.) List the molecules of elements in the sample of air. Give the

One method of determining the location of the center of gravity of a person is to weigh the person as he/she lies on a board of

andrey2020 [161]

Answer:

x = 1.018 m

Explanation:

given,

height of man = 190 cm

= 1.9 m

scale reading on left = 450 N

scale reading on the right = 390 N

Let center of gravity of man be x distance from feet, feet is on right side.

For system to be in equilibrium moment about center should be equal to zero.

∑M = 0

now,

450(1.9 - x ) - 390 × x = 0

450(1.9 - x ) = 390 × x

855 - 450 x = 390 x

840 x = 855

$x = \dfrac{855}{840}$

x = 1.018 m

hence, point of center of gravity from feet is equal to x = 1.018 m

7 0

4 years ago

A falling stone is at a certain instant 90 feet above the ground. two seconds later it is only 10 feet above the ground. if it w

choli [55]

The equation relevant to this is:

S(t) = So + Vot - At²/2

Therefore we can create two equations:
S(t) = 90 = So - 4t - 16.1t² --> eqtn 1
S(t+2) = 10 = So - 4(t+2) - 16.1(t+2)² --> eqtn 2

Expanding eqtn 2:
10 = So - 4t - 8 - 16.1(t² + 4t + 4)
10 = So - 4t - 8 - 16.1t² - 64.4t - 64.4
10 + 8 + 64.4 = So - 68.4t - 16.1t²
82.4 = So - 68.4t - 16.1t² --> eqtn 3

Subtracting eqtn 1 by eqtn 3:

90 = So - 4t - 16.1t²

82.4 = So - 68.4t - 16.1t²

=> 7.6 = 64.4t

t = 0.118 s

Therefore calculating for initial height So:

82.4 = So - 68.4(0.118) - 16.1(0.118)²

So = 90.7 ft

7 0

3 years ago

describe how you would connect a voltmeter to measure the potential difference across a resistor and an ammeter to measure the c

Vesna [10]

A voltmeter needs to be connected in parallel to the two sites in order to measure the potential difference between them.

<h3>What is a voltmeter?</h3>

The electrical potential difference between two locations in an electric circuit is measured using a voltmeter.
The potential difference between two ends of a current-carrying conductor is measured using a voltmeter.
On a scale that is typically measured in volts, millivolts (0.001 volts), or kilovolts, a voltmeter primarily monitors the voltage of direct or alternating electric current (1,000 volts).
Numerous voltmeters are digital and display readings as numbers.

<h3>How is a voltmeter connected?</h3>

A galvanometer with high resistance is a voltmeter.
Any portion of an electrical circuit can be measured.
Due to its high resistance, it is linked in parallel so that it does not impact the circuit's overall net resistance and does not draw any current by itself.
A device's voltage is measured by a voltmeter connected in parallel, and its current is measured by an ammeter connected in series.

Learn more about voltmeters here

brainly.com/question/83321

#SPJ4

3 0

2 years ago

In dim light

faltersainse [42]

The Answer is A, the iris dilates the pupil.

4 0

3 years ago

Read 2 more answers

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

8 0

3 years ago