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ohaa [14]
3 years ago
11

Which statement describes a credible source?

Physics
1 answer:
Assoli18 [71]3 years ago
7 0

Answer: C. It includes factual information supported by other sources

Explanation:

:) its really obvious tbh w you

gamertag: itzmamas1k

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In the simulation above, as the projectile travels upward, how does the vertical velocity change?
alexandr1967 [171]

Your answer should be, "Vertical velocity decreases."

added the simulation after results.

3 0
4 years ago
A small ball is attached to one end of a spring that has an un- strained length of 0.200 m. The spring is held by the other end,
masya89 [10]

Answer:

\Delta x=0.002287\ m=2.287\ mm

Explanation:

Given:

  • un-stretched length of the spring, l=0.2\ m
  • speed of revolution of the ball in horizontal plane, v=3\ m.s^{-1}
  • length stretch during the motion, \Delta l=0.01\ m

<u>Now the radius of revolution of the ball:</u>

r=l+\Delta l

r=0.2+0.01

r=0.21\ m

<u>Now in this case the centrifugal force is equal to the spring force:</u>

F_c=F_s

m.\frac{v^2}{r} =k.\Delta l

where:

m = mass of the ball

k = spring constant

m\times \frac{3^2}{0.21} =k\times 0.01

k=(4285.714\times m)\ N.m^{-1}

<u>Now the extension in the spring upon hanging the ball motionless:</u>

m.g=k.\Delta x

9.8\times m=(4285.714\times m)\times \Delta x

\Delta x=0.002287\ m=2.287\ mm

6 0
3 years ago
Stellar objects moving away from earth________.
Zanzabum
Omg. Ummm. I am pretty sure it is Shift toward the red end of the spectrum
7 0
4 years ago
Read 2 more answers
What is the difference between amplitude modulation and frequency modulation​
elena-s [515]

Answer:

Explanation:

Difference Between AM And FM. ... The main difference between both modulations is that in frequency modulation, the frequency of the carrier wave is modified as per the transmit data, while in amplitude modulation, the carrier wave is modified according to the data.

3 0
3 years ago
Read 2 more answers
When blue light of wavelength 460 nm falls on a single slit, the first dark bands on either side of center are separated by 45.0
lyudmila [28]

Answer:

width of slit =1.23× 10⁻⁶ m  

Explanation:

we know the condition of diffraction minima,

d sin θ = n λ

λ = wavelength     θ = angle between the central maxima and 1st  minima

d = slit width

for first minima  n = 1

now,

d =\dfrac{n \lambda}{sin \theta}

\theta = \dfrac{45^0}{2} = 22.5^0

d =\dfrac{1\times 460 \times 10^{-9}}{sin 22.5^0}

d = 1228 × 10⁻⁹ m  = 1.228× 10⁻⁶ m

d =  1.23× 10⁻⁶ m

width of slit =1.23× 10⁻⁶ m  

6 0
3 years ago
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