1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
Answer:
5437.5 N
Explanation:
from the question we are given the following:
mass = 7.25 kg
initial speed (u) = 10.5 
finial velocity (v) = 0
compression (s) = -7.35 cm = -0.0735 m (the negative sign is because the face is compressed, so there is a decrease)
we know that force = mass x acceleration,
we have our mass but need to find the acceleration. we can get the acceleration by applying the formula below
therefore
0 = 10.5^{2} + (-2)× a × 0.0735[/tex]
a = 
}[/tex]
a = 750
from F = m x a
force = 7.25 x 750 = 5437.5 N
Answer:
y = 54.9 m
Explanation:
For this exercise we can use the relationship between the work of the friction force and mechanical energy.
Let's look for work
W = -fr d
The negative sign is because Lafourcade rubs always opposes the movement
On the inclined part, of Newton's second law
Y Axis
N - W cos θ = 0
The equation for the force of friction is
fr = μ N
fr = μ mg cos θ
We replace at work
W = - μ m g cos θ d
Mechanical energy in the lower part of the embankment
Em₀ = K = ½ m v²
The mechanical energy in the highest part, where it stopped
= U = m g y
W = ΔEm = - Em₀
- μ m g d cos θ = m g y - ½ m v²
Distance d and height (y) are related by trigonometry
sin θ = y / d
y = d sin θ
- μ m g d cos θ = m g d sin θ - ½ m v²
We calculate the distance traveled
d (g syn θ + μ g cos θ) = ½ v²
d = v²/2 g (sintea + myy cos tee)
d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)
d = 1555.85 /7.8145
d = 199.1 m
Let's use trigonometry to find the height
sin 16 = y / d
y = d sin 16
y = 199.1 sin 16
y = 54.9 m
