If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don't have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart
Answers
Complete Question
If you are lying down and stand up quickly, you can get dizzy or feel faint. This is because the blood vessels don’t have time to expand to compensate for the blood pressure drop. If your brain is 0.4 m higher than your heart when you are standing, how much lower is your blood pressure at your brain than it is at your heart? The density of blood plasma is about 1025 kg/m3 and a typical maximum (systolic) pressure of the blood at the heart is 120 mm of Hg (= 0.16 atm = 16 kP = 1.6 × 104 N/m2).
Answer:
The pressure at the brain is
Explanation:
Generally is mathematically denoted as
Substituting for
(the density) ,
for g (acceleration due to gravity) , 0.4m for h (the height )
We have that the pressure difference between the heart and the brain is
But the pressure of blood at the heart is given as
Now the pressure at the brain is mathematically evaluated as
Final answer:
When you stand up quickly, the blood pressure at your brain is lower than at your heart. The decrease in blood pressure can be calculated using the equation ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart.
Explanation:
When you stand up quickly, your blood pressure drops because the blood vessels don't have enough time to expand and compensate for the change in posture. The brain, which is 0.4 m higher than the heart when standing, experiences a decrease in blood pressure. To calculate how much lower the blood pressure is at the brain compared to the heart, we need to use the equation: ΔP = ρgh, where ΔP is the change in pressure, ρ is the density of the blood, g is the acceleration due to gravity, and h is the height difference between the two points. In this case, we can use the height difference of 0.4 m and the density of blood to find the change in pressure.
Using the equation, ΔP = ρgh, we can calculate the change in pressure:
- ρ = density of blood = 1060 kg/m³ (approximately)
- g = acceleration due to gravity = 9.8 m/s² (approximately)
- h = height difference = 0.4 m
Plugging in the values into the equation, we get:
ΔP = (1060 kg/m³)(9.8 m/s²)(0.4 m) = 416.32 Pa
Therefore, the blood pressure at the brain is approximately 416.32 Pa lower than at the heart when standing up quickly.
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A plane monochromatic electromagnetic wave with wavelength ? = 3 cm, propagates through a vacuum. Its magnetic field is described byB? =(Bxi^+Byj^)cos(kz+?t)where Bx = 3.3 X 10-6 T, By = 3.9 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.1)What is f, the frequency of this wave?GHz2)What is I, the intensity of this wave?W/m23)What is Sz, the z-component of the Poynting vector at (x = 0, y = 0, z = 0) at t = 0?W/m24)What is Ex, the x-component of the electric field at (x = 0, y = 0, z = 0) at t = 0?V/m5)Compare the sign and magnitude of Sz, the z-component of the Poynting vector at (x=y=z=t=0) of the wave described above to the sign and magnitude of SIIz, the z-component of the Poynting vector at (x=y=z=t=0) of another plane monochromatic electromagnetic wave propagating through vaccum described by:B? =(BIIxi^?BIIyj^)cos(kz??t)where BIIx = 3.9 X 10-6 T, BIIy = 3.3 X 10-6 T, and i-hat and j-hat are the unit vectors in the +x and +y directions, respectively.SIIz < 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)SIIz < 0 and magnitude(SIIz) = magnitude(Sz)SIIz > 0 and magnitude(SIIz) =/ (does not equal sign) magnitude(Sz)SIIz > 0 and magnitude(SIIz) = magnitude(
In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.
The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.What is the moment of inertia of the propeller?How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.
Answers
Answer:
The skater's speed after she stops pushing on the wall is 1.745 m/s.
Explanation:
Given that,
The average force exerted on the wall by an ice skater, F = 120 N
Time, t = 0.8 seconds
Mass of the skater, m = 55 kg
It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :
The change in momentum is equal to the impulse delivered. So,
So, the skater's speed after she stops pushing on the wall is 1.745 m/s.
Answers
Answer:
-384.22N
Explanation:
From Coulomb's law;
F= Kq1q2/r^2
Where;
K= constant of Coulomb's law = 9 ×10^9 Nm^2C-2
q1 and q2 = magnitudes of the both charges
r= distance of separation
F= 9 ×10^9 × −7.97×10^−6 × 6.91×10^−6/(0.0359)^2
F= -495.65 × 10^-3/ 1.29 × 10^-3
F= -384.22N
Answers
Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77
What is Friction?
According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.
Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.
F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.
Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.
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Answer:
7.77
Explanation:
F=μ*m
n=w which also means n=mg
14.7=0.193*n
n=76.2
76.2=m*9.8
m=7.77
b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.
Answers
Answer:
a) x = ⅔ d, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q
Final answer:
The third charge, -Q, can either be placed at a distance of d/4 to the left of origin or at a distance d/2 to the right of +4Q charge to create an equilibrium among the three charges. Its sign should be negative and its magnitude should be equal to Q.
Explanation:
To create equilibrium among all three charges, our third charge can be placed in two potential positions on the x-axis. One position is a distance of d/4 to the left of the origin (negative x-axis), and the other is a distance of d/2 to the right of the +4Q charge (positive x-axis).
The third charge should be a negative charge, denoted as -Q, in order to balance the positive charges and create equilibrium.
The magnitude of this third charge should be equal to Q, the original charge. The reason is that the forces need to be balanced to create equilibrium and the force is proportional to the charge. Therefore, if -Q is placed at d/4 to the left of the origin or if -Q is placed at d/2 to the right of the +4Q charge, the system will be in equilibrium.
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Answers
And so, the kinetic energy of the object is
(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
And so the new kinetic energy is
So, the work done on the object is the variation of kinetic energy of the object:
Final answer:
The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.
Explanation:
The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.
The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.
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Answers
Answer:
electric flux is 280 Nm²/C
so correct option is D 280 Nm²/C
Explanation:
radius r = 0.50 m
angle = 30 degree
field strength = 713 N/C
to find out
the electric flux through the surface
solution
we find here electric flux by given formula that is
electric flux = field strength × area× cos∅ .......1
here area = πr² = π(0.50)²
put here all value in equation 1
electric flux = field strength × area× cos∅
electric flux = 713 × π(0.50)² × cos60
we consider the cosine of the angle between the direction of the field and the normal to the surface of the disk
so we use cos60
electric flux = 280 Nm²/C
so correct option is D 280 Nm²/C