Answer:
r = 4.44 m
Explanation:
For this exercise we use the Archimedes principle, which states that the buoyant force is equal to the weight of the dislodged fluid
B = ρ g V
Now let's use Newton's equilibrium relationship
B - W = 0
B = W
The weight of the system is the weight of the man and his accessories (W₁) plus the material weight of the ball (W)
σ = W / A
W = σ A
The area of a sphere is
A = 4π r²
W = W₁ + σ 4π r²
The volume of a sphere is
V = 4/3 π r³
Let's replace
ρ g 4/3 π r³ = W₁ + σ 4π r²
If we use the ideal gas equation
P V = n RT
P = ρ RT
ρ = P / RT
P / RT g 4/3 π r³ - σ 4 π r² = W₁
r² 4π (P/3RT r - σ) = W₁
Let's replace the values
r² 4π (1.01 10⁵ / (3 8.314 (70 + 273)) r - 0.060) = 13000
r² (11.81 r -0.060) = 13000 / 4pi
r² (11.81 r - 0.060) = 1034.51
As the independent term is very small we can despise it, to find the solution
r = 4.44 m
Let 
be the direction the swimmer must swim relative to east. Then her velocity relative to the water is
The current has velocity vector (relative to the Earth)
The swimmer's resultant velocity (her velocity relative to the Earth) is then
We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:
which is approximately 41º west of north.
Answer:
the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Explanation:
Since airplane is moving horizontally with constant speed v
so when object is dropped from the plane then the speed of the object will be same as that of the speed of the airplane
so we can say that two object when dropped after some interval of time then they always lie in same vertical line
now we know that they both have same acceleration in vertical line so the motion of two objects relative to each other in vertical direction is always uniform motion because they have no acceleration with respect to each other
So the vertical distance between the two object will increase uniformly when they are dropped after a fixed interval of time
Answer:
1.1 sec
Explanation:
m = mass of the box = 8 kg
k = spring constant of the spring = 69 N/m
v = initial speed of the box = 1.5 m/s
t = time period of oscillation of box in contact with the spring
Time period is given as
Inserting the values
t = 1.1 sec
Answer:
H = 109.14 cm
Explanation:
given,
Assume ,
Total energy be equal to 1 unit
Balance of energy after first collision = 0.78 x 1 unit
= 0.78 unit
Balance after second collision = 0.78 ^2 unit
= 0.6084 unit
Balance after third collision = 0.78 ^3 unit
= 0.475 unit
height achieved by the third collision will be equal to energy remained
H be the height achieved after 3 collision
0.475 ( m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
