All categories

3 years ago

Hydrogen is a pure substance represented by the chemical symbol H. Which of the following best describes hydrogen?

Physics

2 answers:

n200080 [17]3 years ago

8 0

the answer is b hydrogen is an element made of only one kind of atom

Korvikt [17]3 years ago

5 0

Hydrogen is an element made of only one kind of atom

You might be interested in

an object is thrown with an initial speed v near the surface of earth. assume that air resistance is negligible and the gravitat

nydimaria [60]

Answer:

what

Explanation:

Kagaya mo di krin alam

6 0

3 years ago

Read 2 more answers

Green light (λ = 566 nm) strikes a single slit at normal incidence. Part A

VARVARA [1.3K]

Answer:

755 x 10⁻⁷ m

Explanation:

D = distance of screen

=2. m

d is width of slit = ?

wave length λ = 566 nm

Width of central maxima = 2 λD /d

= 3.00 x 10⁻² =2 λD /d

$d=\frac{2\times566\times10^{-9}\times2}{3\times10^{-2}}$

= 754.67 x 10⁻⁷ m .

= 755x 10⁻⁷ m

8 0

3 years ago

Define 1 kg of mass as international standard

Ann [662]

The kilogram is the Standard International System of Units unit of mass. It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures.

8 0

3 years ago

Read 2 more answers

Determine a valid way of finding the wire’s diameter if you know the resistivity of the material, \rho , and can measure the cur

olganol [36]

Answer:

To find the diameter of the wire, when the following are given:

Resistivity of the material (Rho), Current flowing in the conductor, I, Potential difference across the conductor ends, V, and length of the wire/conductor, L.

Using the ohm's law,

Resistance R = (rho*L)/A

R = V/I.

Crossectional area of the wire A = π*square of radius

Radius = sqrt(A/π)

Diameter = Radius/2 = [sqrt(A/π)]

Making A the subject of the formular

A = (rho* L* I)V.

From the result of A, Diameter can be determined using

Diameter = [sqrt(A/π)]/2. π is a constant with the value 22/7

Explanation:

Error and uncertainty can be measured varying the value of the parameters used and calculating different values of the diameters. Compare the values using standard deviation

7 0

3 years ago

When blueshift occurs,the preceived frequency of the wave would be?

LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency $f\; {\rm Hz}$ at the source. In other words, the source of the wave sends out a peak after every $(1/f)\; {\text{seconds}}$ .

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of $(1/f)\; {\text{seconds}}$ . It would appear to the observer that consecutive peaks arrive every $(1/f)\; {\text{seconds}}\!$ . That would correspond to a frequency of $f\; {\rm Hz}$ .

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of $v \; {\rm m \cdot s^{-1}}$ .

Again, the source of this wave would send out a peak after each period of $(1/f)\; {\text{seconds}}$ . However, by the time the source sends out the second peak, the source would have been $v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m}$ closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than $(1/f)\; {\text{seconds}}$ . The observed frequency of this wave would be larger than the original $f\; {\rm Hz}$ .

6 0

3 years ago