Friction is generated when?

What happens to the bulb when the battery changes from 1.5V to 9v?

timofeeve [1]

Answer:

it gets brighter because the volta increases

4 0

3 years ago

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

erastova [34]

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is $Q = 5.866 * 10^9 J$

The power is $P = 5866\ MW$

Explanation:

From the question we are told that

Mass of the water per second is $m = 1917 \ kg$

The initial temperature of the water is $T_i = 35^oC$

The boiling point of water is $T_b = 100^oC$

The final temperature $T_f = 450^oC$

The latent heat of vapourization of water is $c__{L}} = 2256*10^3 J/kg$

The specific heat of water $c_w = 4184 J/kg^oC$

The specific heat of stem is $C_s =1520 \ J/kg ^oC$

Generally the heat needed each second is mathematically represented as

$Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]$

Then substituting the value

$Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]$

$Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]$

$Q = 1917 * [3.05996 * 10^6]$

$Q = 5.866 * 10^9 J$

The power required is mathematically represented as

$P = \frac{Q}{t}$

From the question $t = 1\ s$

So

$P = \frac{5.866 *10^9}{1}$

$P = 5866*10^6 \ W$

$P = 5866\ MW$

6 0

3 years ago

Substances released into the air are known as

irakobra [83]

Your answer is Emissions

3 0

3 years ago

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A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle

Andru [333]

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

7 0

3 years ago

What are the 3 main states of matter in water?/

BigorU [14]

Answer:

ice (solid), water (liquid) and vapor (gas)

Explanation:

7 0

3 years ago

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