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Salsk061 [2.6K]
3 years ago
11

What is an atmospheric window?

Physics
1 answer:
Salsk061 [2.6K]3 years ago
4 0

Answer:

A

Explanation:

Because it is a region in atmosphere that allows an particular wavelength of EMWs to pass through and of the options above only visible light is allowed to pass through

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A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude
OverLord2011 [107]

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

We would calculate the magnitude by applying pythagorean theorem:

\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}

\longrightarrow \sf{Magnitude = 33.12}

\longrightarrow \sf{The \:  vector \: is \: (- 14, 30)}

The angle between two vectors is given by the formula:

\sf{\longrightarrow \small  \cos \emptyset =  \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }

In two dimensional, the x axis of vector form is:

\small\sf{\longrightarrow (b1, b2) = (1, 0) }

\sf{\longrightarrow \small \cos \:  \emptyset =  \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }

\small\longrightarrow \sf{ \dfrac{14}{33.12} }

\small\longrightarrow \sf{\emptyset  \:  = arcCos (\dfrac{ - 14}{33.12} )}

\small\longrightarrow \sf{\emptyset= 115^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{The \: angle  \: between  \: the  \: vector \: }

\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}

6 0
2 years ago
A tightly wound 1000-turn toroid has an inner radius 1.00 cm and an outer radius 2.00 cm, and carries a current of 1.50 A. The t
V125BC [204]

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

<u>Explanation:</u>

Given;

Number of turns, N = 1000

Inner radius, r₁ = 1cm

Outer radius, r₂ = 2cm

Current, I = 1.5A

Magnetic field strength, B = ?

The magnetic field inside a tightly wound toroid is given by B = μ₀ NI / 2πr

where,

a < r < b and a and b are the inner and outer radii of the toroid.

The magnetic field of toroid is

B = \frac{u_oNI}{2\pi r}

Substituting the values in the formula:

B (1.10cm) = \frac{(4\pi X 10^-^7 ) ( 1000)(1.5)}{2\pi (1.10) } \\\\

B (1.10cm) = 27.3mT

Therefore, the magnitude of magnetic field at a distance 1.10cm from the origin is 27.3mT

7 0
3 years ago
Question 4 of 10
ch4aika [34]

Habitat fragmentation is a cost of urban development.

Option: A

Explanation:

Though from the view point or perspective of up gradation and development urban development is much needed but in cost of habitat fragmentation which feels very bitter. As habitat fragmentation leads to the loss of habitat, disruption of ecological cycle and environmental equilibrium.

Actually in the name of urban development we the human use our bread giver environment in a wrong way which causes natural disasters in long run. Animals become endangered , vulnerable and extinct with passage of time. Because they forced to enter into human settlements.

4 0
3 years ago
How Many Days Would a Scientist Have To Wait For The Radioactivity To Be 12.5 The Starting Amount
kiruha [24]
They have to wait 18 days.
3 0
4 years ago
Read 2 more answers
Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that
Greeley [361]

Answer:

The  value  is   m_i =  0.0234 \  kg

Explanation:

Generally from the calorimetry principle we have that

      Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

So here heat gained water is mathematically represented as i.e

      Q_w  = m_w  *  c_w *  (T_w - T )

substituting  0.150 kg for m_w , 4200 J/kg.°C for  c_w , 100°C for   T_w and  75°C for  T

We have  

       Q_w  = 0.150  *  4200 *  (100 - 75 )

         Q_w  =15750 \  J

The Heat loss by the ice is mathematically represented as

      Q_i  = Q_1 + Q_2 +  Q_3

Here     Q_1 is the energy to move the ice to its melting point which is evaluated as  

        Q_1  =  m_i *  c_i * ( T_o -T_i)

Here  m_i is the mass of  ice

       c_i is the specific heat of ice with value  2.05 * 10^3   J/kg.°C

          T_o temperature of ice at melting point with value 0°C

           T_i is the temperature of ice with value  -12°C

Q_2 is the energy to move the ice from its  its melting point to liquid which is evaluated as  

     Q_2  = m_i  *  L

        Here  L  is the Latent heat of melting of ice with value    334 * 10^3   J/kg

Q_3 is the energy to move the ice from  liquid  to  the equilibrium temperature  which is evaluated as        

       Q_1  =  m_i *  c_w * ( T -T_o)

So  

     Q_i  = m_i [ c_i * ( T_o -T_i) + L  + c_w * ( T -T_o) ]

=>   Q_i  = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3  +  4200 * ( 75 - 0) ]

From

 Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\

We have that

  m_i *  673600  =15750

=>     m_i =  \frac{15750}{673600}

=>     m_i =  0.0234 \  kg

8 0
4 years ago
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