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Ahat [919]
3 years ago
6

How much does approximately a cup of water increase in temperature when heated in a microwave for 45 seconds?

Physics
1 answer:
Dovator [93]3 years ago
5 0

Answer:

32 K

Explanation:

A cup of water has a volume of approximately 236 \text{ cm}{}^3. With water having a density of 1 \text{ g/cm}{}^3, the mass of water in a cup is 1 \text{ g/cm}{}^3 \times 236 \text{ cm}{}^3 = 236\text{ g} = 0.236\text{ kg}.

A microwave outputs about 700 W of power. In 45 seconds, the energy expended is 700 \text{ W} \times 45 \text{s} = 31500 \text { J}.

The heat energy supplied to the water is related to the temperature change and the mass of the water by

H = mc\Delta\theta where c is the specific heat capacity of water. This value is c = 4200 \text{ J/kg K}.

\Delta\theta is the temperature change.

\Delta\theta = \dfrac{H}{mc}=\dfrac{31500}{0.236\times4200}=32 \text{ K}

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8 0
3 years ago
Which wave interaction is responsible for echoes?
IrinaVladis [17]

Answer:Reflection

Explanation:

5 0
3 years ago
Read 2 more answers
Find the voltage across the 15 Ω resistor. .<br> 35 Ω<br> 20 Ω<br> 15 Ω<br> 10 V
almond37 [142]

Answer:

35 Ω

Explanation:

3 0
3 years ago
Two identical spheres carry charges of +0.6 coulomb and -0.2 coulomb, respectively. If these spheres touch, the resulting charge
masya89 [10]

Answer:

2. +0.2 C

Explanation:

When the two spheres touch, the total charge will be redistributed such that the two spheres are at same potential:

V_1 = V_2

The potential can be rewritten as ratio between the charge on the sphere (Q) and the capacitance of the sphere (C):

\frac{Q_1}{C_1}=\frac{Q_2}{C_2}

Since the two spheres are identical, they have same capacitance:

C_1 =C_2=C

So we can write

\frac{Q_1}{C}=\frac{Q_2}{C}

Q_1=Q_2

And since the total charge is

Q=+0.6 C-0.2 C=+0.4 C

And this charge will be redistributed equally on the two spheres (Q_1 = Q_2), we have

Q_1 = Q_2 = \frac{Q}{2}=\frac{+0.4 C}{2}=+0.2 C

5 0
4 years ago
A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h
zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

m_{1}u_{1} + m_{2} u_{2} = m_{1}v_{1} - m_{2}v_{2}

m_{1} = 3 kg, u_{1} = 8 m/s, m_{2} = 2 kg, u_{2} = 0 m/s ( since it is at rest), v_{1} = 2 m/s, v_{2} = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x v_{2})

24 + 0 = 16 - 2v_{2}

2v_{2} = 16 - 24

2v_{2} = -8

v_{2} = \frac{-8}{2}

   = -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0
3 years ago
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