All categories

3 years ago

Convert as indicated step by step 1)1 sec into day 2) 1 week into sec (3) 50cm into meter

Physics

1 answer:

Allushta [10]3 years ago

4 0

Answer:

1) 1,157*10⁻⁵ day; 2) 604800 sec.; 3) 0.5 m.

Explanation:

1) 1 second into one day:

one day consists of 24 hours, it means 24*3600=86400 seconds. Then 1 second is 1/86400 day or 1,157*10⁻⁵ day.

2) 1 week into seconds:

1 day consists of 3600*24=86400 seconds, then the week is 7*86400=604800 seconds.

3) 50 cm into meter:

100 cm is one meter, then 50 cm is 0.5 meter.

PS. note, there is not the only way to perform conversion.

You might be interested in

Which statements describe intensity? Check all that apply.

REY [17]

Answer:

it's A, D, E

Explanation:

6 0

3 years ago

Read 2 more answers

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

2 years ago

Consider a large truck and a small car driving up a straight steep hill. The truck is moving at 60 miles per hour and the car at

Oksana_A [137]

Answer:

The correct answer is option A.

Explanation:

Force is defined as push or pull on an object with mass due to which change velocity occurs (acceleration).

$Force=Mass\times Acceleration$

$F=ma$

$a=\frac{dv}{dt}[tex][tex]m\times \frac{dv}{dt}[tex]We are given with large truck and small car at constant speed, which means that acceleration zero.[tex]a=\frac{dv}{dt}=0$ if ,v = constant

Net force on the vehilce

F = Normal - Weight

0 = N = W

N = W

So, the force experienced by the object will be due to its mass, and higher the mass more will be the experienced by an object.So, large truck will experience larger net force.

7 0

3 years ago

PLEASE HELP!!!!

weqwewe [10]

The answer is A. <span>Some work input is used to overcome friction. </span>

6 0

3 years ago

During a field experiment about speed, a scientist created the chart above. The chart shows distance and time measurements for a

BaLLatris [955]

Answer:

yes

Explanation:

its not a good thing for the rest of your life but you have a PS4

6 0

2 years ago

Convert as indicated step by step 1)1 sec into day 2) 1 week into sec (3) 50cm into meter​

Convert as indicated step by step 1)1 sec into day 2) 1 week into sec (3) 50cm into meter