Ask question

Ask question

All categories

3 years ago

13

How is high temperature achieved by concave mirror?

2 answers:

Anika [276]3 years ago

8 0

<u>Answer: </u>

Because <em>"a concave mirror converge the parallel sun rays at a point, so high temperature is achieved"</em>

Example:

It is used in the design of solar furnaces, because they converge the parallel sun rays at a point. This helps to increase the temperature of the furnace.

tensa zangetsu [6.8K]3 years ago

3 0

A concave mirror is used in the design of solar furnaces because they converge the parallel sunrays at a point. This helps to increase the temperature of the furnace.

You might be interested in

Suppose a spring has a relaxed length of 28.3 cm. The simulation refers to this as the natural length. This is the length of the

den301095 [7]

Answer:

Explanation:

Normal length of spring = 28.3 cm

stretched length of spring = 38.2 cm

length of extension = 38.2 - 28.3 = 9.9 cm

= 9.9 x 10⁻² m

force applied to stretch = .55 x 9.8 ( mg )

= 5.39 N

Force constant = force applied / extension

= 5.39 / 9.9 x 10⁻²

= .5444 x 10² N /m

= 54.44 N/m

4 0

3 years ago

Problem #1- In the first scene in which Miss Clark appears, she is supposed to be wearing a BLACK dress.

katrin [286]

Answer:

Blue Lighting

Explanation:

In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.

6 0

3 years ago

Read 2 more answers

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as show

Grace [21]

Answer:

pretty sure its motor 2, with body 1

Explanation:

4 0

3 years ago

Read 2 more answers

(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensiona

bixtya [17]

Answer:

Explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

$1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1
$

$\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0
$

The depth of the well for 100% transmission should be

$V_0=-0.7~{\rm{eV}}$

7 0

3 years ago

A sample contains radioactive atoms of two types, A and B. Initially there are five times as many A atoms as there are B atoms.

victus00 [196]

Answer:

Explanation:

Initially no of atoms of A = N₀(A)

Initially no of atoms of B = N₀(B)

5 X N₀(A) = N₀(B)

N = N₀ $e^{-\lambda t}$

N is no of atoms after time t , λ is decay constant and t is time .

For A

N(A) = N(A)₀ $e^{-\lambda_1 t}$

For B

N(B) = N(B)₀ $e^{-\lambda_2 t}$

N(A) = N(B) , for t = 2 h

N(A)₀ $e^{-\lambda_1 t}$ = N(B)₀ $e^{-\lambda_2 t}$

N(A)₀ $e^{-\lambda_1 t}$ = 5 x N₀(A) $e^{-\lambda_2 t}$

$e^{-\lambda_1 t}$ = 5 $e^{-\lambda_2 t}$

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

half life = .693 / λ

For A

.77 = .693 / λ₁

λ₁ = .9 h⁻¹

$e^{\lambda_2 t}$ = 5 $e^{\lambda_1 t}$

Putting t = 2 h , λ₁ = .9 h⁻¹

$e^{\lambda_2\times 2}$ = 5 $e^{.9\times 2}$

$e^{\lambda_2\times 2}$ = 30.25

2 x λ₂ = 3.41

λ₂ = 1.7047

Half life of B = .693 / 1.7047

= .4065 hours .

= .41 hours .

6 0

3 years ago