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Alexandra [31]
4 years ago
9

As microwave light travels through a liquid, it moves at a speed of 2.2 x 108 m/s. If the frequency of this light wave is 1.1 x

108 Hz, what is the wavelength of this microwave?
Physics
1 answer:
Alexandra [31]4 years ago
7 0

λ = 2 m.

The easiest way to solve this problem is using the equation of frecuency of a wave f = v/λ, where v is the velocity of the wave, and λ is the wavelength.

To calculate the wavelength of a microwave light travels through a liquid, it moves at a speed of 2.2 x 10⁸ m/s. If the frecuency of the light wave is 1.1 x 10⁸ Hz, we have to clear λ from the equation f = v/λ:

f = v/λ -------> λ = v/f

λ = 2.2 x 10⁸ m/s / 1.1 x 10⁸ Hz

λ = 2 m (wavelength of the microwave)

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cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
The graph represents the reaction 2H2 + 02 32H20 as it reaches
Alex

Answer:

C and D

Explanation:

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v = s/t = 300/300 = 1m/s

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