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sergey [27]
4 years ago
7

An object is in simple harmonic motion. The rate at which the object oscillates may be described using the period T, the frequen

cy f, and the angular frequency . If the angular frequency decreases, what is the effect on the period and the frequency?
Physics
1 answer:
Leto [7]4 years ago
3 0

Explanation:

The relation between angular frequency and frequency is directly proportional:

\omega=2\pi f

So, if the angular frequency decreases, the frequency also decreases.

The relation between angular frequency and period is inversely proportional:

\omega=\frac{2\pi}{T}

So, if the angular frequency decreases, the period increases.

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Answer: c. increased sensitivity to ADH

Explanation:

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b. a reduction in the GFR (glomerular filtration rate): The GFR tend to decline in older age even though there is no disease. These people are required to check with the GFR in future.

d. problems with the micturition reflex: With aging people experience problem of bladder control. This leads to leakage or incontinence of urine or urinary retention that is inability to empty the bladder.

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4 years ago
Question 6 of 10
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C (11.7)

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cos 33 ÷ 14 = 11.7

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Animals which only consume other animals are called
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Milwaukee is 121 in Miles Due West of Grand Rapids Maria drives 255 miles in 4.75 hours from Grand Rapids to Milwaukee find her
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                            =  255 miles / 4.75 hours

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6 0
3 years ago
A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t
QveST [7]

Explanation:

(a)   First, we will calculate the number of moles as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

       No. of moles = \frac{\text{mass}}{\text{molar mass}}

                             = \frac{\text{100 g}}{4 g/mol}  

                             = 25 mol

According to the ideal gas equation,

                           PV = nRT

or,       (P_{2} - P_{1})V = nR (T_{2} - T_{1})

          (6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T

          \Delta T = 336.17 K

Hence, temperature change will be 336.17 K.

(b)   The total amount of heat required for this process will be calculated as follows.

                   q = mC \Delta T

                      = 100 g \times 5.193 J/g K \times 336.17 K

                      = 174573.081 J/K

or,                  = 174.57 kJ/K        (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0
3 years ago
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