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rosijanka [135]
1 year ago
6

new generation cordless phones use a 9.00x10^2 MHz frequency and can be operated up to 60.0 m from their base. how many waveleng

ths of the electromagnetic waves can fit between your ear and a base 60.0 m away?
Physics
1 answer:
Musya8 [376]1 year ago
3 0
Λ = 3*10^8 / 9*10^8 = 1/3 m
no. of wavelengths = 60/(1/3) = 180
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A 145-g baseball is thrown so that it acquires a speed of 25 m/s. What was the net work done on the ball to make it reach this s
inysia [295]

When the ball has left your hand and is flying on its own, its kinetic energy is

KE = (1/2) (mass) (speed²)

KE = (1/2) (0.145 kg) (25 m/s)²

KE = (0.0725 kg) (625 m²/s²)

<em>KE = 45.3 Joules</em>

If the baseball doesn't have rocket engines on it, or a hamster inside running on a treadmill that turns a propeller on the outside, then there's only one other place where that kinetic energy could come from:  It MUST have come from the hand that threw the ball.  The hand would have needed to do  <em>45.3 J</em>  of work on the ball before releasing it.

6 0
2 years ago
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

n_{r}=1.572

n_{b}=1.587

We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

Put the value into the formula

1.572\sin37=1\times\sin\theta_{r}

\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})

\theta_{r}=71.0^{\circ}

We need to calculate the angle for blue wavelength

Using Snell's law,

n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}

Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

We need to calculate the angle between the red and blue light

Using formula of angle

\Delta \theta=\theta_{b}-\theta_{r}

Put the value into the formula

\Delta \theta=72.7-71.0

\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

8 0
2 years ago
In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
A fish is 80 cm below the surface of a pond. What is the apparent depth (in cm) when viewed from a position almost directly abov
Rudik [331]

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

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2 years ago
Do the data for the first part of the experiment support or refute the first hypothesis? Be sure to explain your answer and incl
Sphinxa [80]

The data for the first part of the experiment support the first hypothesis. As the force applied to the cart increased, the acceleration of the cart increased. Since the increase in the applied force caused the increase in the cart's acceleration, force and acceleration are directly proportional to each other, which is in accordance with Newton's second law.

5 0
2 years ago
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