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Archy [21]
3 years ago
9

What is the relationship between sunspots and geomagnetic storms?

Physics
1 answer:
shutvik [7]3 years ago
8 0
Sunspots are spot like structure on the surface of sun, they are created due to geomagnetic flux which inhibits the convection of heat waves thus reducing the temperature of spot with some factor
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katen-ka-za [31]
There is many activities you can do like running, sit-ups, push-ups, and squats. 
7 0
3 years ago
Read 2 more answers
NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
What is the purpose of an man made ecosystem
alexandr402 [8]
Usually to preserve the organisms (maybe the organisms are endangered species or need to be protected in some other way). Another reason could be for the benefit of mankind (for example, an artificial river provides recreational purposes)
3 0
3 years ago
A jet with mass m = 5 × 104 kg jet accelerates down the runway for takeoff at 1.9 m/s2. What is the net vertical force on the ai
OLEGan [10]
The first sentence got me all psyched up to answer the question "What
horizontal force do the engines generate in order to accelerate it ?". 

But the actual question, in the second sentence, turned out to be
a completely different one.

When the plane levels off and continues on at a constant altitude, it's
not accelerating up or down, so the net vertical force on it is zero. 
The lift generated by the wings is exactly balancing the downward
force of gravity on the airplane.
3 0
3 years ago
You place a box weighing 268.3 N on an inclined plane that makes a 42◦ angle with the horizontal.
vivado [14]

Answer:

Wy = - 268.3*cos(42) = - 199.38 [N]

Wx = 268.3*sin(42) = 179.52 [N]

Explanation:

To solve this problem we must make a free body diagram, in the attached image we can see a free body diagram.

Taking the inclined X & y axes with the same angle of 42°, we can see that the weight can be decomposed on both axes.

Since the angle is adjacent to the y-axis, we can use the cosine function

Wy = - 268.3*cos(42) = - 199.38 [N]

Wx = 268.3*sin(42) = 179.52 [N]

5 0
2 years ago
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