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2 years ago

12

A thunderclap sends a sound wave through the air and the ocean below. The

1 answer:

marysya [2.9K]2 years ago

8 0

Answer:

C. 14.93 m

Explanation:

The given frequency of the wave, f = 100 Hz

The given equation for the wave speed, <em>v</em>, is presented as follows;

v = f × λ

The speed of sound in water, v = 1,493 m/s

Therefore, we get;

The wavelength, λ = v/f

∴ λ = 1,493 m/s/(100 Hz) = 14.93 m

The wavelength, λ = 14.93 m.

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What will happen if the two plungers are pressed together firmly? Explain your answer.

steposvetlana [31]

Answer:

This air being forced out causing the air pressure inside to be much lower than that on the outside. As higher air pressure always pushes, it keeps the two plungers together.

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2 years ago

Two circular loops of wire, each containing a single turn, have the same radius of 5.10 cm and a common center. The planes of th

Fofino [41]

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2 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

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$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

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2 years ago

The theoretical line perpindicular to the surface where a light ray hits a mirror is called the

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That's called the "normal" to the surface at that point.

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2 years ago

A motor keep a Ferris wheel (with moment of inertia 6.97 × 107 kg · m2 ) rotating at 8.5 rev/hr. When the motor is turned off, t

Talja [164]

Answer:

$P = 133.13 Watt$

Explanation:

Initial angular speed of the ferris wheel is given as

$\omega_i = 2\pi f$

$\omega_i = 2\pi(8.5/3600)$

$\omega_i = 0.015 rad/s$

final angular speed after friction is given as

$\omega_f = 2\pi f$

$\omega_f = 2\pi(7.5/3600)$

$\omega_f = 0.013 rad/s$

now angular acceleration is given as

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

$\alpha = \frac{0.015 - 0.013}{15}$

$\alpha = 1.27 \times 10^{-4} rad/s^2$

now torque due to friction on the wheel is given as

$\tau = I \alpha$

$\tau = (6.97 \times 10^7)(1.27 \times 10^{-4})$

$\tau = 8875.3 N m$

Now the power required to rotate it with initial given speed is

$P = \tau \omega$

$P = 8875.3 \times 0.015$

$P = 133.13 Watt$

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2 years ago