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Gemiola [76]
4 years ago
6

A distance of 200 ft must be taped in a manner to ensure a standard deviation smaller than ± 0.04 ft. What must be the standard

deviation per 100 ft tape length to achieve the desired precision?Express your answer to three significant figures and include appropriate units.
Physics
1 answer:
Nutka1998 [239]4 years ago
3 0

Answer:

The error in tapping is ±0.02828 ft.

Explanation:

Given that,

Distance = 200 ft

Standard deviation = ±0.04 ft

Length = 100 ft

We need to calculate the number of observation

Using formula of number of observation

n=\dfrac{\text{total distance}}{\text{deviation per tape length}}

Put the value into the formula

n=\dfrac{200}{100}

n=2

We need to calculate the error in tapping

Using formula of error

E_{series}=\pm E\sqrt{n}

E=\dfrac{E_{series}}{\sqrt{n}}

Put the value into the formula

E=\dfrac{0.04}{\sqrt{2}}

E=\pm 0.02828\ ft

Hence, The error in tapping is ±0.02828 ft.

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You are standing on a street corner with your friend. You then travel 14.0 m due west across the street and into your apartment
Margarita [4]

Answer:

Explanation:

We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .

14 m due west = - 14 i

22.0 m upward in the elevator = 22 k

12 m north = 12 j

6.00 m east = 6 i

Total displacement = - 14 i + 22 k + 12 j + 6 i

D = - 8 i + 12 j + 22 k

magnitude = √ ( 8² + 12² + 22² )

= √ ( 64 + 144 + 484 )

= √ 692

= 26.3 m

Net displacement from starting point = 26.3 m .

5 0
3 years ago
Wha is the frequency of a wave having a period equal to 18 seconds?
Rainbow [258]

Answer:

5.5 × 10-2 hertz

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

= 0.055 per second          (1 cycle per second = 1 Hertz)

Thus, we can conclude that the frequency of the wave is 5.5 X 10^{-2} hertz.

Hopes this helps, love <3

5 0
3 years ago
Please answer for me thanks
Vinvika [58]

I believe one of these are called Elliptical, Spiral, and Irregular Galaxies.

6 0
3 years ago
Una bola de 1 kg gira alrededor de un circulovrtical en el extremo de un cuerda. El otro extremo de la cuerda esta fijo en el ce
Vladimir79 [104]

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}

Donde:

T_{min}, T_{max} - Tensiones mínima y máxima, medidas en newtons.

m - Masa de la bola, medida en kilogramos.

g - Constante gravitacional, medida en metros por segundo al cuadrado.

L - Distancia con respecto al eje de rotación, medida en metros.

v - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

T_{min} + m\cdot g = T_{max} - m\cdot g

Ahora, la diferencia entre las tensiones máxima y mínima es:

T_{max} - T_{min} = 2\cdot m \cdot g

Si m = 1\,kg y g = 9.807\,\frac{m}{s^{2}}, entonces:

T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

T_{max}-T_{min} = 19.614\,N

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

5 0
3 years ago
In a control system, an accelerometer consists of a 4.70-g object sliding on a calibrated horizontal rail. A low-mass spring att
irga5000 [103]

Answer:

7.37 N/m

Explanation:

Given:

Mass of the object = 4.70 g = 0.0047 kg

acceleration = 0.800 g = 0.800 × 9.81 = 7.848 m/s²

Displacement = 0.500 cm = 0.005 m

Now,

Force by the object = Mass × Acceleration

= 0.0047 × 7.848

= 0.0368856 N

and,

Force by the spring = kx

where, x is the displacement

k is the spring constant

thus,

0.0368856 = k × 0.005

or

k = 7.37 N/m

6 0
3 years ago
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