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evablogger [386]
3 years ago
13

If mass is measured in kg and acceleration is measured in m/s^2, what units would force be measured in? This unit is also know a

s a newton (N).
Physics
1 answer:
Vlada [557]3 years ago
5 0
  • Mass=kg
  • Acceleration=m/s^2

Using Newtons second law

\\ \sf\longmapsto Force=Mass\times Acceleration

\\ \sf\longmapsto Force=kgm/s^2

\\ \sf\longmapsto Force=Newton(N)

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A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 43.0 vibrations
blsea [12.9K]

Answer:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

7 0
3 years ago
Which demonstrates simple harmonic motion?
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3 years ago
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Which is heavier?<br> A kilogram of steel? <br> or<br> A kilogram of feathers?
Dahasolnce [82]

Answer: below

Explanation: 1kg of steel is slightly heavier than 1 kg of feathers. 1 kg of feather will displace more air as the density of feather is very less comparitively. More the volume displaced more is the upthrust and less the apparent weight.

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3 years ago
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Which layer of the sun is seen during a total solar eclipse?
maks197457 [2]

Answer:

The Corona.

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The Corona is the outermost part of the Sun. Usually, it is nto visible due to the light emitted by the Sun. During a total solar eclipse, that is the moon comes between the Earth and the Sun, the glowing white edge around the eclipse is the Corona.

4 0
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A woman is going to a friends house to discuss opening a business. At 11am she starts from rest and accelerates at a constant 2.
bagirrra123 [75]

Answer:

a) Since the woman starts from rest hence her initial velocity equals 0 m/s.

b) We can calculate the final velocity as follows

1) Distance covered while accelerating for 9 seconds is

s_1=\frac{1}{2}\times at^{2}\\\\s=0.5\times 2.5\times 9^{2}=101.25m

2) Cruising speed =v_{c}=at=v_{c}=2.5\times 9=22.5m/s

Distance covered while travelling for 15 minutes at cruising speed =

s_{2}=22.5\times 15\times 60\\\\s_{2}=20.250km

Since the woman cover's 27 km in total hence the speed she travles with traffic speed can be calculated as

v_{traffic}=\frac{Distance}{Time}\\\\v_{traffic}=\frac{27000-101.25-20250}{4500-9-900}\\\\v_{traffic}=1.85m/s

hence the final speed of woman = 1.85 m/s

Average velocity = V_{avg}=\frac{Distance}{Time}\\\\V_{avg}=\frac{27000}{4500}m/s\\\\V_{avg}=6m/s

Her velocity 9 seconds into trip is calculated earlier as 22.5 m/s

5 0
3 years ago
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