Answer:
Option D is correct: 170 µW/m²
Explanation:
Given that,
Frequency f = 800kHz
Distance d = 2.7km = 2700m
Electric field Eo = 0.36V/m
Intensity of radio signal
The intensity of radial signal is given as
I = c•εo•Eo²/2
Where c is speed of light
c = 3×10^8m/s
εo = 8.85 × 10^-12 C²/Nm²
I = 3×10^8 × 8.85×10^-12 × 0.36²/2
I = 1.72 × 10^-4W/m²
I = 172 × 10^-6 W/m²
I = 172 µW/m²
Then, the intensity of the radio wave at that point is approximately 170 µW/m²
<span>160 Joules
For this problem, we can ignore the vertical component of the applied force and focus on only the horizontal component of 80 N and since work is defined as force over distance, let's multiply the force by the distance:
80 N * 2.0 m = 160 Nm = 160 kg*m^2/s^2 = 160 Joules.
So the cart has a final kinetic energy of 160 Joules.</span>
C. meter per second
Velocity and speed share the same SI unit.
Answer:
34.8 and 55.2º
Explanation:
This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation
The scope equation is
R = v₀² sin 2θ /g
sin 2 θ = R g / v₀²
sin 2 θ = 31 9.8 / 18²
2 θ = sin⁻¹ 0.93765
θ = 34.8º
At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º
Answer:250.03m
Explanation:
Initial velocity =u=51mph
Angle=β=37
Acceleration due to gravity=g=10
The distance covered by the football is the range
Range=(u^2xsin2β)/g
Range=(51^2xsin(2x37))/10
Range=(51x51xsin74)/10
Range=(2601xsin74)/10
Range=(2601x0.9613)/10
Range=(2500.3)/10
Range=250.03m
