Why does gravity on Earth have a stronger attraction with you than the sun has with you? Plsssssss help I really need this rn

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

A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass o

luda_lava [24]

Answer:

Explanation:

We shall apply law of conservation of momentum to know velocity after collision . Let it be v .

total momentum before collision = total momentum after collision

15 x 1.5 - 12 x .75 = ( 15 + 12 ) v

v = .5 m /s

kinetic energy before collision

1/2 x 15 x 1.5² + 1/2 x 12 x .75²

= 16.875 + 3.375

= 20.25 J

kinetic energy after collision

= 1/2 x ( 15 + 12 ) x .5²

= 3.375 J

Loss of energy = 16.875 J

This energy appear as heat and sound energy that is produced during collision .

4 0

4 years ago

What is the value of x in the triangle? a 45-45-90 triangle with leg length x and hypotenuse length 4

mrs_skeptik [129]

Answer:

$x = 2\sqrt 2$

Explanation:

Given

$Hypotenuse = 4$

Required

Find x

Since the triangle is a 45-45-90 triangle, the following relationship exists

$x^2 + x^2 = 4^2$ --- i.e. the other legs are equal

So, we have:

$2x^2 = 16$

Divide both sides by 2

$x^2=8$

Take square roots of both sides

$x = \sqrt 8$

Simplify

$x = 2\sqrt 2$

5 0

3 years ago

Read 2 more answers

A beaker of water rests on an electronic balance that reads 875.0 g. A 2.75-cm-diameter solid copper ball attached to a string i

Firdavs [7]

Answer:

a

The tension in the string is $T = 0.85 N$

b

The new balance reading is $M_b = 885.86 g$

Explanation:

From the question we are told that

The mass of the beaker of water is $m = 875 .0g$

The diameter of the copper ball is $d = 2.75 cm = \frac{2.75}{100} = 0.0275 m$

There are two forces acting on the copper ball

The first is the Buoyant force of the water pushing it up which is mathematically represented as

$F = \rho V g$

Where $\rho$ is the density of water which has value of $\rho = 1000 kg/m^3$

g is the acceleration due to gravity $g= 9.8 \ m/s^2$

$V$ is the volume of water displaced by the copper ball which is mathematically evaluated as

$V = \frac{4}{3} \pi r^3$

The radius r is $r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m$

Substituting value

$V = \frac{4}{3} * 3.142 * (0.01375)^3$

$V = 1.08 9 *10^{-5 } m^3$

Substituting for F

$F = 1000 * 1.089 *10^{-5} * 9.8$

$F = 0.1067 N$

The second force is the weight of the copper ball which is mathematically represented as

$W_c = mg$

Now m is the mass which can be mathematically evaluated as

$m = \rho_c * V$

Where is the density of copper with value of $\rho_c = 8960 kg /m^3$

So $m = 8960 * 1.089 *10^{-5}$

$m = 0.0976$

So the weight of copper is

$W_c = 0.09756 * 9.8$

$W_c = 0.956 N$

Now the tension the string would be mathematically evaluated as

$T = W_c - F$

So $T = 0.956 - 0.1067$

$T = 0.85 N$

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

$m_z = \frac{1.0645 }{9.8 }$

$m_z = 0.01086 kg$

Converting to grams

$m_z = 10.86 g$

So the new balance reading is

$M_b = 875.0 +10.86$

$M_b = 885.86 g$

5 0

3 years ago

2. (Work-Energy Theorem) A watermelon of mass m is dropped from rest from the roof of a building of height h and feels no air re

andreyandreev [35.5K]

Answer:

SORRY SISO REALLY DON'T KNOW ANS SORRY!........☹☹☹

5 0

3 years ago