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algol13
3 years ago
7

The Sun's declination is 0° at the _________. A. summer and winter solstice B. summer and winter equinox C. vernal and autumnal

solstice D. vernal and autumnal equinox
Physics
2 answers:
Olegator [25]3 years ago
5 0

-- "Declination zero" means the object is in the sky at some point directly over the Earth's equator.  

-- If it's the sun and it appears to be over the equator, then that tells us that the Earth's axis is not tilted toward or away from it.  

-- That in turn tells us that the Earth is at one of the two equinoxes in its orbit, either the Spring one or the Autumn one. <em> (D)</em>

-- (The first days of Summer and Winter coincide with solstices, not equinoxes.)  

tigry1 [53]3 years ago
3 0

Answer:

The sun's declination is 0 degrees at the vernal and autumnal equinox hopefully this helps!

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As you drive in your car at 22.5 mi/h, you see a child’s ball roll into the street ahead of you. You hit the brakes and stop as
evablogger [386]

Answer:

Explanation:

Speed of car =22.5miles/hr

U=22.5miles/hour

Applied brake and come to rest

Final velocity, =0

t, =2sec

Given that,

Speed=distance /time

Then,

Distance, =speed, ×time

Converting mile/hour to m/s

Given that

Use: 1 mile= 1600 m, 1 h= 3600s

22.5miles/hour × 1600m/mile × 1hour/3600s

Therefore, 22.5mile/hour=10m/s

Using speed =10m/s

Distance =speed ×time

Distance=10×2

Distance, =20m

The distance travelled before coming to rest is 20m.

5 0
3 years ago
Calculate the mass of an ice block of volume 12m³. the density of the ice is 920kg/m³.​
beks73 [17]

Mass = density • volume

so (12)(920)(kg/m^3•m^3)

11040 kg

4 0
3 years ago
A rectangular plate has a length of (21.7 ± 0.2) cm and a width of (8.2 ± 0.1) cm. Calculate the area of the plate, including it
Grace [21]

Answer:

(177.94 ± 3.81) cm^2

Explanation:

l + Δl = 21.7 ± 0.2 cm

b + Δb = 8.2 ± 0.1 cm

Area, A = l x b = 21.7 x 8.2 = 177.94 cm^2

Now use error propagation

\frac{\Delta A}{A}=\frac{\Delta l}{l}+\frac{\Delta b}{b}

\frac{\Delta A}{A}=\frac{0.2}{21.7}+\frac{0.1}{8.2}

\Delta A=177.94 \times \left ( 0.0092 + 0.0122 \right )=3.81

So, the area with the error limits is written as

A + ΔA = (177.94 ± 3.81) cm^2

8 0
3 years ago
If you were to walk at a constant speed 20m/s for 30 seconds, how far would you walk?
lana [24]

Answer:

600m

Explanation:

30×20 at a constant speed is 600m.

6 0
3 years ago
Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?
vovangra [49]
The kinetic energy of a moving object is given by
K= \frac{1}{2}mv^2
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
K_i =  \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J

while the final kinetic energy is:
K_f =  \frac{1}{2}mv_f^2 =  \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J

So, the kinetic energy lost by Lucy and her bike is
\Delta K = K_i - K_f = 1215 J - 735 J = 480 J
7 0
3 years ago
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