If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
a) (95.4 i^ + 282.6 j^) N
, b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º
Explanation:
a) Let's use trigonometry to break down Jennifer's strength
sin θ = Fjy / Fj
cos θ = Fjx / Fj
Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be
T = 90 -32 = 58º
Fjy = Fj sin 58
Fjx = FJ cos 58
Fjx = 180 cos 58 = 95.4 N
Fjy = 180 sin 58 = 152.6 N
Andrea's force is
Fa = 130.0 j ^
We perform the summary of force on each axis
X axis
Fx = Fjx
Fx = 95.4 N
Axis y
Fy = Fjy + Fa
Fy = 152.6 + 130
Fy = 282.6 N
F = (95.4 i ^ + 282.6 j ^) N
b) Let's use the Pythagorean theorem and trigonometry
F² = Fx² + Fy²
F = √ (95.4² + 282.6²)
F = √ (88963)
F = 298.27 N
tan θ = Fy / Fx
θ = tan-1 (282.6 / 95.4)
θ = tan-1 (2,962)
θ = 71.3º
c) To avoid the movement they must apply a force of equal magnitude, but opposite direction
F' = 298.27 N
θ' = 180 + 71.3
θ = 251.4º
Answer:
The answer is 26/98 how i did this is i divided them mulitiplyed well i cant really explain it but im pretty dure its right
Explanation:
There are two stages to the flight: acceleration stage and deceleration stage.
m₁ = 200 kg, mass of the rocket
m₂ = 100 kg, mass of fuel
a₁ = 30.0 m/s², upward acceleration when burning fuel
Ignore air resistance and assume g = 9.8 m/s².
Acceleration stage:
The rocket starts from rest, therefore the initial vertical velocity is zero.
The distance traveled is given by
s₁ = (1/2)*(30.0 m/s²)*(30.0 s)² = 13500 m
Deceleration stage (due to gravity):
The initial velocity is u = (30.0 m/s²)*(30 s) = 900 m/s
The initial height is 13500 m
At maximum height, the vertical velocity is zero.
Let s₂ = the extra height traveled. Then
(900 m/s)² - 2*(9.8 m/s²)*(s₂ m) = 0
s₂ = 900²/19.6 = 41326.5 m
The maximum altitude is
s₁+s₂ = 54826.5 m
Answer: 54,826.5 m
