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Anettt [7]
3 years ago
12

A person pushes a refrigerator across a horizontal floor. The mass of the refrigerator is 110 kg, the coefficient of static fric

tion between the refrigerator and the floor is 0.85, and the coefficient of kinetic friction between the refrigerator and the floor is 0.59.
(a) If the refrigerator is initially at rest, determine the minimum magnitude of the force that the person must exert horizontally on the refrigerator to cause it to move.
N

(b) Once the refrigerator is moving, determine the magnitude of the force that the person must exert horizontally on the refrigerator to keep it moving at a constant speed.
N

(c) Determine the acceleration of the refrigerator if the person applies a horizontal force of magnitude 950 N on the refrigerator.
Physics
1 answer:
Law Incorporation [45]3 years ago
4 0

Answer:

Explanation:

mass of refrigerator, m = 110 kg

coefficient of static friction, μs = 0.85

coefficient of kinetic friction, μk = 0.59

(a) the minimum force required to just start the motion in refrigerator

F = μs x mg

F = 0.85 x 110 x 9.8

F = 916.3 N

(b) The force required to move the refrigerator with constant speed

F' = μk x mg

F' = 0.59 x 110 x 9.8

F' = 636.02 N

(c) Let a be the acceleration.

Net force = Applied force - friction force

F net = 950 - 636.02

F net = 313.98 N

a = F net / mass

a = 313.98 / 110

a = 2.85 m/s²

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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

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Answer:

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In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The
Ratling [72]

Answer:

Electric force, F=-3.59\times 10^6\ N

Explanation:

Given that,

Electric charge 1, q_1=+40\ C

Electric charge 2, q_2=-40\ C

Distance, d=2\ km=2\times 10^3\ m

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The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

F=k\dfrac{q_1q_2}{d^2}

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F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}

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