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Molodets [167]
3 years ago
7

At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is allowed from a speed of 20

m/s to a speed of 5 m/s over a distance of 20 meters. Determine the braking force required to slow the train of cars by this amount.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
6 0

Answer:

56250 N

Explanation:

mass, m = 6000 kg

initial speed, u = 20 m/s

final speed, v = 5 m/s

distance, s = 20 m

Use third equation of motion

v^{2}=u^{2}+2as

5 x 5 = 20 x 20 + 2 a x 20

25 = 400 + 40 a

a = - 9.375 m/s^2

Braking force, F = mass x acceleration

F = 6000 x 9.375

F = 56250 N

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The moon orbits Earth approximately once a month. <br><br> True<br><br> False
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Answer:true

Explanation:

It takes the moon approximately a month to complete an orbit round the earth

5 0
4 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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