Please i need answer for part d

Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0

3 years ago

Which scientist was the first to use the telescope in astronomy

Mashutka [201]

Answer:

Galileo Galilei

Explanation:

although Galileo was not the scientist who invented the telescope, he was the first to use it to observe celestial objects. he used the telescope in 1609. his discovery included more accurate information about the moon, the sun and some of the planets.

4 0

3 years ago

a ladybug crawls vertically on a leaf. its motion is shown on the following graph of vertical position y vs time t what is the a

Sladkaya [172]

Explanation:

The attached figure shows the position-time graph of a ladybug. We need to find the average speed of the ladybug between t = 4 s to t = 7 s.

We know that, the slope of a position-time graph gives velocity of an object. It can be given by :

$v=\dfrac{y_2-y_1}{t_2-t_1}$

In this case, the position of a ladybug at t = 4 s and at t = 7 s is the same i.e. 2 m.

It means its velocity is equal to at this time or we can say that ladybug is at rest.

8 0

4 years ago

Read 2 more answers

Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinet

evablogger [386]

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

8 0

4 years ago

Two identical strings, A and B, have nearly the samtension. When they both vibrate in their fundamental resonantmodes, ther is a

prohojiy [21]

Answer:

A. Before and after tightening B has a higher frequency than that of A

Explanation:

Since, the beat frequency is 3 Hz before tightening. Therefore, the difference between the frequencies of strings, A and B is 3 Hz. As, we tighten the string B, its tension increases. From the following relation it is clear that the increase in tension increases the fundamental frequency.

f₁ = (1/2L)√[T/μ]

where,

f₁ = fundamental frequency

L = length of string

T = Tension in string

μ = linear mass density of string

Therefore, frequency of B increases slightly. And the beat frequency or the difference between frequencies of A and B, becomes 6 Hz.

Now, there can be two cases, which are:

Either, the frequency of B was 3 Hz less than frequency of A earlier and after tightening frequency of B became 6 Hz greater than frequency of A.

Or, the frequency of B was 3 Hz greater than the frequency of A, and after tightening it became 6 Hz greater than frequency of A.

The second option seems more reasonable because it requires 3 Hz of increase in frequency of B, while the first option requires increase of 9 Hz in frequency of B. Since, the there is slight tightening. Therefore, the frequency will also increase by a slight value. Thus, the correct option is:

<u>A. Before and after tightening B has a higher frequency than that of A</u>

3 0

3 years ago