All categories

2 years ago

A cellular telephone transmits electromagnetic waves at a frequency of 835 mhz. what is the wavelength of these waves?

1 answer:

4 0

HI!

instead of just giving you the answer, i will help you to understand it in simple terms.

As electromagnetic waves (read: light) move forwards it 'vibrates' up and down. The frequency of light tells us how many up-down vibrations it makes each second (1hz = 1 up-down vibration per second & 1mhz = 1 million up-down vibrations per second). Each individual vibration is called one wave.

Since we also know how many meters light travels in 1 second (290,000,000 meters), we can easily see that (in the case of this specific question) every '290,000,000' meters, there are '835,000,000' vibrations or waves.

It's then just a simple division to work out what is the length of just one wave! - which is what we call the wavelength of light at that frequency.

290,000,000 / 835,000,000 = 0.34m (or 34cm)

You might be interested in

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago

What is the mass of an object that creates 33,750 joules of energy by traveling at 30 m/sec?

nikklg [1K]

The Energy is Kinetic Energy.

Kinetic Energy = 1/2*mv², Where m is mass in kg, v is velocity in m/s

Energy is 33750 Juoles, v = 30m/s

1/2*mv² = E

1/2*m*30² = 33750

m = (2*33750) / (30²) Using a calculator

m = 75 kg

Mass of object is 75 kg.

5 0

2 years ago

Read 2 more answers

A sample of tendon 3.00 cm long and 4.00 mm in diameter is found to break under a minimum force of 128 N. If instead the sample

Hatshy [7]

Answer:128 N

Explanation:

Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .

If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.

3 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

where,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago

Your low-flow showerhead is delivering water at 1.2×10−4m3/s, about 2.0 gallons per minute.

Oksanka [162]

To solve this problem it is necessary to apply the fluid mechanics equations related to continuity, for which the proportion of the input flow is equal to the output flow, in other words:

$Q_1 = Q_2$