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3 years ago

If there are 1.609 km in a mile, convert 135 miles/hour into meters per second. There are 1000 m in a kilometer.

Physics

2 answers:

Effectus [21]3 years ago

4 0

Answer:

97.1037936

Explanation:

lukranit [14]3 years ago

4 0

Answer:

60.3375 m/ per sec

Explanation:

first multiply 1609 (the amount of meters in a mile) by 135( the total number of miles). this brings you to conclude there are a total of 217,215 meters in 135 miles.

So 135 mph = 217,215 meters per hour.

From here I took a simple route so as to not need a calculator.

217,215 meters = 60mins. (divide both sides by 2)

108,607.5 = 30mins. (divide both sides by 2)

54,303.75 = 15mins. (divide both sides by 15)

3620.25 = 1min

3620.25 = 60secs. (divide both sides by 60)

60.3375 = 1sec

60.3375 meters per second = 135miles per hr.

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g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

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Lostsunrise [7]

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~Deceptiøn

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Jack tries to place magnets on his refrigerator at home, but they won’t stick. What could be the reason?

saul85 [17]

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Artyom0805 [142]

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