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Semenov [28]
3 years ago
7

A huge cannon is assembled on an airless planet having insignificant axial spin. The planet has a radius of 5.00 × 106 m and a m

ass of 1.46 × 1023 kg. The cannon fires a projectile straight up at 2000 m/s. An observation satellite orbits the planet at a height of 1000 km. What is the projectile's speed as it passes the satellite? (G= 6.67 × 10-11 N · m2/kg2)
Physics
1 answer:
Nata [24]3 years ago
8 0

Answer:

1830 m /s

Explanation:

The potential energy of the projectile at  the surface of planet

- G x 1.46 x 10²³ m  / 5 x 10⁶

Kinetic energy at surface

1/2 m v²

Total energy = - G x 1.46 x 10²³ m  / 5 x 10⁶ + 1/2  m v²

- 6.67 X 10⁻¹¹ x 1.46 x 10²³ /5 x 10⁶  m + 1/2 m x 2000²

- 1.947 x 10⁶ m + 2m x 10⁶

=  .053  x 10⁶ m

Potential energy of projectile at 1000 km height

= - G x 1.46 x 10²³ m  / 6 x 10⁶

= -  1.6225 x 10⁶  m J  

Total energy

= - 1.6225 x 10⁶ + 1/2 m V ²

Applying conservation of energy in gravitational field at surface and height

- 1.6225 x 10⁶  m + 1/2 m V ² = .053  x 10⁶ m

1/2 V ²  = ( .053 + 1.6225 ) x 10⁶

1/2 V² = 1.6755  x 10⁶ m /s

V = 1.83 X 10³ m/s

1830 m /s

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Rashid [163]

Answer: The inconvienence is it is hurting our ecosystem because all of the plastic is being thrown away and it takes a long time to break down. On the other hand it is handy to have a lot of water supply. I thing water bottles are worth it.

Explanation:

4 0
4 years ago
A person jumps from the roof of a house 3.5-m high. When he strikes the ground below, he bends his knees so that his torso decel
Korvikt [17]

The force exerted on his torso by his legs during the deceleration is 4365 N.

<u>Explanation:</u>

Mass of the torso m=45kg

Height of the building s=3.5 m

Decelerating distance=0.71 m

when he jumps to the ground, the only acceleration is acceleration due to gravity g

<u>motion1 from top to ground </u>

initial velocity u=0

we have to calculate final velocity v using the following equation of motion.

v^2-u^2=2gs\\v^2-0^2=2\times 9.8\times3.5=68.6\\v=\sqrt{68.6} \\=8.3

use height of the building as the distance s as the jump from top to the ground is only described here.

<u>Motion 2 on the ground</u>

v=0

u=8.3(final  velocity of motion 1)

The deceleration after striking the ground can  be calculated from the equation of motion

v^2-u^2=2as\\\\a=v^2-u^2/2\times 0.71\\=0^2-8.3^2/0.71=97 m/s^2

The decelerating distance is used in the place of s since since the motion after hitting the ground is described in this case.

The equation of force is

F=ma\\=45\times 97=4365 N

6 0
3 years ago
2(a)
Tamiku [17]

Answer:

2a.

a=1.13ms^-2

2b.

S=277m

2c.

V=27.7ms-¹

Explanation:

Initial Velocity (U)=22m/s¹

Final Velocity (V)=43m/s²

Time(t) =18.6s

a. a=V-U/t

a=43-22/18.6

a=1.129

a=1.13m/s²

2b.

S=ut+1/2 at²

s=22(10)+1/2×1.13(10)²

s= 220+0.57(10)²

s= 220+0.57(100)

s= 220+57

s=277m

2c.

V=U+AT

V=22+1.13(5)

V=22+5.65

V=22+5.7

V=27.7m/s¹

6 0
3 years ago
Which of the following changes would make a heat engine less efficient
DIA [1.3K]
A heat engine would be less efficient due to many factors 
For instance, a heat engine is more efficient when it uses in cold weather because there is a greater temperature difference ( Carnot Efficient )
A heat engine could be less efficient because of friction
Hope it helps I am a beginner
8 0
4 years ago
The difference in heights of the liquid in the two sides of the manometer is 43.4 cm when the atmospheric pressure is 755 mm hg.
Schach [20]

The atmospheric P is greater than the P in the flask, since the Hg level is lacking down lower on the side open to the atmosphere. 

43.4 cm x (10 mm / 1 cm) = 435 mm 

the density of Hg is 13.6 / 0.791 = 17.2 times better than the liquid in the manometer. This means that 1 mmHg = 17.2 mm of manometer liquid. 

435 mm manometer liquid x (1 mm Hg / 17.2 mm manometer liquid) = 25.3 mm Hg 

The pressure in the flask is 755 - 25.3 = 729.7 mmHg. 

729.7 mmHg x (1 atm / 760 mmHg ) = 0.960 atm.

4 0
3 years ago
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