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1 year ago

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A formula for the normal systolic blood pressure for a man age a , measured in mmhg, is given as p=0. 006a2−0. 02a 120. Find the

age of a man whose normal blood pressure measures 123 mmhg

1 answer:

djyliett [7]1 year ago

5 0

The age of a man whose normal blood pressure measures 123 mm of hg

9 years

<h3>What is Quadratic equation ?</h3>

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = a $x^{2}$ + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 $a^{2}$ - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 $a^{2}$ - 0.02a - 3

you can use the quadratic equation formula to solve for the man's age.

A = (-b ± ( $\sqrt{b^{2} - 4*a*c}$ ) ) / (2a)

A = (0.02 ± $\sqrt{(-0.02)^{2} - 4*0.006*(-3)}$ / (2*0.006)

A = (0.02 ± $\sqrt{0.0076}$ ) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation here

brainly.com/question/17177510?referrer=searchResults

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A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,

n200080 [17]

Answer:

a. $t = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$ b. D = v₀²/2g

Explanation:

Here is the complete question

A ball is thrown straight up from the ground with speed v₀ . At the same instant, a second ball is dropped from rest from a height D , directly above the point where the first ball was thrown upward. There is no air resistance

Find the time at which the two balls collide.

Express your answer in terms of the variables D ,v₀ , and appropriate constants..

t = ?!

Part B

Find the value of D in terms of v₀ and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

Express your answer in terms of the variables v₀ and g .

D =?!

Solution

The distance moved by the ball dropped from distance,D with velocity v₀, H₁ = D - (v₀t - gt²/2) = D + v₀t + gt²/2.

The distance moved by the ball thrown straight upward with velocity v₀ is H₂ = v₀t - gt²/2.

The two balls collide when their vertical distances are equal. That is H₁ = H₂

So, D - v₀t + gt²/2 = v₀t - gt²/2

Collecting like terms

D + gt²/2 + gt²/2 = v₀t + v₀t

D +gt² = 2v₀t

gt² - 2v₀t + D = 0.

Using the quadratic formula,

$t = \frac{-(-2v_{0} ) +/- \sqrt{(-2v_{0} )^{2} - 4 X g XD} }{2g} = \frac{2v_{0} +/- \sqrt{4v_{0} ^{2} - 4gD} }{2g} = \frac{v_{0} +/- \sqrt{v_{0} ^{2} - gD} }{g}$

B. At its highest point, the velocity of the first ball, v = 0. Using v² = u² - 2gs where s = highest point of first ball when they collide and u = v₀.

0 = v₀² - 2gs

s = v₀²/2g.

Also, the time it takes the first ball to reach its highest point is gotten from v = u - gt. At highest point, v = 0 and u = v₀. So,

0 = v₀ - gt₀

t₀ = v₀/g

Also H = s₁ + s where s₁ = distance moved by second ball in time t₀ for collision = v₀t₀ - gt₀²/2.

So, H = v₀t₀ - gt₀²/2 + v₀²/2g = v₀(v₀/g) - g(v₀/g)²/2 + v₀²/2g = v₀²/2g - v₀²/2g + v₀²/2g = v₀²/2g

6 0

2 years ago

All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certa

masha68 [24]

Answer:

R= 20 ohm

Explanation:

Given that

Current ,I = 6 A

Voltage difference ,ΔV = 120 V

Lets take resistance of the stem iron = R

We know that ,the relationship between current ,voltage difference and resistance is given as

ΔV = I R

$R=\dfrac{\Delta V}{I}$

Now by putting the values in the above equation we get

$R=\dfrac{120}{6}\ ohm$

R= 20 ohm

Therefore the resistance of the steam will be 20 ohm.

3 0

2 years ago

Which statement is true of a mechanical wave?

svet-max [94.6K]

It must have a medium. It must travel in empty space. Mechanical waves are waves which needs medium of propogation.

7 0

1 year ago

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

2 years ago

A ball moving at positive 3.0 m per s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table

MA_775_DIABLO [31]

consider the motion along the horizontal direction :

v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s

X = horizontal displacement of the ball = 2.0 m

a = acceleration along the horizontal direction = 0 m/s²

t = time taken to land = ?

using the kinematics equation

X = v₀ t + (0.5) a t²

2.0 = 3.0 t + (0.5) (0) t²

t = 2/3

consider the motion of the ball along the vertical direction

v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s

Y = vertical displacement of the ball = height of the table = h

a = acceleration along the vertical direction = 9.8 m/s²

t = time taken to land = 2/3

using the kinematics equation

Y = v₀ t + (0.5) a t²

h = 0 t + (0.5) (9.8) (2/3)²

h = 2.2 m

C 2.2 m

3 0

2 years ago