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Harman [31]
3 years ago
15

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the

mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent?
Physics
1 answer:
LiRa [457]3 years ago
8 0

Answer:

a)   M = 2,939  10¹⁷ kg , b)   M_{e} / M  = 2 10⁷

Explanation:

a) The equation for gravitational force is

         F = G m M / r²

Where G is the gravitational constant that is worth 6.67 10⁻¹¹ N m² / kg², m the mass epa person. M the mass of the Mountain and r the distance between them.

The value of this force is 2% of the person's weight

          F = 0.02 W = 0.02 mg

we replace

         0.02 mg = G m M / r²

         M = 0.02 g r² / G

         r = 10 km = 10 10³ m = 1.0 10⁴ m

         M = 0.02  9.8  (10⁴)² / 6.67 10⁻¹¹

         M = 2,939  10¹⁷ kg

b) to compare the masses we find their relationship

       M_{e} / M = 5.98 1024 / 2,939 1017

       M_{e} / M  = 2 10⁷

c) treating the mountain as a point object

d) The mountain is not spherical so the distance changes depending on the height of the mountain

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Answer:

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7 0
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What is the sound intensity level in decibels? Use the usual reference level of I0 = 1.0×10−12 W/m2.
jek_recluse [69]

Answer:

L = 130 decibels

Explanation:

The computation of the sound intensity level in decibels is shown below:

According to the question, data provided is as follows

I = sound intensity = 10 W/m^2

I0 = reference level = 1 \times 10-12 W/m^2

Now

Intensity level ( or Loudness)is

L = log10 \frac{I}{10}

L = log10 \frac{10}{1\times 10^{-12}}

L = log10 \times 1013

= 13 \times 1 ( log10(10) = 1)

Therefore  

L = 13 bel

And as we know that

1 bel = 10 decibels

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The  Sound intensity level is

L = 130 decibels

5 0
3 years ago
Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets
PolarNik [594]

 Step by step solution :

standard deviation is given by :

\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}

where, \sigma is standard deviation

\bar{x} is mean of given data

n is number of observations

From the above data, \bar{x}=24.88

Now, if x=24.8, then (x-\bar{x})^2=0.0064

If  x=23.9, then (x-\bar{x})^2=0.9604

if x=26.1, then (x-\bar{x})^2=1.4884

If x=25.1, then (x-\bar{x})^2=0.0484

If x=24.5, then (x-\bar{x})^2=0.1444

so, \sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}

\sum (x-\bar{x})^2 =2.648

\sqrt{\sum \frac{(x-\bar{x})^2}{n}}

\sigma =0.7277

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0
3 years ago
When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

V_{0} = A_{1} V_{1} + A_{2} V_{2} \\

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |

CMRR = |\frac{48}{0.4} |\\CMRR = 120

4 0
3 years ago
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