A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbiting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.

A) 6,800 m/s
B) 7,200 m/s
C) 7,500 m/s
D) 7,900 m/s

The force by the pitcher's hand on the ball during the acceleration phase is found using Newton's Second Law of Motion and is calculated to be 116 Newton.

The force exerted by the pitcher's hand can be found using Newtons Second Law of Motion which states that the force acting on an object is equal to the mass of the object times its acceleration.

it can be expressed as

F = m * a

Here, the mass (m) is 0.145 kg, and the acceleration (a) can be found using the formula a = Δv/Δt, where Δv is the change in velocity (40 m/s) and Δt is the change in time (50 ms or 0.05 s).

So, a = 40/0.05 = 800 m/s², and then the force F = 0.145 * 800 = 116 N. Therefore, the force of the pitcher's hand on the ball during this acceleration phase is 116 Newton.

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The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion.

The force of the pitcher's hand on the ball can be calculated using Newton's second law of motion, which states that force equals mass times acceleration. In this case, the mass of the ball is 0.145 kg and the acceleration is the change in velocity divided by the time interval. The change in velocity is 40 m/s (the final velocity) minus 0 m/s (the initial velocity), and the time interval is 50 ms (or 0.05 s). Therefore, the force can be calculated as:

Force = mass × acceleration = 0.145 kg × (40 m/s - 0 m/s) / 0.05 s = 0.116 N.

So, the force of the pitcher's hand on the ball during this acceleration phase is approximately 0.116 Newtons.

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Answer: A. force with which air rushes across the vocal folds

Explanation:

The human voice is produced in the larynx, whose essential part is the glottis. This is how the air coming from the lungs is forced during expiration through the glottis, making its two pairs of vocal folds to vibrate.

It should be noted that this process can be consciously controlled by the person who speaks (or sings), since the variation in the intensity of the sound of the voice depends on the strength of the breath.

The loudness of a person's voice mainly depends on the force of airflow across their vocal folds. This causes the vocal folds to vibrate and the sound to be produced. The volume of this sound is determined by the amplitude of the resulting sound pressure wave.

The loudness of a person's voice is primarily dependent on the force with which air rushes across the vocal folds. Sound is created when air is pushed up from the lungs through the throat, causing the vocal folds to vibrate. When air flow from the lungs increases, the amplitude of the sound pressure wave becomes greater, resulting in a louder voice. Changes in pitch are related to muscle tension on the vocal cords.

Vocal cord vibration and sound pressure wave amplitude are thus key factors in determining the loudness of a person's voice.

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A) The magnetic field doubles as well

The relationship between rms value of the electric field and rms value of the magnetic field for an electromagnetic wave is the following:

[tex]E_{rms}=cB_{rms}[/tex]

where

E_rms is the magnitude of the electric field

c is the speed of light

B_rms is the magnitude of the magnetic field

From the equation, we see that the electric field and the magnetic field are directly proportional: therefore, if the rms value of the electric field is doubled, then the rms value of the magnetic field will double as well.

B) The intensity will quadruple

The intensity of an electromagnetic wave is given by

[tex]I=\frac{1}{2}c\epsilon_0 E_0^2[/tex]

where

c is the speed of light

[tex]\epsilon_0[/tex] is the vacuum permittivity

[tex]E_0[/tex] is the peak intensity of the electric field

The rms value of the electric field is related to the peak value by

[tex]E_{rms}=\frac{E_0}{\sqrt{2}}[/tex]

So we can rewrite the equation for the intensity as

[tex]I=c\epsilon_0 E_{rms}^2[/tex]

we see that the intensity is proportional to the square of the rms value of the electric field: therefore, if the rms value of the electric field is doubled, the average intensity of the wave will quadruple.

Answer:

Part a)

[tex]L = 18 kg m^2/s[/tex]

Part b)

[tex]v = 0.6 m/s[/tex]

Part c)

[tex]R_{max} = 6.04 kg[/tex]

[tex]R_{min} = 5.96 kg[/tex]

Explanation:

As we know that Ferris wheel start from rest with angular acceleration

[tex]\alpha = 0.001 rad/s^2[/tex]

time taken = 2 min

so here we have its angular speed after t = 2min given as

[tex]\omega = \alpha t[/tex]

[tex]\omega = (0.001)(2\times 60)[/tex]

[tex]\omega = 0.12 rad/s[/tex]

Part a)

Angular momentum of the Penguine about the center of the wheel is given as

[tex]L = I\omega[/tex]

[tex]L = (6\times 5^2)(0.12)[/tex]

[tex]L = 18 kg m^2/s[/tex]

Part b)

tangential speed is given as

[tex]v = r\omega[/tex]

[tex]v = (5)(0.12)[/tex]

[tex]v = 0.6 m/s[/tex]

Part c)

Maximum reading of the scale at the lowest point is given as

[tex]R_{max} = \frac{m\omega^2 r + mg}{g}[/tex]

[tex]R_{max} = \frac{6(0.12^2)(5) + 6(9.81)}{9.81}[/tex]

[tex]R_{max} = 6.04 kg[/tex]

Minimum reading of the scale at the top point is given as

[tex]R_{min} = \frac{mg - m\omega^2 r}{g}[/tex]

[tex]R_{min} = \frac{6(9.81) - 6(0.12^2)(5)}{9.81}[/tex]

[tex]R_{min} = 5.96 kg[/tex]

Answer:

Magnetic field shall be zero at exactly in between the wires.

Explanation:

We can find the magnetic field by biot Savart law as follows

[tex]\overrightarrow{dB}=\frac{\mu _{0}I}{4\pi }\int \frac{\overrightarrow{dl}\times \widehat{r}}{r^{2}}[/tex]

For current carrying wire in positive y direction we have

[tex]\overrightarrow{dB_{1}}=\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}[/tex]

Similarly for wire carrying current in -y direction we have [tex]\overrightarrow{dB_{2}}=\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}[/tex]

Thus the net magnetic field at any point in space is given by

[tex]\overrightarrow{dB_{1}}+\overrightarrow{dB_{2}}[/tex]

[tex]\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}+\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}=0\\\\\Rightarrow \overrightarrow{r_{1}}=\overrightarrow{r_{2}}[/tex]

For points with same position vectors from the 2 wires we have a net zero magnetic field. These points are exactly midway between the 2 wires