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klio [65]
2 years ago
14

A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floo

r and that its period is 27.5 s. (a) How tall is the tower
Physics
1 answer:
wariber [46]2 years ago
4 0

Answer: 187.68 m

Explanation:

Given

Time period T=27.5\ s

The time period of a pendulum is given by

\Rightarrow T=2\pi \sqrt{\dfrac{L}{g}}

where L=height of tower

Put values

\Rightarrow 27.5=2\pi \sqrt{\dfrac{L}{9.8}}\\\\\Rightarrow L=\dfrac{27.5^2\times 9.8}{(2\pi )^2}=187.68\ m

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Answer:

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Explanation:

Let the distance from spotlight to wall be 15m, and distance from the man to the building be x.

#Therefore the height of the shadow as a function of the above is s=(15-x )m

Hence, height of the shadow is expressed as s=(15-x)m

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3 years ago
Which type of electromagnetic waves has highest frequency​
vesna_86 [32]

Answer:

Gamma rays have the highest energies.

Explanation:

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4. What is the momentum of a 70 kg object traveling at 20 m/s?
Soloha48 [4]

Answer:

1400 units of momentum.

Explanation:

Using the formula p=mv. We can get the momentum using 70*20 =1400 units of momentum

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3 years ago
A 45.0 g hard-boiled egg moves on the end of a spring with force constant 25.0 N/m. Its initial displacement 0.500 m. A damping
leva [86]

Answer:

0.02896 kg/s

Explanation:

A_1 = Initial displacement = 0.5 m

A21 = Final displacement = 0.1 m

t = Time taken = 0.5 s

m = Mass of object = 45 g

Displacement is given by

x=Ae^{-\dfrac{b}{2m}t}cos(\omega t+\phi)

At maximum displacement

cos(\omega t+\phi)=1

\\\Rightarrow A_2=A_1e^{-\dfrac{b}{2m}t}\\\Rightarrow \dfrac{A_1}{A_2}=e^{\dfrac{b}{2m}t}\\\Rightarrow ln\dfrac{A_1}{A_2}=\dfrac{b}{2m}t\\\Rightarrow b=\dfrac{2m}{t}\times ln\dfrac{A_1}{A_2}\\\Rightarrow b=\dfrac{2\times 0.045}{5}\times ln\dfrac{0.5}{0.1}\\\Rightarrow b=0.02896\ kg/s

The magnitude of the damping coefficient is 0.02896 kg/s

6 0
3 years ago
Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0
2 years ago
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