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Anuta_ua [19.1K]

1 year ago

11

Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 12,500kg. The thrust of its engines is 2

8,000 N.
(a) Calculate its the magnitude of acceleration in a vertical takeoff from the Moon in meter per square second, assuming the acceleration due to gravity on the Moon is 1.67m/s^2.

1 answer:

Nimfa-mama [501]1 year ago

7 0

Answer:

The acceleration due to gravity on the surface of the Moon is approximately 1.625 m/s2, about 16.6% that on Earth's surface or 0.166 ɡ.

gravity of the moon is 1.62 m/s²

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A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the stri

san4es73 [151]

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

$T - mg cos40 = F_c$

$T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}$

$T = 3 + \frac{0.40\times 5^2}{1.8}$

$T = 3 + 5.55$

$T = 8.55 N$

6 0

2 years ago

Suppose the sun were to suddenly shrink in size but its mass remained the same. according to the law of conservation of angular

Ronch [10]

Since the angular momentum is conserved, and the moment of inertia has decreased, so the angular velocity of the sun will rapidly increase and the sun will spin faster

8 0

2 years ago

Please don’t troll. I need someone to actually answer these questions.

mash [69]

Answer:

-3+3 i think this is the answer

Explanation:

i think you can ask someone else sorry

4 0

2 years ago

forces of 3.0 N (west) and 4.0 N (33° S of E) act on an object. What is the net force of the object? Calculate both magnitude an

alexandr402 [8]

The net force is 2.2 N at 80.9 degrees south of east

Explanation:

In order to find the net force, we have to resolve each force along the x-y direction, and then add the components in each direction.

Taking east as positive x-direction and north as positive y-direction, we have:

- First force:

$F_{1x} = -3.0 N\\F_{1y} = 0$

- Second force:

$F_{2x} = (4.0)(cos (-33^{\circ})=3.35 N\\F_{2y} = (4.0)(sin (-33^{\circ})=-2.18 N$

Therefore, the components of the net force are

$F_x = F_{1x}+F_{2x}=-3.0+3.35=+0.35 N\\F_y = F_{1y}+F_{2y}=0+(-2.18)=-2.18 N$

And so, the magnitude of the net force is

$F=\sqrt{F_x^2+F_y^2}=\sqrt{(0.35)^2+(-2.18)^2}=2.2 N$

And the direction is

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-2.18}{0.35})=-80.9^{\circ}$

which means 80.9 degrees south of east.

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

6 0

2 years ago

A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the

Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g= $\frac{530}{1000}=0.530 kg$

1kg=1000 g

Area=A= $935 cm^2=935\times 10^{-4} m^2$

$1 cm^2=10^{-4} m^2$

Height,h=4.4 m

C=1

$\rho=1.21 kg/m^3$

Width of rectangular prism,b=11.6 cm= $\frac{11.6}{100}=0.116 m$

1 m=100 cm

Length,l=23.2 cm= $0.232 m$

Area= $l\times b=0.116\times 0.232=0.0269 m^2$

Terminal velocity, $v_t=\sqrt{\frac{2mg}{\rho CA}}$

Where $g=9.8 m/s^2$

Using the formula

$v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}$

$v_t=17.86 m/s$

The velocity of person, $v=\sqrt{2gh}$

Using the formula

$v=\sqrt{2\times 9.8\times 4.4}$

$v=9.29 m/s$

4 0

2 years ago