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Sergio [31]
3 years ago
10

Calculate the heat energy required to increase the temperature of 3 kg of water by 20ºC. The specific heat capacity of water is

4200 J/ kg °C
Physics
1 answer:
Marina86 [1]3 years ago
6 0

Answer:

Heat energy required = 252000J or 252KJ.

Explanation:

<u>Given the following data;</u>

Mass = 3kg

Temperature = 20ºC

Specific heat capacity of water = 4200 J/kg°C

To find the heat energy required;

Heat capacity is given by the formula;

Q = mct

Where;

  • Q represents the heat capacity or quantity of heat.
  • m represents the mass of an object.
  • c represents the specific heat capacity of water.
  • t represents the temperature.

Substituting into the equation, we have;

Q = 3*4200*20

Q = 252000 Joules or 252 Kilojoules.

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Exercise 1 Electric Fields In this exercise, you will use a digital multimeter to collect voltage data to graph electric fields.
Lyrx [107]

Answer:

The answers are in the explanation section below

Explanation:

1) The generalization that can be made from the exploration is that as we move away from the positive electrode, the potential energy gets lower. If we move away from the negative electrode, then the potential energy becomes higher.

2) The positive test charge will have the least potential energy when it gets to the negative electrode point.

3) To move one electron 1m in a direction along one of the equal potential lines, there is no energy needed. Zero work will be required for a charge to move on the equipotential line.

4) If lightning strikes a tree 20m away, it would be better to face the tree or have our back facing the tree. This is because the equipotential line will be present at the point where our body stands, this will protect from electric shock.

The pattern to be sketched is attached.

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4 0
3 years ago
The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5
KiRa [710]

Answer:

0.68 s

Explanation:

We are given that

Initial velocity of box=u=13m/s

Final velocity of box=v=11.5 m/s

Distance=d=8.5 m

We have to find the time taken by box to slow by this amount.

We know that

v^2-u^2=2as

Substitute the values

(11.5)^2-(13)^2=2a(8.5)

132.25-169=17a

-36.75=17a

a=\frac{-36.75}{17}=-2.2m/s^2

We know that

Acceleration=a=\frac{v-u}{t}

Substitute the values

-2.2=\frac{11.5-13}{t}

-2.2=\frac{-1.5}{t}

t=\frac{1.5}{2.2}=0.68 s

Hence, the time taken by box to slow by this amount=0.68 s

8 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
In pea plants Round (R) is a dominant trait and wrinkled nose (r) is a recessive trait. Which combination would result in wrinkl
bixtya [17]
The answer is would be number 3 rr
7 0
3 years ago
Read 2 more answers
A 6.00-mH solenoid is connected in series with a 5.0-μF capacitor and an AC source. The solenoid has internal resistance 3.0 Ω w
son4ous [18]

Answer:

5773.50269 Hz

23 A

Explanation:

L = Inductance = 6 mH

C = Capacitance = 5 μF

R = Resistance = 3 Ω

\epsilon = Maximum emf = 69 V

Resonant angular frequency is given by

\omega=\dfrac{1}{\sqrt{LC}}\\\Rightarrow \omega=\dfrac{1}{\sqrt{6\times 10^{-3}\times 5\times 10^{-6}}}\\\Rightarrow \omega=5773.50269\ Hz

The resonant angular frequency is 5773.50269 Hz

Current is given by

I=\dfrac{\epsilon}{R}\\\Rightarrow I=\dfrac{69}{3}\\\Rightarrow I=23\ A

The current amplitude at the resonant angular frequency is 23 A

7 0
3 years ago
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