Answer: 6,400 km
Explanation:
The weight of a person is given by:
where m is the mass of the person and g is the acceleration due to gravity. While the mass does not depend on the height above the surface, the value of g does, following the formula:
where
G is the gravitational constant
M is the Earth's mass
r is the distance of the person from the Earth's center
The problem says that the person weighs 800 N at the Earth's surface, so when r=R (Earth's radius):
(1)
Now we want to find the height h above the surface at which the weight of the man is 200 N:
(2)
If we divide eq.(1) by eq.(2), we get
By solving the equation, we find:
which has two solutions:
--> negative solution, we can ignore it
--> this is our solution
Since the Earth's radius is 
, the person should be at 
above Earth's surface.
Answer:
35.2 ps
Explanation:
Hello!
A US short ton is equal to 2 000 pounds, therefore:
2.25 tons = 4500 pounds
Now lets convert the area of each tire footprint:
1 cm = 0.393701 in
12.5 cm = 4.92126 in
16.5 cm = 6.49607 in
Therefore, the area of the footprint is: 31.9688 in^2
Then, the pressure exerted by the car is given by the weigth devided by the total area, that is 4 times the area of the footprint :
4500 /(4 * 31.9688) psi = 35.1906 psi
Rounding to the first tenth
35.2 psi
Answer:
a = 616850.28 m/s²
Explanation:
Given that,
The radius of the neutron star, r = 10 Km
= 10,000 m
The time period of the neutron star, T = 0.8 s
The centripetal acceleration is given by the formula,
a = v²/r
The linear velocity is given by the relation,
v = rω
The time taken to complete one complete rotation is given by the relation
T = 2π /ω
Where,
ω = 2π / T
Substituting v and ω into the equation for centripetal acceleration. It becomes
a = 4π²r/T²
Substituting the given values in the above equation
a = 4π² x 10000 / 0.8²
= 616850.28 m/s²
Hence, the centripetal acceleration of this person is, a = 616850.28 m/s²
Answer:
xcritical = d− m1
/m2
( L
/2−d)
Explanation: the precursor to this question will had been this
the precursor to the question can be found online.
ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)
. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces
smallest possible value of x such that the bar remains stable (call it xcritical)
∑τA = 0 = m2g(d− xcritical)− m1g( −d)
xcritical = d− m1
/m2
( L
/2−d)
