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cupoosta [38]
3 years ago
12

Different between cell and dynamo short and sweet ​

Physics
1 answer:
Phantasy [73]3 years ago
4 0

Answer:

Cell is the unit which has two terminals one positive and one negative. A battery is a group of cells in which one negative terminal is attached to one positive terminal.

Cell:- Cell is the smallest unit of a battery.

Battery:-A battery is combination of cells.

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Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
3 years ago
Which two points on the wave shown in the diagram below are in phase with each other?
NNADVOKAT [17]

Answer:

4. B and D

Explanation:

Two points along a transverse wave (such as the one in the figure) are said to be in phase when:

- the vertical position of the two points is the same

- The oscillation of the wave is going in the same way for both points

Basically, we say that two points are in phase when they are separated by a complete cycle (one complete oscillation) of the wave.

For this wave, we see that point B and C have same displacement, but they are not in phase since in B the oscillation is going down while in C is going up.

Instead, B and D are in phase, because they are separated by one complete cycle: both points have same displacement and the oscillation is going in the same way for both of them.

8 0
3 years ago
g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and
cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

  = 20(11.8)

  = 236N

4 0
4 years ago
HELPPpPpPpPpPp!!!! ASAP !!!
DENIUS [597]

Answer:

50 N

Explanation:

8 0
2 years ago
What two factors determine how bright a star appears to be in the sky?
Delicious77 [7]

Answer:

1. Luminosity

2.Apparent brightness

Explanation:

There are two factors on which brightness of star appear to be in the sky

The two factors are

1. Luminosity

2.Apparent brightness

1.Luminosity :It is defined as the total energy emitted by the object in a given time.Luminosity vary with the distance of observer from the star.Luminosity is a intrinsic property which depends on the fundamental chemical composition and structure of the  material.Luminosity is depends on the size of star.Lager the star luminosity will be more.

2.Apparent brightness: It is defined as how bright a star appears from an observer on the earth and the amount of starlight reaching the earth.if the distance is large then the brightness decreases.When the distance of star from us small then the   brightness of star increases.Distance is inversely proportional to brightness of the star.

7 0
3 years ago
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