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Nata [24]
2 years ago
5

You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kg

K. (a) How much energy did it take to heat the pot of water
Physics
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

<em>765,000Joules or 765kJ</em>

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

<em>Q = 765,000</em>

<em>Hence the amount of energy needed is 765,000Joules or 765kJ</em>

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Dimension of density <br>​
Rom4ik [11]

Answer:

See the explanation below.

Explanation:

We know that density is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

Therefore Ro is given in:

[kg/m^{3} ]

8 0
2 years ago
What are three types of electromagnetic waves that are used to transmit information? What types of devices are used to receive e
MrMuchimi

Answer:

Three types of electromagnetic waves, used to transmit various information

Explanation:

A form of energy waves having both electric & magnetic fields are Electromagnetic waves. Three types -

Radio Waves - These have longest wavelengths & transmit data through radio, satellites, radar .

Micro Waves - These have shorter wavelengths & are used in cooking appliances & predicting weather.

X rays - These have more short wavelength and can penerate soft tissues like skin & muscle, hence are used for medical examining

3 0
2 years ago
Which of the following are known to exist on the moon? select all that apply.
denis-greek [22]
B. ice and d basalt............
3 0
3 years ago
Read 2 more answers
what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?
zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

R = ρL/A ......... (1)

A = πr² (since the wire is circular in shape)

r = d/2

A = πr² = π(d/2)²

A = πd²/4

Substitute the value of A into equation 1

R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

R = 4ρL /πd²

Where:

R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

I is the current

From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

5 0
3 years ago
Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the
zlopas [31]

Answer:

\dfrac{I_1}{I_2}=\dfrac{4}{9}

Explanation:

c = Speed of wave

\rho = Density of medium

A = Area

\nu = Frequency

\nu_1=\dfrac{2}{3}\nu_2

Intensity of sound is given by

I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2

So,

I\propto \nu^2

We get

\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}

The ratio is \dfrac{I_1}{I_2}=\dfrac{4}{9}

8 0
3 years ago
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