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3 years ago

A girl rides her bike at 2.52 m/s on the way to her friends how which is 628 m away.

Physics

1 answer:

Harrizon [31]3 years ago

7 0

Answer: 249 seconds

Explanation:

Speed of bike = 2.52 m/s

Distance = 628 m

Time taken = ? ( let the unknown value be Z)

Recall that speed is the distance covered per unit time.

Hence, speed = Distance / Time taken

Time taken = Distance / speed

Z = 628m / 2.52 m/s

Z = 249.2 seconds (Round to nearest tenth i.e 249 seconds)

Thus, it takes 249 seconds for the girl to get to her friends house

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A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t

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Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

$f_n=\frac{nv_s}{4L}$

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

$f_1=\frac{v_s}{4L}$

B) For the 3rd harmonic you have:

$f_3=\frac{3v_s}{4L}$

C) For the 2nd harmonic:

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Julia performs an experiment to measure the wavelength of four different waves and records her data in the table below.

Svet_ta [14]

The correct order they go in is "1-4-2-3" The correct answer is D.

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3 years ago

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The periodic table is an arrangment of the chemical elements, ordered by atomic number, into families, and by doing so illustrat

olya-2409 [2.1K]

Answer:

Option B) reactive to nonreactive

Explanation:

A) Solid to gas

This might apply, because most of the gases are to the right portion of the periodic table

There are just a few elements that are gases at room temperature; a total of 11: six noble gases (group 18), two halogens (group 17), oxygen (group 16), nitrogen (group 15), and hydrogen (group 1).

Hydrogen is an exception to this trend because they are in the left part of the periodic table.

B) Reactive to nonreactive

The most reactive metals are in group 1 (alkali metals) and the most reactive nonmentals are in the group 18 (halogens). Hence, it is false that the from left to right the preperties go from reactive to nonreactive.

Hence, this is the one that cannot apply.

C) Metal to metalloid to nonmetal

It is true that the metallic character of the elements decreases from left to right; the metals are to the left, going to the right you find some metalloids, and further to the right you find the nonmetals.

Hence, this trend is correct.

D) Increasing number of outer-shell electrons

The number of outer-shell electrons definetly increase from left to right.

The number of electrons in neutral atoms equals the number of protons (to balance the positive and negative charges). Since, the atomic number (the number of protons) increase from left to right, the number of electrons also increase.

In a period (a row of the periodic table) the electrons add to the same outer-shell, then you will find that the number of electrons of the outer-shell increase from left to right. Hence, this is also a correct trend.

In conclusion, the only trend that does not apply is the described by the option B).

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3 years ago

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If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0

Alex73 [517]

Answer:

a) v = 54.7m/s

b) v = (58 - 1.66a) m/s

c) t = 69.9 s

d) v = -58.0 m/s

Explanation:

Given;

The height equation of the arrow;

H = 58t - 0.83t^2

(a) Find the velocity of the arrow after two seconds. m/s;

The velocity of the arrow v can be given as dH/dt, the change in height per unit time.

v = dH/dt = 58 - 2(0.83t) ......1

At t = 2 seconds

v = dH/dt = 58 - 2(0.83×2)

v = 54.7m/s

(b) Find the velocity of the arrow when t = a. m/s

Substituting t = a into equation 1

v = 58 - 2(0.83×a)

v = (58 - 1.66a) m/s

the time when H = 0

Substituting H = 0, we have;

H = 58t - 0.83t^2 = 0

0.83t^2 = 58t

0.83t = 58

t = 58/0.83

t = 69.9 s

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from equation 1;

v = dH/dt = 58 - 2(0.83t)

Substituting t = 69.9s

v = 58 - 2(0.83×69.9)

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