Answer:
the time Joshua travels 1 mile is 12.5 min
Explanation:
Let's start by finding the distance traveled on each lap,
Let's reduce everything to the SI system
R = 400 m
d = 1 mile (1609 m / 1 mile) = 1609 m
L = 2 pi R
L = 2 pi 400
L = 2513 m
Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took
lap Julian = 2/3 turn Joshua
Let's calculate what distance is the same for both of them since they are on the same track
1 lap = 2513 m
d. Julian = 2/3 2513 m
d Julian = 1675 m distance Joshua
Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m
t Julian = 1675/1609 12
t = 12.5 s
Since this is the distance Joshua travels, this is the time Joshua travels 1 mile
Given:
m(mass of the box)=10 Kg
t(time of impact)=4 sec
u(initial velocity)=0.(as the body is initially at rest).
v(final velocity)=25m/s
Now we know that
v=u+at
Where v is the final velocity
u is the initial velocity
a is the acceleration acting on the body
t is the time of impact
Substituting these values we get
25=0+a x 4
4a=25
a=6.25m/s^2
Now we also know that
F=mxa
F=10 x6.25
F=62.5N
Answer:
Explanation:
velocity of ship with respect to water = 6.5 m/s due north

velocity of water with respect to earth = 1.5 m/s at 40° north of east

velocity of ship with respect to water = velocity of ship with respect to earth - velocity of water with respect to earth



The magnitude of the velocity of ship relative to earth is
= 5.66 m/s
Answer:
Earth's water is always in movement, and the natural water cycle, also known as the hydrologic cycle, describes the continuous movement of water on, above, and below the surface of the Earth. Water is always changing states between liquid, vapor, and ice, with these processes happening in the blink of an eye and over millions of years.
Hope this helped!! :))
Explanation:
Answer:
vₐ = v_c
Explanation:
To calculate the escape velocity let's use the conservation of energy
starting point. On the surface of the planet
Em₀ = K + U = ½ m v_c² - G Mm / R
final point. At a very distant point
Em_f = U = - G Mm / R₂
energy is conserved
Em₀ = Em_f
½ m v_c² - G Mm / R = - G Mm / R₂
v_c² = 2 G M (1 /R - 1 /R₂)
if we consider the speed so that it reaches an infinite position R₂ = ∞
v_c =
now indicates that the mass and radius of the planet changes slightly
M ’= M + ΔM = M (
)
R ’= R + ΔR = R (
)
we substitute
vₐ =
let's use a serial expansion
√(1 ±x) = 1 ± ½ x +…
we substitute
vₐ = v_ c (
)
we make the product and keep the terms linear
vₐ = v_c