During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
Answer: 0.98m
Explanation:
P = -74 mm Hg = 9605 Pa = 9709N/m^2
= 9605 kg m/s^2/m^2
density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3
Pressure equation: P = rho g h
h = P/(rho g)
h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)
h = 0.98 m
0.98m is the maximum depth he could have been.
Answer:
Explanation:
<u>Free Fall Motion</u>
A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.
The speed vf of the object when a time t has passed is given by:
Where 
Similarly, the distance y the object has traveled is calculated as follows:
If we know the height h from which the object was dropped, we can solve the above equation for t:
The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:
The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:
Therefore, it has traveled down a distance:
Thus, the height of the pen is:
Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa
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.
