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Grace [21]
2 years ago
5

Which of the following statements is an accurate description of vibrations?

Physics
2 answers:
dexar [7]2 years ago
6 0
<span>So we want to know what statement is an accurate description of vibrations. So humans can hear sound frequencies from 20-20000 Hz. Below 20 Hz is infra sound and above 20000 Hz is ultra sound. Humans cant hear both infra sound and ultra sound so the correct answer is A.</span>
Daniel [21]2 years ago
6 0

Infrasonic and ultrasonic vibrations can’t be heard by humans because they’re outside of the range of frequencies that can be detected. Infrasonic vibrations have a lower frequency than can be heard by the human ear; ultrasonic vibrations have a higher frequency than can be heard by the human ear. Infrasonic vibrations can be felt if the amplitude is great enough. Ultrasonic vibrations are utilized in sonar equipment and in applications to detect flaws in steel castings and to remove grease or foreign material from machine-part surfaces.

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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
3 years ago
Energy can come from electricity, but it<br> can also come from water<br> false<br> true
Hitman42 [59]

Answer:

the answer is truth

Explanation:

6 0
3 years ago
Read 2 more answers
Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche
DaniilM [7]

Answer:

R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}

Explanation:

Using kinematics equations:

\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t

Use \Delta y = 0 due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

t=\frac{2v_{0y}}{g}

Use the expression in the first equation to have

R = \frac{2v^2 \cos\theta\sin\theta}{g}

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

5 0
3 years ago
How do you find the range of a data set?
Helga [31]

The answer is A.

The answer is A.

The answer is A.

6 0
3 years ago
If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east
Montano1993 [528]

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta

F_{3}=2mg\cos\theta

Put the value into the formula

F_{3}=2\times0.225\times10\cos20^{\circ}

F_{3}=4.228\ N

We need to calculate the voltage

Using formula of voltage

Voltage =sensitivity\times F_{3}

Put the value into the formula

Voltage=0.5\times4.228

Voltage =2.114\ V

Hence, The voltage is 2.114 V.

7 0
3 years ago
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