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dangina [55]
3 years ago
9

Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con

clude is also 8 N?
Physics
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

<h2>Frictional force</h2>

Explanation:

For the block placed on the table, there are several force acting on the body along the horizontal and vertical axis. All this forces tends to keep the body in a state of equilibrium.The forces acting along the horizontal are the moving force (Fm) and the frictional force (Ff).

Frictional force are forces that acts opposite to the force that causes the body to move (moving force).

If Lien uses a spring scale to pull a block toward the right across the lab table and the scale reads 8 N, this means that the force that causes the body to move is the 8N force (moving force).

Taking the sum of force along the horizontal;

\sum fx = ma_x

Since the body is static, max = 0

\sum fx = 0\\fm+(-Ff)= 0

Note that the frictional force acts is the force of opposition acting in the negative x direction.

fm = 0+Ff\\fm = Ff

Since Fm = 8N, Ff will also be equal to 8N.

Based on the above proof, Lien can also conclude that 8 N is a frictional force

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a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

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Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

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