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Gnoma [55]
3 years ago
10

Suppose a blood vessel's radius is decreased to 87% of its original value by plaque deposits and the body compensates by increas

ing the pressure difference along the vessel to keep the flow rate constant.
By what factor must the pressure difference increase?

Physics
1 answer:
dexar [7]3 years ago
3 0

Answer:

The pressure difference will increase by the factor of 1.75

Explanation:

For constant flow rate, coefficient of viscosity, length of the vessel and the pressure difference is inversely proportional to the fourth power of the radius of the blood vessel

Apply the principle of Poiseuille’s law.

Q = (P2 - P1)/R

Pls check the attached file for step by step solution of the question. It is submitted in this way as typing the equation may not be explanatory.

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Plz help no links
Kaylis [27]
I pretty sure it’s the mantle
8 0
3 years ago
Read 2 more answers
An ordinary flashlight uses two D cells 1.50 V batteries connected in series. The bulb draws 380 m???? when turned on.
melamori03 [73]

Answer:

a. 7.89 Ω, 1.14 W

b. By a factor of 2; The batteries would heat up.

Explanation:

a. Using Ohm's law, we have that Voltage, V, is directly proportional to Current, I, with the constant of proportionality being Resistance, R. Mathematically:

V = IR

The voltage of the two batteries would be:

V = V1 + V2 = 1.5 + 1.5 = 3.0 V

The current is 380 mA = 0.38 A

Hence, the Resistance, R, will be:

R = \frac{V}{I}

R = \frac{3}{0.38}

R = 7.89 Ω

Power is given as:

P = IV

P = 3 * 0.38

P = 1.14 W

b. If the batteries are now 4, the new voltage will be:

V = 4 * 1.5

V = 6 V

Power becomes:

P = 0.38 * 6

P = 2.28 W

Comparing with Power in (a) above, we see that the new Power is double the value of the former Power. Hence, the Power has increased by a factor of 2.

It is not advisable to use double the number of batteries to power a flashlight because it would cause the batteries to heat up and thereby, leak or in the worst case scenario, blow up. This could be damaging to the flashlight.

3 0
3 years ago
What power (in kW) is supplied to the starter motor of a large truck that draws 260 A of current from a 25.5 V battery hookup
joja [24]

Answer:

P = 6.63 kW

Explanation:

Given that,

Current, I = 260 A

Voltage of the battery, V = 25.5 V

We need to find the power supplied to the starter motor. We know that,

P = VI

Put all the values,

P = 25.5 × 260

P = 6630 W

or

P = 6.63 kW

So, the power supplied to the motor is 6.63 kW.

7 0
3 years ago
Read 2 more answers
Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a
FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

d_1 = 1* t

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

d_2 = 2*(t - 120)

now it is given that Blaise wins by 10 m distance

d_1 - d_2 = 10

1* t - 2*(t - 120) = 10

t - 2t + 240 = 10

t = 230 s

now the distance moved by Blaise is given by

d_1 = 1*230 = 230 m

6 0
3 years ago
What is the resistance of a 2 m long tungsten wire whose cross-sectional area of 0.15 mm2?​
nadya68 [22]

The resistance of a 2 m long tungsten wire whose cross-sectional area of 0.15 mm² will be 0.74 ohm.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

ρ is the resistivity of tungsten = 5.6×10⁻⁸ (ohm m)

The relation of resistance with length and thickness is given by ;

\rm R= \frac{\rho L}{A} \\\\ \rm R= \frac{5.6 \times 10^{-8}\times 2}{0.15 \times 10^{-6}} \\\\ R=0.74 \ ohm

Hence, the resistance of tungsten wire will be 0.74 ohm.

To learn more about the resistance, refer to the link;

brainly.com/question/20708652

#SPJ1

3 0
2 years ago
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