γ = 1/2 mr²α
α = -2f /mr
α = -2(3.92) /95.2 (0.399)
α = - 7.84 / 37.9848 = -0.2063
so angular acceleration is -0.2063 rad/s²
Answer:
- <u>77.8 m/s, downward</u>
Explanation:
For uniform acceleration motion, the average speed is equal to half the soum of the initial velocity, Vi, and the final velocity, Vf
- Average speed = (Vf + Vi)/2
Also, by definition, the average speed is the distance divided by the time:
- Average speed = distance / time
Then:
Other kinematic equation for uniform acceleration is:
Since the window is falling and the air resistance is ignored, a = g (gravitational acceleration ≈ 9.8m/s²)
Replacing the known values we can set a system of two equations:
From (Vf + Vi)/2 = 300m/6.62s
(Vf + Vi) = 2 × 300m/6.62s
- Vf + Vi = 90.634 equation 1
From Vf = Vi + a×t
Vf - Vi = 9.8 (6.62)
- Vf - Vi = 64.876 equation 2
Adding the two equations:
- Vf = 77.8 m/s downward (velocities must be reported with their directions)
Wave speed = frequency * wavelength
Wave speed = 4 * 25
Wave speed = 100 m/s
Answer:
Highest point reached = 3.37 m
Explanation:
Initial velocity, = 7.1 m/s
Initial vertical velocity = 7.1 sin 61 = 6.21 m/s
Consider the vertical motion of skateboarder,
We have equation of motion, v² = u² + 2as
Initial velocity, u = 6.21 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = 0 m/s
Substituting
v² = u² + 2as
0² = 6.21² + 2 x -9.81 x s
s = 1.97 m
So from ramp the position it goes up by 1.97 m
Highest point reached = 1.97 + 1.4 = 3.37 m
