Nichrome is used in electric heater cooker because________

Exercício:

zysi [14]

Answer:

a

Explanation:

a

a is the answer

7 0

3 years ago

The two metallic strips that constitute some thermostats must differ in:_______

bazaltina [42]

Answer:

E. Coefficient of linear expansion

7 0

3 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

$W_{o}=\frac{hc}{\lambda }$

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }$

$W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV$

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

5 0

3 years ago

You are investigating the report of a UFO landing in an isolated portion of New Mexico, and encounter a strange object that is r

SIZIF [17.4K]

Answer: 5 m

Explanation:

We have the following data:

$I_{1}=0.11 W/m^{2}$ is the intensity of the sound at 7.5 m from the source

$r_{1}=7.5 m$ is the distance at which the intensity $I_{1}$ was measured

$I_{2}=1 W/m^{2}$ is the intensity of the sound at $r_{2}$ from the source

We have to find $r_{2}$

Since the object is radiating the signal uniformly in all directions, we can use the <u>Inverse Square Law for Intensity:</u>

$\frac{I_{1}}{I_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}$

Isolating $r_{2}$ :

$r_{2}=r_{1}\sqrt{\frac{I_{1}}{I_{2}}}$

$r_{2}=7.5 m\sqrt{\frac{0.11 W/m^{2}}{1 W/m^{2}}}$

$r_{2}=2.48 m$ This is the distance at which the intensity is the "threshold of pain"

Now, we have to substract this value to $r_{1}$ to find how much closer to the source can we move:

$r_{1}-r_{2}=7.5 m - 2.48 m=5.02 m \approx 5 m$

3 0

3 years ago

Read 2 more answers

1.

Kruka [31]

Answer:

The weight of the object X is approximately 3.262 N (Acting downwards)

The weight of the object Y is approximately 8.733 N (Acting downwards)

Explanation:

The question can be answered based on the principle of equilibrium of forces

The given parameters are;

The weight of Z = 12 N (Acting downwards)

The weight of the pulleys = Negligible

From the diagram;

The tension in the in the string attached to object Z = The weight of object Z = 12 N

The tension in the in the string attached to object X = The weight of the object X

The tension in the in the string attached to object Y = The weight of the object Y

Given that the forces are in equilibrium, we have;

The sum of vertical forces acting at a point, $\Sigma F_y$ = 0

Therefore;

$T_{1y} + T_{2y} + T_{3y} = 0$

$T_{1y} = -( T_{2y} + T_{3y} )$

Where;

$T_{1y}$ = The weight of object Z = 12 N

$T_{1y}$ = 12 N

$T_{2y}$ = The vertical component of tension, T₂ = T₂ × sin(24°)

∴ $T_{2y}$ = T₂ × sin(156°)

Similarly;

$T_{3y}$ = T₃ × sin(50°)

From $T_{1y} = -( T_{2y} + T_{3y} )$ , and $T_{1y}$ = 12 N, we have;

12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)

Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0

Therefore at point B, we have;

T₁ₓ + T₂ₓ + T₃ₓ = 0

The tension force, T₁, only has a vertical component, therefore;

∴ T₁ₓ = 0

∴ T₂ₓ + T₃ₓ = 0

T₂ₓ = -T₃ₓ

T₂ₓ = T₂ × cos(156°)

T₃ₓ = T₃ × cos(50°)

From T₂ₓ = -T₃ₓ, we have;

T₂ × cos(156°) = - T₃ × cos(50°)...(2)

Making T₃ the subject of equation (1) and (2) gives;

Making T₃ the subject of equation in equation (1), we get;

12 = -(T₂ × sin(156°) + T₃ × sin(50°))

∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))

Making T₃ the subject of equation in equation (2), we get;

T₂ × cos(156°) = - T₃ × cos(50°)

∴ T₃ = T₂ × cos(156°)/(-cos(50°))

Equating both values of T₃ gives;

(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))

-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))

∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283

∴ T₂ ≈ -8.02 N

From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;

T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717

∴ T₃ ≈ -11.4 N

The weight of the object X = -T₂ × sin(156°)

∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N

The weight of the object X ≈ 3.262 N (Acting downwards)

The weight of the object Y = -(T₃ × sin(50°))

∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N

The weight of the object Y ≈ 8.733 N (Acting downwards)

4 0

3 years ago

Nichrome is used in electric heater cooker because_________.