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Sliva [168]
3 years ago
8

( Pennfoster plz help )

Physics
2 answers:
Talja [164]3 years ago
7 0

<em>Choice-C </em>is nonsense.

Electrons positioned closer to the nucleus are closer to the protons in the nucleus and more strongly attracted to them.  Therefore these electrons are LESS likely to be discharged from the atom than electrons farther away from the nucleus are.

Mnenie [13.5K]3 years ago
4 0

I would say C i'm not 100% sure

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Newton’s empirical law of cooling/warming of an object is given by ( ), T Tm k dt dT = − where k is a constant of proportionalit
pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let T_{0} = 70^{0}F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

\frac{DT}{dt} = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

\frac{DT}{t-70} = <em>kdt</em>

In [T-70]= <em>kt </em>+C_{1}

    T   =  70  + C_{2} e^{kt}

Finding he value for C_{2} using the initial value of T (0)= 300, therefore we get:

300=70+C_{2}

C_{2} = 230 therefore

T= 70+ 230 e^{kt}

Finding the value for <em>k </em>using T (3)  = 200, therefore we get

T (3) = 200

e^{3k} = \frac{13}{23}

<em>K </em>= \frac{1}{3} in \frac{13}{23}

= -0.19018

Therefore

T(t) = 70+230e^{-0.19018}

4 0
3 years ago
ANSWERRR PLEASEEEE!!!
GuDViN [60]

Answer:

50N to the left

Explanation:

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6 0
2 years ago
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A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled
GaryK [48]

The boat's position x relative to its starting point x_0=0 is determined by

x=x_0+v_0t+\dfrac12at^2

where v_0 is its initial velocity, a is its acceleration, and t is time. After t=6\,\mathrm s, the boat has traveled

x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

\implies x=91\,\mathrm m

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
How does`ezzzzz321erfq23c2f
Nataliya [291]
Lol what???? i don’t understand
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