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3 years ago

Which feature or process is unique to nuclear power plants when compared to conventional coal-burning power plants?

Physics

2 answers:

Mazyrski [523]3 years ago

5 0

Answer:

Control rods

Explanation:

Control rods is unique to nuclear power plants as they are mainly used in nuclear reactors to stabilize the rate of fission of plutonium or uranium. They are usually composed of chemical elements such as cadmium, silver, boron, and indium.

Maru [420]3 years ago

4 0

Answer:

Control rods

Explanation:

Control rods are used to control the rate of nuclear chain reaction in nuclear reactors. The reactivity of the nuclear reactor is greatly affected by the distance of control rods inserted and number of control rods inserted. Control rods control the rate of fission of uranium, etc., in a nuclear reactor

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Use your understanding of heat loss to ESTIMATE the cost of the lost energy through one standard window during an average summer

miskamm [114]

Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34

3 0

3 years ago

A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h

Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is $m_1=1.6\times 10^4 kg$

speed of bus towards North $v_1=15.2 km/h\approx 4.22\ m/s$

mass of bus travelling in South direction is $m_2=1.578\times 10^4 kg$

speed of bus $v_2=12.2 km/h\approx 3.38\ m/s$

mass of each Passenger in south moving bus $m_0=64.8 kg$

Momentum of North moving bus

$P_1=m_1\times v_1$

$P_1=1.6\times 10^4\times 4.22$

$P_1=6.768\times 10^4 kg-m/s$

Momentum with south moving bus

$P_2=m_2\times v_2+n\cdot m_0\times v_2$

$P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38$

For total momentum to be towards south

$P_2-P_1$ should be greater than 0

thus for least value of n

$P_2=P_1$

$(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4$

$1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4$

$n=\frac{4243.6686}{64.8}=65.48\approx 66$

5 0

3 years ago

The absolute magnitude of a star________.

serg [7]

<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us. So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>

8 0

3 years ago

The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.

tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

$W=K_f -K_i$

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

$K_f =\frac{1}{2}mv^2$

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

$W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J$

5 0

3 years ago

Read 2 more answers

Which of the following would have the greatest force of gravitational attraction?

Mekhanik [1.2K]

Answer:

two 5kg objects 2 meters apart✨

8 0

3 years ago