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zhuklara [117]
3 years ago
12

What is use of lemon Squeezer​

Physics
2 answers:
AysviL [449]3 years ago
7 0

Answer:

A lemon squeezer is a small kitchen utensil designed to extract juice from lemons or other citrus fruit such as oranges, grapefruit, or lime. It is designed to separate and crush the pulp of the fruit in a way that is easy to operate.

shusha [124]3 years ago
3 0

Answer:

A lemon squeezer is a small kitchen utensil designed to extract juice from lemons or other citrus fruit such as oranges, grapefruit, or lime. It is designed to separate and crush the pulp of the fruit in a way that is easy to operate.

Explanation:

stay safe

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Friction will cause a moving object to slow down and finally stop.

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The data table for the decomposition reaction of hydrogen peroxide H2 O2 shows how the reaction rate changes over time which sta
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Where should clouds form and why
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Explanation:

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8 0
2 years ago
A 1000-kilogram car traveling with a velocity of 20. meters per second decelerates uniformly at -5.0 meters per second2 until it
Elis [28]

Answer:

-20000 kgm/s

Explanation:

Impulse: This can be defined as the product of the mass of a body and its change in velocity. The S.I unit of impulse is kgm/s.

Mathematically, impulse can be expressed as

I = m(v-u).............. Equation 1.

Where I = impulse applied to the car to bring it to rest, m = mass of the car, u = initial velocity of the car, v = final velocity of the car.

Given: m = 1000 kg, u = 20 m/s, v = 0 m/s ( to rest)

Substitute into equation 1

I = 100(0-20)

I = 1000(-20)

I = -20000 kgm/s

Hence the impulse applied to the car to bring it to rest = -20000 kgm/s

3 0
3 years ago
A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 V. It is used to charge two storage batt
Natali [406]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 \Omega. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300\Omega . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is R_z =  4.4 \Omega

b

The rate at which internal energy increase at the supply is Z_1 = 32 W

c

The rate at which internal energy increase in the battery  is  Z_1 = 32 W

d

The rate at which internal energy increase in the added series resistance is  Z_3 = 70.4 W

e

the increase rate of the chemically energy in the battery is C =  48 W

Explanation:

From the question we are told that

    The  open circuit voltage is  V =  40.0V

     The internal resistance is R = 2 \Omega

     The emf of each battery is e =  6.00 V

      The internal resistance of the battery is  r = 0.300V

      The  charging current is  I = 4.00 \ A

Let assume the the additional resistance to to added to the circuit is  R_z

 So this implies that

        The total resistance in the circuit is

                              R_T =  R + 2r +R_z

Substituting values

                             R_T = 2.6 +R_z

And  the difference in potential in the circuit is  

                         E = V -2e

                 =>   E =  40 - (2 * 6)

                        E =  28 V

Now according to ohm's law

            I = \frac{E}{R_T}

Substituting values

           4 = \frac{28}{R_z + 2.6}        

Making R_z the subject of the formula

So    R_z =  \frac{28 - 10.4}{4}

           R_z =  4.4 \Omega

The  increase rate of   internal energy at the supply is mathematically represented as

        Z_1  = I^2 R

Substituting values

     Z_1  = 4^2 * 2

     Z_1 = 32 W

The  increase rate of   internal energy at the batteries  is mathematically represented as

         Z_2 = I^2 r

Substituting values

         Z_2 = 4^2 * 2 * 0.3

         Z_2 = 9.6 \ W

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        Z_3 = I^2 R_z

Substituting values

       Z_3 = 4^2 * 4.4

      Z_3 = 70.4 W

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         C = 2 * e * I

Substituting values

       C =  2 * 6  * 4

      C =  48 W

6 0
3 years ago
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