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lutik1710 [3]
3 years ago
6

Assuming you are in the Northern Hemisphere, how would you expect the location of the Sun in your local sky at noon to differ fr

om season to season?
Physics
1 answer:
topjm [15]3 years ago
3 0

Answer:

It should be higher during the summer.

Explanation:

Assuming that youlive in the Northern Hemisphere, the sun's location through the season from winter to summer at noon, should be different. The sun would be observed at a higher position over the horizon in the summer season than it is in the winter season. The reasons are equinoxes, solstices and the ecliptic circle.

I hope this answer helps.

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When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking
postnew [5]

Answer:

d=2.4\ miles

Explanation:

Given:

  • average walking speed, v_w=3\ mph
  • average biking speed, v_b=12\ mph

<u>According to given condition:</u>

t_w=t_b+\frac{36}{60}

where:

t_w= time taken to reach the building by walking

t_b= time taken to reach the building by biking

We know that,

\rm time=\frac{distance}{speed}

so,

\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}

\frac{d}{3}=\frac{d}{12} +\frac{3}{5}

d=2.4\ miles

7 0
3 years ago
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Is the amplitude just 2? Or do I combine all of them and do 6?​
oksian1 [2.3K]

Answer:

combine them all

6 m

Explanation:

7 0
2 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
3 years ago
How is carbonic acid responsible for the formation of some caves? ASAP PLEASE!!!!
butalik [34]
It desolves calcium carbonate causing plates to bend and faults to occur. Faults and plates moving cases the formation of caves.

-Hopes this helps :)
4 0
3 years ago
A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,
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<span>The velocity would be 54.2 m/s We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
5 0
3 years ago
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