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Ymorist [56]
3 years ago
10

Define 1ohm resistance​

Physics
1 answer:
algol133 years ago
7 0

Answer:

1 ohm is equal to one volt (V)/ one ampere (1A) 1 Ohm is defined as the resistance of a conductor with a potential difference of 1 volt applied to the ends through which 1-ampere current flows. Ohms is the SI unit of electrical resistance.

Explanation:

I hope it's help u

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A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi
EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is,\rm d_1 = 2.2 \ cm

initial radius,

r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm

The initial crossection area;

\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times  (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2

The final crossection area;

\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2

The initial flow rate is;

R = density ×velocity ×area

\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1  = 3.947 \ m/sec

The speed of the water in the wider part will be;

From the continuity equation;

\rm A_1 V_1 = A_2V_2  \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec

Hence, the speed of the water in the wider part will be 1.194 m/sec.

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

brainly.com/question/7359669

#SPJ1

7 0
2 years ago
A body of volume 100cc immersed completely in water contained in a jar. The weight of water and jar before immersion of the body
BlackZzzverrR [31]
<h2>Answer:</h2>

800gm

<h2>Explanation:</h2>

Archimedes principle states that when an object is immersed in a liquid there is an apparent loss of weight of the object. This apparent loss of weight is also the upthrust experienced by the liquid. The upthrust is equal to the weight of the liquid displaced.

Following from the above statement, when the body of volume 100c.c is immersed in the water contained in the jar, the upthrust experienced is equal to the weight of the water displaced.

<em>Note: In the question, weight is measured just using the mass.</em>

Mass (m) is the product of density (ρ) of liquid (which is water in this case) and volume (v) of body immersed. i.e

m = ρ x v

Where;

ρ = 1 gm/cm³

v = 100c.c = 100cm³

=> m = 1 gm/cm³ x 100cm³

=> m = 100gm

Therefore the weight of water displaced is 100gm

Now, the weight of the water and jar after immersion is the sum of the weight of water and jar before immersion, and the weight of the water displaced. i.e

Weight of water and jar after immersion = 700gm + 100gm = 800gm

8 0
3 years ago
In a physics experiment, a ball is released from rest, and it falls toward the ground. The timer was not paying attention but es
tigry1 [53]

Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

4 0
3 years ago
The current in a stream runs at 5 miles per hour. if a boat can go 15 miles per hour on still water, how fast can the boat go do
arsen [322]
Down stream it would be going 20 mph and up stream 10 mph
3 0
4 years ago
ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly
Igoryamba
The period T of a pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the length of the pendulum while g=9.81 m/s^2 is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is T=0.200 s. Using this data, we can solve the previous formula to find L:
L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm
4 0
3 years ago
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