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dmitriy555 [2]
3 years ago
13

Can lamp that works on a 2.5 v work on a 1.12 v ?​

Physics
1 answer:
12345 [234]3 years ago
5 0

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

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Calculate the electric potential energy in a capacitor that stores 4.0 10-10 C of charge at 250.0 V
Arada [10]
Q = C.v
v = Q/C
v = 4 × 10^(-10)/250
 = 4 × 10^(-10)/2.5 × 10^2
 = 1.6 × 10^(-12) volt
7 0
3 years ago
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Help with a length problem
Andru [333]
It would be B

Explanation:
Because if you're not measuring in inches you want to go the next one down other than inches which would be millimeters!(: hope this helps.
6 0
3 years ago
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A particle moves at a constant speed of 34 m/s in a circular path of radius 6.3 m. From this information, what can be calculated
Yakvenalex [24]

Answer:

Centripetal acceleration

Explanation:

  • The centripetal acceleration is the motion inwards towards the center of a circular path.
  • <em><u>Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path. </u></em>
  • That is;

         ac = v²/r

         Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the  radius, m

3 0
3 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
Help! <br><br><br>I'm not sure my answer is correct.<br><br><br>Problem is attached!​
kaheart [24]
  • Let G be xN

\\ \sf\longmapsto x-20=30

\\ \sf\longmapsto x=30+20

\\ \sf\longmapsto x=50N

Option B

3 0
3 years ago
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