Ii. Which of these has a heating element with a low melting point?

Why the formation of sand is a physical change.

Neko [114]

This is a physical change because there isn’t a change in composition of the substance, sand. It is only changing the shape, not having a chemical reaction of any sort.
I hope this helps!

7 0

3 years ago

How can you double the frequency of a wave if you have control over both the wavelength and the wave velocity?

sammy [17]

Answer:

Change the wavelength to half while keeping velocity constant or change the wave velocity to 2 times while keeping wavelength constant.

Explanation:

The frequency of a wave can be defined as the rate of change of wave speed with respect to the wavelength of the wave.

Mathematically,

$f=\frac{v}{\lambda}$

Here, v is the velocity, lambda is the wavelength, and f is the frequency of the wave.

If the observer want to double the frequency of the wave, then he should increase the wave velocity two times and take wavelength constant for this case or either he should half the wavelength of the wave to take the velocity of the wave constant.

5 0

3 years ago

Read 2 more answers

A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the

Advocard [28]

Answer:

a

The total distance traveled is $D = 26760 \ m$

b

The average velocity is $v_{avg} = 6.8 \ m/s$

Explanation:

From the question we are told that

The time taken for first part $t_1 = 22 \ minutes = 22*60 = 1320 \ s$

The speed for the first part is $v_1 = 7.2 \ m/s$

The time taken for second part is $t_2 = 36 \ minutes = 2160 \ s$

The speed for the second part is $v_2 = 5.1 \ m/s$

The time taken for third part is $t_3 = 8 \ minutes = 480 \ s$

The speed for the third part is $v_3 = 13 m/s$

Generally

$distance(D) = velocity * time$

Therefore the total distance traveled is

$D = (v_1 * t_1) + (v_2 * t_2 ) + (v_3 * t_3)$

substituting values

$D = (7.2 * 1320) + (5.1 * 2160) + (13 * 480)$

$D = 26760 \ m$

Generally the average velocity is mathematically represented as

$v_{avg} = \frac{D}{t_{total}}$

Where $t_{total}$ is the total time taken which is mathematically represented as

$t_{total} = 1320 + 2160 + 480$

$t_{total} =3960\ s$

The average velocity is

$v_{avg} = \frac{26760}{3960}$

$v_{avg} = 6.8 \ m/s$

8 0

3 years ago

What is the acceleration along the ground of a 10 kg wagon when it is pulled with a force of 44 N at an angle of 35Â° above the

Olegator [25]

The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

Fcos∅ = ma................. Equation 1

⇒ Where:

F = Force
∅ = angle above the horizontal
m = mass of the wagon
a = acceleration of the wagon

⇒ make a the subject of equation 1

a = Fcos∅/m..................... Equation 2

From the question,

⇒ Given:

F = 44 N
∅ = 35°
m = 10 kg

⇒ Substitute these values into equation 2

a = 44(cos35°)/10
a = 44(0.8191)/10
a = 3.6 m/s²

Hence, The acceleration of the wagon along the ground is 3.6 m/s²

Learn more about acceleration here: brainly.com/question/9408577

3 0

2 years ago

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

5 0

3 years ago