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Softa [21]
2 years ago
11

PLEASE ANSWER ASAP

Physics
1 answer:
neonofarm [45]2 years ago
6 0

Answer:

2.55sec

Explanation:

time = distance/speed = (87 m)/(34 m/s) = 2.55sec

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Answer <br> A)<br> B <br> C <br> please
babymother [125]

Answer:

<h2>a is b is c is</h2>

Explanation:

brash this so hard

5 0
1 year ago
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
2 years ago
What are the parts of a wedge and their function?​
Svet_ta [14]

Answer:

A wedge is a machine that consists of two inclined planes, giving it a thin end and thick end. A wedge is used to cut or split apart objects. Force is applied to the thick end of the wedge, and the wedge applies force to the object along both of its sloping sides. This force causes the object to split apart

5 0
2 years ago
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What Organ filters the blood to remove the waste
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The answer to your question is kidneys
4 0
2 years ago
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A bobsled team accelerates the sled to go to 150 m/s in 3 seconds from rest.(a)What is the
Sergeeva-Olga [200]

Answer:

Explanation:

From the equation of Newton's laws of motion

v = u + at where v is final velocity , u is initial velocity and t is time.

150 = 0 + a x 3

a = 50 m / s ²

s = ut + 1/2 at²  ; s is distance travelled

s = 50 x 3 + .5 x 50 x 3²

= 150 + 225

= 375 m .

7 0
2 years ago
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