The stiffness constant of the spring is 68,290.3 N/m
<h3>
Stiffness constant of the spring</h3>
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
- v is speed = 60 km/h = 16.67 m/s
- x is the distance
k = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
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188m - 556m is the wavelength range for AM radio.
2.77 - 3.40m is the wavelength range for FM radio.
1) For AM radio
f(max) = 1600 × 10³ Hz
f(min) = 540 ×10³Hz
c = fλ
λ = c/f where,
c = speed of light
f = frequency
λ= wavelength
So,
λ(min) = 3 ×
/ 1600 × 10³ = 188 m
λ(max) = 3 × 10⁸/540 ×10³ = 556 m
So wavelength range of AM radio is 188 m - 556m
2) For FM radio
f(max) = 108 × 10⁶ Hz
f(min) = 88 × 10⁶ Hz
λ(min) = 3 × 10⁸ / 108 × 10⁶ = 2.77 m
λ(max) = 3 × 10⁸ / 88 × 10⁶ = 3.40 m
So wavelength range of FM radio is 2.77m - 3.40m
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Answer:

Explanation:
<u>Accelerated Motion</u>
It occurs when an object changes its speed over time. If the changes in speed are uniform, then the acceleration is constant, positive if the speed increases, negative if the speed decreases.
The acceleration is calculated as follows:

The aeroplane starts with a speed of vo=62 m/s and reaches a speed of vf=6 m/s in t=35 s.
The acceleration is:


Answer:
First calculate the deceleration making use of the formulation: V(very last speed= U(initial speed) -a(acceleration) t(ime) so 10=20- 4a. unfavourable a as decelerating. making a the acceleration the challenge by using rearranging the elements 4a= 20-10 = 10 so a=10/4 making use of Newton's 2d regulation P(stress) = M (mass) x A (acceleration). so P = 800 x 10/4 = 8000 /4 = 2000 newtons
Explanation: