What are the units for mass and volume

What is the cause of atmospheric pressure.

marusya05 [52]

Explanation:

hope it's help you ok have a good day

6 0

3 years ago

John wants a new video tape and a new shirt, but he only has $12.00. He buys the video tape and gets $0.52 change. What was his

mestny [16]

Answer:

You have the answer in your comments. I will be copying it so your question doesn't get deleted.

The answers is $0.58

$11.48

the video tape

the new shirt

7 0

2 years ago

The average speed during any time interval is equal to the total distance of travel divided by the total time. Let d represent t

Tanya [424]

Answer:

Average speed = 3.63 m/s

Explanation:

The average speed during any time interval is equal to the total distance travelled divided by the total time.

That is,

Average speed = distance/ time

Let d represent the distance between A and B.

Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,

5.15 = d/t1.

Make d the subject of formula

d = 5.15t1

Let t2 represent the longer time for the return trip at 2.80 m/s . That is,

2.80 = d/t2.

Then the times are t1 = d/5.15 5 and

t2 = d/2.80.

The average speed vavg is given by the following equation.

avg speed = Total distance/Total time

Avg speed = d + d/t1 + t2

Where

Total distance = 2d

Total time = t1 + t2

Total time = d/5.15 + d/2.80

Total time = (2.8d + 5.15d)/14.42

Total time = 7.95d/14.42

Total time = 0.55d

Substitute total distance and time into the formula above.

Avg speed = 2d / 0.55d

Avg Speed = 3.63 m/s

7 0

3 years ago

A motorcycle at a stoplight takes 4.22 seconds to accelerate 28.0 m/s after the light turns green. what is the acceleration?

stira [4]

Acceleration=velocity/time
acceleration=28/4.22
therefore, acceleration=6.64

4 0

2 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0

3 years ago

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