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sukhopar [10]
3 years ago
6

If the coefficient alpha for a stress scale was computed to be 0.80, the scale would be:

Physics
1 answer:
Lina20 [59]3 years ago
5 0

If the coefficient alpha for a stress scale was computed to be 0.80, the scale would be strongly reliable. A coefficient alpha that is at least 0.70 and above is considered to have a strong internal consistency, which means the items in the scale are closely related as a group.

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In the fusion reaction occurring in the Sun, how does the mass of helium atom formed compare to the mass of the hydrogen atoms?
PolarNik [594]
B bc there is a lot of hydrogen used

3 0
3 years ago
The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values
Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level  β  = 10 log(I/I_{0})        ......................a

put here value (I/I_{0}) = 10^−12 W/m² and  β = 115

115  = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r²    ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m    = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0
4 years ago
(a) Calculate the absolute pressure at the bottom of a freshwater lake at a point whose depth is 30.0 m. Assume the density of t
Law Incorporation [45]

Answer:

(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b) The force exerted by the water on the window is 36101.5 N

Explanation:

(a)

The absolute pressure is given by the formula

P = P_{o} + \rho gh

Where P is the absolute pressure

P_{o} is the atmospheric pressure

\rho is the density

g is the acceleration due to gravity (Take g = 9.8 m/s^{2} )

h is the height

From the question

h = 30.0 m

\rho = 1.00 × 10³ kg/m³ = 1000 kg/m³

P_{o} = 101.3 kPa = 101300 Pa

Using the formula

P = P_{o} + \rho gh

P = 101300 + (1000×9.8×30.0)

P = 101300 + 294000

P =395300 Pa

P = 395.3 kPa

Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa

(b)

For the force exerted

From

P = F/A

Where P is the pressure

F is the force

and A is the area

Then, F = P × A

Here, The area will be area of the window of the underwater vehicle.

Diameter of the circular window = 34.1 cm = 0.341 m

From Area = πD²/4

Then, A = π×(0.341)²/4 = 0.0913269 m²

Now,

From F = P × A

F = 395300 × 0.0913269

F = 36101.5 N

Hence, the force exerted by the water on the window is 36101.5 N

5 0
3 years ago
A 65-kg person stands on a scale in a moving elevator while holding a 5.0 kg mass suspended from a massless spring with spring c
disa [49]

Answer:

The question is incomplete. However, I believe, it is asking for the acceleration of the elevator. This is 3.16 m/s².

Explanation:

By Hooke's law, F = ke

F is the force on a spring, k is the spring constant and e is the extension or compression.

From the question,

F = (1.08\text{ kN/m}) \times (6.0 \times 10^{-2}\text{ m}) = 64.8 \text{ N}

This is the force on the mass suspended on the spring. Its acceleration, a, is given by

F = ma

a = \dfrac{F}{m}

a = \dfrac{64.8 \text{ N}}{5\text{ kg}} = 12.96\text{ m/s}^2

This acceleration is more than the acceleration due to gravity, g = 9.8 m/s². Hence the elevator must be moving up with an acceleration of

12.96 - 9.8 m/s² = 3.16 m/s²

7 0
3 years ago
A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur
Firlakuza [10]

Answer:

α =18.75  rad/s²

Explanation:

Given that

Acceleration a = 0.15 g

We know that   g =10 m/s²

a= 0.15 x 10 = 1.5  m/s²

d= 16 cm

Radius r= 8 cm

Lets take angular acceleration =α rad/s²

As we know that

a= α r

Now by putting the values

1.5 = α  x 0.08

α =18.75  rad/s²

6 0
4 years ago
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