Bromine vapour is heavier than air. even so it's spreads upwards in the experiment above. Why?

Explain how to measure volume using a graduated cylinder

Reil [10]

Put object into cylinder and notice the volume increased. Difference of volume is the object's volume

6 0

3 years ago

A 35 n object is on a 25 degree incline. the force of friction up the incline is 8.0 N.

luda_lava [24]

The downwards component of weight of the object would be 35sin25 degrees = 14.8

F=14.8 - 8 = 6.8 N                                         m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2

F(max)= (U)R                       where (U)= coefficient of friction
                                                   and R= Normal reaction force

R= 35cos25
   = 31.72 N
Since, F(max)= 8
          8= (U)* 31.72
      =>(U)= 8/31.72
      =>(U)= 0.25

4 0

3 years ago

A. How can a spinning ball have more lift than one that is not spinning?

sergiy2304 [10]

Answer:

It is spinning, making the ball a little more airborn than the one that isn't spinning

6 0

3 years ago

Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. Elect

Karolina [17]

Answer:

C. D.C.

Explanation:

The current that is being described here is D.C. or direct current. It is the D.C. that moves in one direction from negative to positive. May be created by a battery. It is different from the A.C.( alternating current whose polarity changes regularly). It is A.C. that is used in electrical outlets and not D.C.So, the current option is C.

3 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