Answer:
Speed of the satellite V = 6.991 × 10³ m/s
Explanation:
Given:
Force F = 3,000N
Mass of satellite m = 500 kg
Mass of earth M = 5.97 × 10²⁴
Gravitational force G = 6.67 × 10⁻¹¹
Find:
Speed of the satellite.
Computation:
Radius r = √[GMm / F]
Radius r = √[(6.67 × 10⁻¹¹ )(5.97 × 10²⁴)(500) / (3,000)
Radius r = 8.146 × 10⁶ m
Speed of the satellite V = √rF / m
Speed of the satellite V = √(8.146 × 10⁶)(3,000) / 500
Speed of the satellite V = 6.991 × 10³ m/s
A mutation in the DNA ca happen during growth, when the DNA splits and becomes RNA. Mutations can also occur from other invasive cells, such as bacteria or viruses. A mutation in your DNA can also be there from when you were born. These mutations can cause variations in your appearance, such as eye color, ability to see, taste, smell and the complexion of your face. The mutations can also result in variations in your growth, causing immature or late growth. These mutations are dangerous, and can result in failure of the organisms cells , to complete and utter death of an organism.
a)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = final position of stone = 20.0 meters
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = ?
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
v² = 30² + 2 (- 9.8) (20 - 0)
v = 22.5 m/s
b)
Y₀ = initial position of the stone at the time of launch = 0 m
Y = maximum height gained
a = acceleration = - 9.8 m/s²
v₀ = initial speed of stone at the time of launch = 30.0 m/s
v = final speed = 0 m/s
Using the equation
v² = v₀² + 2 a (Y - Y₀)
inserting the values
0² = 30² + 2 (- 9.8) (Y - 0)
Y = 46 m
Answer:
Explanation:
<u>2-D Projectile Motion</u>
In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are
The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:
Using the formulas for 
Simplifying
Dividing by 
Rearranging
