2 years ago

A sharp raises a note by

Physics

2 answers:

vivado [14]2 years ago

8 0

The answer should be B. A half step

-Dominant- [34]2 years ago

6 0

The flat lowers a note by a half step while the sharp raises a note by a half step. So your answer is B. A half step

You might be interested in

PLEASE HELP!!

dedylja [7]

Pretty sure it’s A. Hope this helps.

4 0

2 years ago

light of a certain frequency has a wavelength of 438 nm in water.What is the wavelength of this light in benzene

sveticcg [70]

Answer:

388.97 nm

Explanation:

The computation of the wavelength of this light in benzene is shown below:

As we know that

n (water) = 1.333

n (benzene) = 1.501

$\lambda (water) \times n(water) = \lambda (benzene) \times n(benzene)$

And, the wavelength of water is 438 nm

$\lambda (benzene) = \lambda (water) [\frac{n(water)}{n(benzene}]$

Now placing these values to the above formula

So,

$= 438 \times \frac{1.333}{1.501}$

= 388.97 nm

We simply applied the above formula so that we can easily determine the wavelength of this light in benzene could come

5 0

2 years ago

A 9,100-microfarad [muF] capacitor is charged to 22 volts [V]. If the capacitor is completely discharged through an iron rod

MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0

3 years ago

A truck on the freeway originally moving at 6.6 meters/second accelerates uniformly with acceleration a = 2.8 meters/second2 for

Sonja [21]

Answer:

139.514 metres

Explanation:

Initial velocity of the truck = 6.6 m/s

Acceleration of the truck = 2.8 m/s^2

Time interval = 7.9 s

Therefore we use the formula,

s = ut + 1/2 at^2

*where s(the distance travelled)...u(the initial velocity)...t(the time period)

; s = 6.6(7.9) + 1/2 (2.8)(7.9)^2

; s = 52.14 + 87.374

The distance moved by the truck = 139.514m

8 0

2 years ago

A body travels 30m in 5s, 45m in 7s and then 65m in the last 5s. Find the average speed of the body

JulijaS [17]

We can find the average speed of the body by finding the total distance covered, and then dividing it by the total time of the motion.

The total distance covered is:
$S=30 m + 45m+65m=140 m$

while the total time of the motion is
$t=5 s+7s+5 s=17 s$

So, the average speed of the body is:
$v= \frac{S}{t}= \frac{140 m}{17 s}=8.24 m/s$

8 0

3 years ago