The frequency of the radio station is
For radio waves (which are electromagnetic waves), the relationship between frequency f and wavelength
is
where c is the speed of light. Substituting the frequency of the radio station, we find the wavelength:
Answer:
= 3289.8 m / s
Explanation:
This exercise can be solved using the definition of momentum
I = ∫ F dt
Let's replace and calculate
I = ∫ (at - bt²) dt
We integrate
I = a t² / 2 - b t³ / 3
We evaluate between the lower limits I=0 for t = 0 s and higher I=I for t = 2.74 ms
I = a (2,74² / 2- 0) - b (2,74³ / 3 -0)
I = a 3,754 - b 6,857
We substitute the values of a and b
I = 1500 3,754 - 20 6,857
I = 5,631 - 137.14
I = 5493.9 N s
Now let's use the relationship between momentum and momentum
I = Δp = m - m v₀o
I = m - 0
= I / m
= 5493.9 /1.67
= 3289.8 m / s
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
