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2 years ago

10

When 1kg of a substance changes from solid to liquid at its melting point, it must absorbed an amount of heat equal to which of

the following?
A. specific heat of phase changes
B. specific heat capacity
C. latent heat of fusion
D. latent heat of vaporization

1 answer:

quester [9]2 years ago

4 0

Answer is (c), the latent heat of fusion. That is by definition the heat that 1 kg of a substance must absorb to melt in the vicinity of its melting point.

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What are 3 wether instuerments i know anemometer and barometer but the third one and what there job!!

BaLLatris [955]

Well, those are good ones. Now how about a <u><em>thermometer</em></u> to <em>measure the temperature</em> ?

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2 years ago

The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons

bulgar [2K]

Answer: $\lambda =248.99 nm$

Explanation:

Given

Work function $\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98$

$h=6.626\times 10^{-34} J$

$c=2.998\times 10^8$

$\phi =\frac{hc}{\lambda }$

$\lambda =\frac{hc}{\phi }$

$\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}$

$\lambda =248.99 nm$

3 0

3 years ago

nataly862011 [7]

The density of a material is constant and it is given by the ratio of the mass to the volume of the material

The mass of the liquid and the full bottle ae

The mass of the liquid is <u>450 g</u>

The mass of the filled bottle is <u>530 grams</u>

<u></u>

The reason the above values are correct are as follows:

The given parameters are;

Volume of the bottle, V = Half litre

Mass of the bottle, $m_b$ = 80 g

The volume of liquid in the bottle when filled, V = 500 cm³

The density of the olive oil with which the bottle is filled, ρ = 0.9 g/cm³

a. Required:

To calculate the mass of oil in the bottle

Solution:

The volume of oil in the bottle when the bottle is filled, V = 500 cm³

$Density, \ \rho = \dfrac{Mass}{Volume} = \dfrac{m}{V}$

The mass of the liquid, m = ρ × V

∴ m = 0.9 g/cm³ × 500 cm³ = 450 g

The mass of the liquid, m = <u>450 g</u>

<u></u>

b. The mass of the oil in the bottle, m = grams

The mass of the full bottle, $m_{filled}$ = m + $m_b$

∴ $m_{filled}$ = 450 g + 80 g = 530 g

The mass of the full bottle, $m_{filled}$ = <u>530 grams</u>

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2 years ago

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the

Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0

3 years ago

How to Answer questions 12 and 13

zubka84 [21]

12) The weight of the plane is $7.35\cdot 10^5 N$

13) The lift provided by the wing is $7.35\cdot 10^5 N$

Explanation:

12)

The weight of an object is equal to the force of gravity acting on the object. Near the Earth's surface, it can be calculated as

$W=mg$

where

W is the weight

m is the mass of the object

g is the acceleration of gravity

In this problem, we know that the mass of the plane is

m = 75 tonnes

Since 1 ton = 1000 kg, the mass in kg is

m = 75,000 kg

Also, the acceleration of gravity is

$g=9.8 m/s^2$

Therefore, the weight of the plane is:

$W=(75,000)(9.8)=7.35\cdot 10^5 N$

13)

We can solve this question by applying Newton's second law along the vertical direction of motion. In fact, the net force acting along the vertical direction must be equal to the product between the mass of the plane and the vertical acceleration:

$\sum F_y = ma_y$

where

$\sum F_y$ is the net force in the y-direction

m is the mass of the plane

$a_y$ is the acceleration in the y-direction

There are two forces acting in the vertical direction:

The lift L, acting upward
The weight W, acting downward

So the vertical net force is

$\sum F_y = L-W$

And the equation becomes

$L-W = ma_y$

Also, we know that the plane is travelling in level flight, which means that the vertical acceleration is zero:

$a_y = 0$

Therefore, we get

$L-W=0$

And so the lift is

$L=W=7.35\cdot 10^5 N$

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4 0

3 years ago