Velocity = displacement/time
I hope i helped you <33
Answer:
d = 142.5 m
Explanation:
This is a vector exercise. Let's calculate how much the boat travels in the 40s
d₀ =
t
d₀ = 0.75 40
d₀ = 30 m
Let's write the kinematic equations
Boat
x = d₀ +
t
x = 0 +
t
At the meeting point the coordinate is the same for both
d₀ +
t =
t
t (
-
) = d₀
t = d₀ / (
-
)
The two go in the same direction therefore the speeds have the same sign
t = 30 / (0.95-0.775)
t = 150 s
The distance traveled by man is
d =
t
d = 0.95 150
d = 142.5 m
Answer:
F = -6472.9 N
F= -6.47 kN
Explanation:
First of all you have to convert the data to SI units
so for the velocity you have :
Vi = 43km/h *(1000m/1km)*(1h/3600s) ---> using conversion factors
Vi= 11.9444 m/s
dX : distance the passanger moves
dX = 54cm*(1m/100cm) --> using conversion factors
dX = 0.54 m
Now to calculate the force we are going to use the sum of focers equals to mass for acceleration:
Sum F = m*a
We have to find a so we are going to use the velocity's formula as follows to solve a:
Vf ^2 = Vi^2 +2*a*dX
Vf=0 --> the passenger does not move after the airbag inflates.
a= -(Vi^2)/(2*dX)
you solve de acceleration with the data you hae and you will find
a = -132.1 m/ s^2
Now you can solve the Sum F equation
Sum F = 49 Kg * (-132.1 m/s^2)
F = -6472.9 N
F= -6.47 kN
Answer:
W = 2.3 10² 
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V =
π r³
V =
π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10² 
therefore the weight of the drop is much greater than the electric force
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