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almond37 [142]
2 years ago
9

When used as an energy source in a nuclear power plant, uranium is burned in a similar way as one would burn wood or coal for en

ergy.
a. True
b. False
Physics
1 answer:
zavuch27 [327]2 years ago
3 0
False because nuclear fission is used on uranium in nuclear power station and this reaction is continuous since a neutron hits a uranium atom and splits it. Then 2-3 neutrons are released which continues the cycle or hitting other uranium atoms. If this reaction isn't contained, it could get out of control and become a nuclear bomb. So it can't be burned like wood nor coal.
Hope this is helpful. <span />
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2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

7 0
3 years ago
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves
pantera1 [17]

Answer:

It's A

Explanation:

As the waves progress through the medium, the particles they are made of move perpendicular to the direction in which the waves move. The particles do not move with the wave. So waves transmit energy but not matter as they progress through a medium.

6 0
3 years ago
Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1). Graph △XYZ and its image after a rotation of 180° about (2, –3).
Alla [95]

The image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

<h3>What is Geometry?</h3>

It deals with the size of geometry, region, and density of the different forms both 2D and 3D.

Triangle XYZ has vertices X(0, 2), Y(4, 4), and Z(3, –1).

If the triangle is ΔXYZ. Then the image of the triangle is to be formed by rotating ΔXYZ 180 degrees about the (2, -3) as shown in the graph.

More about the geometry link is given below.

brainly.com/question/7558603

#SPJ1

6 0
2 years ago
B. The silica cylinder of a radiant wall heater is 0.6 m long
SIZIF [17.4K]

So,  If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.

<h3>Power radiated by the radiant wall heater</h3>

The power radiated by the radiant wall heater is given by P = εσAT⁴ where

  • ε = emissivity = 1 (since we are not given),
  • σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
  • A = surface area of cylindrical wall heater = 2πrh where
  • r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
  • h = length of heater = 0.6 m, and
  • T = temperature of heater

Since P = εσAT⁴

P = εσ(2πrh)T⁴

Making T subject of the formula, we have

<h3>Temperature of heater</h3>

T = ⁴√[P/εσ(2πrh)]

Since P = 1.5 kW = 1.5 × 10³ W

Substituting the values of the variables into the equation, we have

T = ⁴√[P/εσ(2πrh)]

T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]

T = ⁴√[1.5 × 10³ W/(43.2π  × 10⁻¹¹ W/K⁴)]

T = ⁴√[1.5 × 10³ W/135.72  × 10⁻¹¹ W/K⁴)]

T = ⁴√[0.01105 × 10¹⁴ K⁴)]

T = ⁴√[1.105 × 10¹² K⁴)]

T = 1.0253 × 10³ K

T = 1025.3 K

So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K

Learn more about temperature of radiant wall heater here:

brainly.com/question/14548124

6 0
2 years ago
What happens to the image when u make the hole bigger in a pinhole camera?​
marta [7]

Answer:

more light enters and disturbs the formation of the image.

7 0
2 years ago
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