Answer:
F = 1.07 x 10⁻⁷ N
Explanation:
The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

where,
F = Gravitational Force of attraction = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
m₁ = m₂ = mass of spheres = 20 kg
r = distance between the objects = 50 cm = 0.5 m
Therefore,

<u>F = 1.07 x 10⁻⁷ N</u>
14 km equals 14,000 metres
<u>Answer:</u> The Young's modulus for the wire is 
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:

where,
Y = Young's Modulus
F = force exerted by the weight = 
m = mass of the ball = 10 kg
g = acceleration due to gravity = 
l = length of wire = 2.6 m
A = area of cross section = 
r = radius of the wire =
(Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm = 
Putting values in above equation, we get:

Hence, the Young's modulus for the wire is 
Answer:
27.5 m/s
Explanation:
applying motion equations we can find the answer,
v = u + a*t
Let assume ,
u = starting speed(velocity)
v = Final speed (velocity)
t = time taken for the motion
a = acceleration
by the time of reaching the highest point subjected to the gravity , the speed should be equal to zero (only a vertical speed component is there)
for the complete motion it takes 5.5 s. that means to reach the highest point it will take 5.5/2 =2.75 seconds
we consider the motion upwards , in this case the gravitational acceleration should be negative in upwards (assume g=10 m/s2)
that is,
v = 0 , a = -10
, t =2.75
v = u + at
0 = u -10*2.75
u = 27.5 
Answer:
24,000 m
Explanation:
First find the rocket's final position and velocity during the first phase in the y direction.
Given:
v₀ = 75 sin 53° m/s
t = 25 s
a = 25 sin 53° m/s²
Find: Δy and v
Δy = v₀ t + ½ at²
Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²
Δy = 7736.8 m
v = at + v₀
v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)
v = 559.0 m/s
Next, find the final position of the rocket during the second phase (as a projectile).
Given:
v₀ = 559.0 m/s
v = 0 m/s
a = -9.8 m/s²
Find: Δy
v² = v₀² + 2aΔy
(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy
Δy = 15945.5 m
The total displacement is:
7736.8 m + 15945.5 m
23682.2 m
Rounded to two significant figures, the maximum altitude reached is 24,000 m.