Answer:
Average speed of car in the first trip is 10 km/hr
Explanation:
It is given that first the car drives 6 hours to the east
Then travels 12 km to west in 3 hours
Average speed for the entire trip = 8 km/hr
Total time = 3+6 = 9 hour
So distance traveled in 9 hour = 9×8 = 72 km
As the car travel 12 km in west so distance traveled in east = 72-12 = 60 km
Time by which car traveled in east = 6 hour
So speed 
So average speed of car in the first trip is 10 km/hr
Answer:
3.44 W/m²
1.134 J
Explanation:
E₀ = Intensity of electric field = 50.9 V/m
I = Intensity of electromagnetic wave
Intensity of electromagnetic wave is given as
I = (0.5) ε₀ E₀² c
I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)
I = 3.44 W/m²
A = Area = 0.0277 m²
t = time interval = 11.9 s
Amount of energy is given as
U = I A t
U = (3.44) (0.0277) (11.9)
U = 1.134 J
Answer:
17. h = l − l cos θ
18. 1.40 m
Explanation:
Let's call d the height of the triangle. We can then say:
h = l − d
Using trig, we can write d in terms of l and θ:
d = l cos θ
h = l − l cos θ
If l = 6 m and l cos θ = 40°:
h = 6 − 6 cos 40
h ≈ 1.40
Answer: please find the answer in the explanation.
Explanation:
Harmonic can be experienced by any body that repeats itself. The pattern can be sinusoidal, square, tooth etc.
The fundamental differences between the harmonic oscillator dynamics and the simple pendulum dynamics are:
1.) The harmonic oscillator dynamics can be sinusoidal or square wave so far the motion is periodic while the simple pendulum dynamics is always sinusoidal.
2.) In simple pendulum dynamics, the period of oscillation is independent of the amplitude. While the period in harmonic oscillator dynamics depends on the amplitude.
3.) Differential equation is only one method to analyze the simple pendulum dynamics where there are several methods to analyze the harmonic oscillator dynamics.
Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y =
t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
=
- gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch