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3 years ago

10

A charge q produces an electric field of strength 2E at a distance of d away. Determine the electric field strength at a distanc

e of 1/2d away.

2 answers:

Alja [10]3 years ago

3 0

The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.

If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .

tigry1 [53]3 years ago

3 0

Answer:

8 E

Explanation:

The formula for the electric field at point P which is at a distance r from the charge Q due to the charge Q is given by

E = k Q / r^2

If a charge is at a distance d from the charge q, the electric field is given by

2E = k q / d^2 .... (1)

Let the electric field at a distance 1/2 d is E'

E' = k q / (d/2)^2

E' = 4 k q / d^2 = 4 x 2E (From equation (1)

E' = 8 E

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A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago

The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot

KiRa [710]

Answer:

$T_{f}$ = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

Q₁ = 2.00 10⁻³ 2.26 10⁶

Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

Q coffee = M $c_{e}$ ( $T_{f}$ - $T_{i}$ )

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

Qc = - Q₁

M ce ( $T_{f}$ - $T_{i}$ ) = - Q₁

The coffee dough left in the cup after evaporation is

M = 250 -2 = 248 g = 0.248 kg

$T_{f}$ -Ti = -Q1 / M $c_{e}$

$T_{f}$ = Ti - Q1 / M $c_{e}$

Since coffee is essentially water, let's use the specific heat of water,

$c_{e}$ = 4186 J / kg ºC

Let's calculate

$T_{f}$ = 90.0 - 4.52 103 / (0.248 4.186 103)

$T_{f}$ = 90- 4.35

$T_{f}$ = 85.65 ° C

$T_{f}$ = 85.7 ° C

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3 years ago

A living thing that feeds on Another living thing and may kill it eventually is called

Murljashka [212]

Answer:

(D) parasite........................

3 0

2 years ago

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

Westkost [7]

Answer:

24 N

Explanation:

$m$ = mass of the cube = $6.0 kg$

Consider the three cubes together as one.

$M$ = mass of the three cubes together = $3 m = 3 (6.0) = 18 kg$

$a$ = acceleration of the combination = 2 ms⁻²

$F$ = Force applied on the combination

Using Newton's second law

$F = ma = (18) (2) = 36 N$

$F_{L}$ = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

$F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N$

4 0

3 years ago

The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875

Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

$F_m=m.a\\F_m=0.875kg*a$

The acceleration is given by:

$a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2$

So the force exerted by the motor is:

$F_m=0.875kg*31.9m/s^2\\F_m=27.9N$

The work done by the motor is given by:

$W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m$

$W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J$

And finally, the power is given by:

$P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W$

8 0

3 years ago