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Brums [2.3K]
4 years ago
8

what is the position of the earth when the sun rises to its lowest midday position in the sky in the northern hemisphere and the

southern hemisphere?​
Physics
1 answer:
Sergeu [11.5K]4 years ago
5 0

Answer:

During this time period, the axis of the earth is tilted by 23.5°. As our earth revolves around the sun and completes its one rotation in 365 days. During this rotation in the month of June the Northern Hemisphere “leans into” the sun and experiences more radiations coming from the sun.

That is why it is more illuminated and hot. While in December Northern Hemisphere leans away and the Southern Hemisphere leans into the Sun.

Whereas in September and march the earth the hemispheres are equally favored by sunshine and the earth neither leans into the sun nor away from the sun.

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What are some thermal insulators you can find in a normal house
adoni [48]

Body heat :)

i'm going du.mb here

6 0
3 years ago
A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0
3 years ago
if you could see the electric field coming off a proton you would notice that electric field lines are
lbvjy [14]
<span>you would notice that they're going out from the proton.</span>
8 0
4 years ago
The table shows data for four planetary bodies. If your mass is 68.05 kg, how
scZoUnD [109]

Answer:

252

Explanation:

just did it

3 0
3 years ago
A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the he
maksim [4K]

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 \frac{m}{s}.

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 \frac{m}{s}).

V_{f} ^{2}=V_{0} ^{2} - 2 * g * h

Where:

V_{f}: final speed at maximum height.

V_{0}: initial speed when it falls from the helicopter.

g: gravity taken at 9.8 \frac{m}{s^{2} }

h: height reached from 60 m when leaving the helicopter

as V_{f}=0

0=(12,5\frac{m}{s}) ^{2} - 2 * 9,8\frac{m}{s^{2} } * h

clear h:

h=(12,5 \frac{m}{s^{2} } )^{2} / (2 * 9,8 \frac{m}{s^{2} })

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

V_{f}=V_{0} - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as V_{f}=0

0=12.5 \frac{m}{s} - 9.8 \frac{m}{s^{2} } * t

clearing t

t=12.5 \frac{m}{s} / 9.8 \frac{m}{s^{2} }

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y=v_{0}*t+(\frac{g*t^{2} }{2} )

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

v_{0}: initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 \frac{m}{s^{2} }

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

v_{0}=0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y=\frac{(g*t^{2})}{2}

Y*2=(g*t^{2})

\frac{Y*2}{g}=t^{2}

\sqrt{\frac{Y*2}{g} }=t

Step 3: I replace the values with the incognites and get "t".

t=\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

3 0
4 years ago
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