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weqwewe [10]
1 year ago
11

Conservation of Energy-not sure how to do problem-confused on how to find the speed and how to figure out energy bar graph

Physics
1 answer:
nika2105 [10]1 year ago
6 0

Given data

*The given mass of the rock is m = 2 kg

*The given potential energy is U_p = 407 J

(a)

The diagram of the energy bar graph is drawn below

(b)

If an object is at rest and has potential energy, once it starts to fall from its rest state then this potential energy is completely transferred to kinetic energy. This means that the magnitude of the kinetic energy is equal to the potential energy of the object.

The change in kinetic energy of the rock while falling to the ground is given as

\begin{gathered} U_k=U_p \\ =407\text{ J} \end{gathered}

(c)

The formula for the speed of the block is given as

\begin{gathered} U_k=\frac{1}{2}mv^2 \\ v=\sqrt[]{\frac{2U_k}{m}} \end{gathered}

Substitute the known values in the above expression as

\begin{gathered} v=\sqrt[]{\frac{2\times407}{2}} \\ =20.17\text{ m/s} \end{gathered}

Hence, the speed of the object is v = 20.17 m/s

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Answer:

D.

R increases

V is constant

I decreases

Explanation:

The resistance of a wire is given by the following formula:

R = \frac{(Resistivity)(L)}{A}

It is clear from this formula that resistance is directly proportional to the length of wire. So, when length of wire is increased, <u>the resistance of circuit increases</u>.

The <u>voltage in the circuit will be constant</u> as the voltage source remains same and it is not changed.

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it means that current is inversely proportional to resistance. Hence, the increase of resistance causes <u>the current in circuit to decrease.</u>

Therefore, the correct option will be:

<u>D.</u>

<u>R increases </u>

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2 years ago
What does the Nucleolus do?
SCORPION-xisa [38]

Answer:

C

Explanation:

6 0
3 years ago
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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

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