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Bond [772]
3 years ago
8

50POINTS! What is the escape velocity for lunar module? Lunar module mass 15,200 kg radius of moon 1.74x106m, mass of moon 7.34x

1022kg
Physics
2 answers:
Airida [17]3 years ago
7 0

Answer:

2.73 km/s

Explanation:

The escape velocity of an object in the gravitational field of the moon is (on the surface of the planet)

v=\sqrt{\frac{2GM}{r}}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M=7.34\cdot 10^{22} kg is the mass of the Moon

r=1.74\cdot 10^6 m is the radius of the Moon

As we can see, the escape velocity does not depend on the mass of the lunar module.

Substituting the numbers into the formula, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(7.34\cdot 10^{22}kg)}{(1.74\cdot 10^6 m)}}=2732 m/s=2.73 km/s

mash [69]3 years ago
6 0

Answer:

2.73 km/s

Explanation:

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At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

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v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

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h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

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A block lies on a horizontal frictionless surface. A horizontal force of 100 N is applied to the block giving rise to an acceler
KIM [24]

Answer:

(a) m = 33.3 kg

(b) d = 150 m

(c) vf = 30 m/s

Explanation:

Newton's second law to the block:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Data

F= 100 N

a= 3.0 m/s²

(a) Calculating of the  mass of the block:

We replace dta in the formula (1)

F = m*a

100 =  m*3

m = 100 / 3

m = 33.3 kg

Kinematic analysis

Because the block  moves with uniformly accelerated movement we apply the following formulas:

d= v₀t+ (1/2)*a*t² Formula (2)

vf= v₀+a*t   Formula (3)

Where:  

d:displacement in meters (m)  

t : time interval in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Data

a= 3.0 m/s²

v₀= 0

t = 10 s

(b) Distance the block will travel if the force is applied for 10 s

We replace dta in the formula (2):

d= v₀t+ (1/2)*a*t²

d = 0+ (1/2)*(3)*(10)²

d =150 m

(c) Calculate the speed of the block after the force has been applied for 10 s

We replace dta in the formula (3):

vf= v₀+a*t

vf= 0+(3*(10)

vf= 30 m/s

4 0
3 years ago
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