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trasher [3.6K]
4 years ago
13

Why would it appear that the sun does not rise during the winter solstice at 66 degrees latitude? (Consider the tilt of the eart

h during that season)
Physics
1 answer:
patriot [66]4 years ago
3 0

<span>This is because the sun never goes beyond the Tropical of Capricorn  and the Tropical of Cancer </span><span><span>whcih are approximately 23.5 degrees S and N respectively</span>. The 66 degrees latitude N and S is the Arctic and Antarctic regions. Because if you add 23.5 and 66 = 90 degrees (approximately), you realize that the sun rays are unable to reach these regions, during winters, because light cannot bend over 90 degrees (light moves in straight lines and only diffracts). </span>







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A projectile is launched from the surface of a planet (mass = 15 x 1024 kg, radius = R = 9.6 x 106 m). What minimum launch speed
Serjik [45]

Answer:

The minimum launch speed is 13366.40 m/s.

Explanation:

Given that,

Mass of planet M=15\times10^{24}\ kg

Radius R= 9.6\times10^{6}\ m

Height = 6R

We need to calculate the speed

Using relation between gravitational force and total energy

-\dfrac{GMm}{6R+R}=\dfrac{1}{2}mv^2-\dfrac{GMm}{R}

\dfrac{}{}mv^2=\dfrac{GMm}{7R}-\dfrac{GMm}{R}

\dfrac{1}{2}mv^2=\dfrac{6GMm}{7R}

v^2=\dfrac{2\times6GM}{7R}

Put the value into the formula

v=\sqrt{\dfrac{2\times6\times6.67\times10^{-11}\times15\times10^{24}}{7\times9.6\times10^{6}}}

v=13366.40\ m/s

Hence, The minimum launch speed is 13366.40 m/s.

6 0
3 years ago
Read 2 more answers
A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca
DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa

Calculate the first moment of area below point C

Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm

Now calculate shear stress at point C

=\frac{FQ}{It}

=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa

Calculate the principal stress at point C

\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa

Calculate the maximum shear stress at piont C

\tau=\frac{\sigma_1-\sigma_2}{2}\\\\=\frac{669.61-(-2.95)}{2}  \\\\=336.28KPa

6 0
3 years ago
Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse
777dan777 [17]

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

      y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

8 0
3 years ago
1. If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possibl
NeX [460]

Answer:

(a)If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision?

No. Because if you have initial momentum P⃗ ≠0 , if both of the objects were at rest after the collision the total momentum of the system would be P⃗ =0 , which violates conservation of momentum

(b)Is it possible for only one to be at rest after the collision?

Yes, that is perfectly possible. It characteristically, happens when both objects are of the same mass. When two objects of the same mass collide and Kinetic energy is conserved (Perfectly Elastic collision) then the two objects interchange velocities.

8 0
2 years ago
Consider a lawnmower of mass m which can slide across a horizontal surface with a coefficient of friction μ. In this problem the
inna [77]

Answer:

Fh = u*m*g / (cos(θ) - u*sin(θ))

Explanation:

Given:

- The mass of lawnmower = m

- The angle the handle makes with the horizontal = θ

- The force applied along the handle = Fh

- The coefficient of friction of the lawnmower with ground = u

Find:

Find the magnitude, Fh, of the force required to slide the lawnmower over the ground at constant speed by pushing the handle.

Solution:

- Construct a Free Body Diagram (FBD) for the lawnmower.

- Realize that there is horizontal force applied parallel to ground due to Fh that drives the lawnmower and a friction force that opposes this motion. We will use to Newton's law of motion to express these two forces in x-direction as follows:

                                     F_net,x = m*a

- Since, the lawnmower is to move with constant speed then we have a = 0.

                                     F_net,x = 0

- The forces as follows:

                                     Fh*cos(θ) - Ff = 0

Where, Ff is the frictional force:

                                     Fh = Ff /cos(θ)

Similarly, for vertical direction y the forces are in equilibrium. Using equilibrium equation in y direction we have:

                                    - W - Fh*sin(θ) + Fn = 0

Where, W is the weight of the lawnmower and Fn is the contact force exerted by the ground on the lawnmower. Then we have:

                                     Fn = W + Fh*sin(θ)

                                     Fn = m*g + Fh*sin(θ)

The Frictional force Ff is proportional to the contact force Fn by:

                                     Ff = u*Fn

                                     Ff = u*(m*g + Fh*sin(θ))

Substitute this expression in the form derived for Fh and Ff:

                                     Fh*cos(θ) = u*(m*g + Fh*sin(θ))

                                     Fh*(cos(θ) - u*sin(θ)) = u*m*g

                                     Fh = u*m*g / (cos(θ) - u*sin(θ))

5 0
3 years ago
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