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3 years ago

Choose two features of the sun and describe them.

1 answer:

8 0

Okay, I am going to use solar prominence and solar flares.

So, Solar Prominence is a large, bright, gaseous feature extending outward from the Sun's surface, often in a loop shape. Prominence are anchored to the Sun's surface in the photo-sphere, and extend outwards into the Sun's corona.

A solar flare is a sudden flash of brightness observed near the Sun's surface. It involves a very broad spectrum of emissions, requiring an energy release of up to 6 × 10²⁵ joules of energy (roughly the equivalent of 160,000,000,000 megatons of TNT, over 25,000 times more energy than released from the impact of Comet Shoemaker–Levy 9 with Jupiter).

Hope this helps

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

Andrews [41]

Explanation:

It is given that,

Mass of the runner, m = 70 kg

Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, $A=110\ mm^2=0.00011\ m^2$

Part A,

Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, $Y=0.15\times 10^{10}\ N/m^2$

The formula of the Young modulus is given by :

$Y=\dfrac{F/A}{\dfrac{\Delta L}{L}}$

$0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}$

$\Delta L=0.0049\ m$

Part B,

The fraction of the tendon's length does this correspond is given by :

$\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}$

$\dfrac{\Delta L}{L}=0.0326$

Hence, this is the required solution.

6 0

3 years ago

Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce

Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

$\omega=\frac{v}{r}$

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

$|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}$

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0

3 years ago

Read 2 more answers

Why are hurricanes considered more damaging than tornadoes when tornadoes have stronger winds

chubhunter [2.5K]

-- A tornado follows a path that's a few miles wide, for a few hours.
   Then it's all over.

-- A hurricane follows a path that's several hundred miles wide,
   for a week or two, before it's over.
   Then comes the rain, continuing on the same path, for another week.

8 0

3 years ago

Read 2 more answers

Suppose an empty grocery cart rolls downhill in a parking lot. The cart has a maximum speed of 1.3 m/s when it hits the side of

Ket [755]

The cart comes to rest from 1.3 m/s in a matter of 0.30 s, so it undergoes an acceleration a of

a = (0 - 1.3 m/s) / (0.30 s)

a ≈ -4.33 m/s²

This acceleration is applied by a force of -65 N, i.e. a force of 65 N that opposes the cart's motion downhill. So the cart has a mass m such that

-65 N = m (-4.33 m/s²)

m = 15 kg

4 0

3 years ago