According to American Heart Association, your target heart rate is__________. Group of answer choices

A .all of the above

B. 85% - 95% of your max heart rate

C. 20%-30% of your max heart rate

D. 50%-85% of your max heart rate

Answers

Answer 1

Answer:

Answer:

Explanation:

A normal heart rate is from 85-95% the other heart rate is not normal because your heart is beating more than normal

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Tarik winds a small paper tube uniformly with 163 turns of thin wire to form a solenoid. The tube's diameter is 6.13 mm and its length is 2.49 cm . What is the inductance, in microhenrys, of Tarik's solenoid?

Answers

Answer:

The inductance is $L = 40\mu H$

Explanation:

From the question we are told that

The number of turns is $N = 163 \ turns$

The diameter is $D = 6.13 \ mm = 6.13 *10^(-3) \ m$

The length is $l = 2.49 \ cm = 0.0249 \ m$

The radius is evaluated as $r = (d)/(2)$

substituting values

$r = (6.13 *10^(-3))/(2)$

$r = 3.065 *10^(-3) \ m$

The inductance of the Tarik's solenoid is mathematically represented as

$L = (\mu_o * N^2 * A )/(l )$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi *10^(-7) \ N/A^2$

A is the area which is mathematically evaluated as

$A = \pi r^2$

substituting values

$A = 3.142 * [ 3.065*10^(-3)]^2$

$A = 2.952*10^(-5) \ m^2$

substituting values into formula for L

$L = ( 4\pi *10^(-7) * [163]^2 * 2.952*10^(-5) )/(0.0249 )$

$L = 40\mu H$

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

Answers

Answer:

0.97566 m/s

Explanation:

$m_1$ = Mass of cannon = 2260 kg

$v_1$ = Velocity of cannon

$m_2$ = Mass of ball = 21 kg

$v_2$ = Velocity of ball = 105 m/s

As the momentum of the system is conserved we have

$m_1v_1=m_2v_2\n\Rightarrow v_1=(21* 105)/(2260)\n\Rightarrow v_1=0.97566\ m/s$

The velocity of the cannon is 0.97566 m/s

The gravitational force law, deduced by Newton in the 1660's, is remarkably similar to Coulomb's law. Recall that the universal law of gravitation states that the magnitude of the gravitational force between two masses M1 and M2 separated by a distance R is given by the following equation:________. F = G (M1 x M2) / R2
G = 6.67 x 10-11 Nm2/kg2
a. Calculate the value of the gravitational force between an electron (mass = 9.11 x 10-31 kg) and a proton (mass is 1836 times greater than the mass of an electron) if the two particles are separated by 3.602 nanometers. (1 nanometer or 1 nm = 1 x 10-9 m)
F= ______ N
b. The force created in the above question is:
1. repulsive
2. attractive

Answers

Answer:

a. $F=7.83* 10^(-51) N$

b.Attractive

Explanation:

We are given that

$F=(GM_1M_2)/(R^2)$

$G=6.67* 10^(-11) Nm^2/kg^2$

Mass of an electron, $M_1=9.11* 10^(-31) kg$

Mass of proton, $M_2=1836* 9.11* 10^(-31) kg$

Distance between electron and proton,R= $3.602nm=3.602* 10^(-9) m$

$1nm=10^(-9) m$

a.Substitute the values then we get

$F=(6.67* 10^(-11)* 9.11* 10^(-31)* 1836* 9.11* 10^(-31))/((3.602* 10^(-9))^2)$

$F=7.83* 10^(-51) N$

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5, and 5.93. Even though you cannot see the line they actually measured, which of the recorded measurements are possible valid measurements for this instrument, according to its precision? Select all that apply. 1. 5.52 2. 6.63 3. 5.5 4. 5.93

Answers

Answer:

correct answer is 1 and 3

Explanation:

In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm

Based on this appreciation, the valid measurements are 5.52 and 5.5.

the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.

The correct answer is 1 and 3

A Venturi tube may be used as the inlet to an automobile carburetor. If an inlet pipe with a diameter of 2.0 cm diameter narrows to diameter of 1.0 cm, determine the pressure drop in the constricted section for an initial airflow of 3.0 m/s in the 2-cm section? (Assume air density is 1.25 kg/m

Answers

The pressure drop is equal to 80.99 Pa

Given information:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

Here we use Bernoulli's principle for the Venturi Tube:

Calculation of pressure drop:

$P1 - P2 = ((v^2* p)/ 2)* ((A1^2/ A2^2)-1)\n\nP1 - P2 = \Delta P = ((v1^2* p)/ 2)* ((A1^2/ A2^2)-1)$

Now the following formula for area calculation should be used:

$A1 = (\pi* d1^2)/ 4 = (\pi* (0.02 m)^2)/ 4 = 0.00031 m^2\n\nA2 = (\pi* (0.01 m)^2)/ 4 = 0.000079 m^2\n\n\Delta P = ((3 m/s)^2 *1.25 kg/m^3)/ 2) * ((0.00031 m^2)^2/(0.000079 m^2)^2)-1)$

= 80.99

Find out more information about the Pressure here: brainly.com/question/356585?referrer=searchResults

Answer:

the pressure drop is equal to 80.99 Pa

Explanation:

we have the following data:

d1 = 2 cm = 0.02 m

d2 = 1 cm = 0.01 m

v = 3 m/s

p = 1.25 kg/m^3

ΔP = ?

For the calculation of the pressure drop we will use Bernoulli's principle for the Venturi Tube:

P1 - P2 = ((v^2*p)/2)*((A1^2/A2^2)-1)

where A = area

P1 - P2 = ΔP = ((v1^2*p)/2)*((A1^2/A2^2)-1)

for the calculation of the areas we will use the following formula:

A1 = (pi*d1^2)/4 = (pi*(0.02 m)^2)/4 = 0.00031 m^2

A2 = (pi*(0.01 m)^2)/4 = 0.000079 m^2

ΔP = ((3 m/s)^2*1.25 kg/m^3)/2)*((0.00031 m^2)^2/(0.000079 m^2)^2)-1) = 80.99 N/m^2 = Pa

Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the heart, which can be modeled as a 500 ohm resistance. The peak amplitude of the pulses is required to be 5 V. However, the battery delivers only 2.5 V. Therefore, we decide to charge two equal value capacitors in parallel from the 2.5V battery and then switch the capacitors in series with the heart during the 1ms pulse. What is the minimum value of the capacitances required so the output pulse amplitude remains between 4.9 V and 5.0 V throughout its 1ms duration

Answers

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

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