Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
Answer:
3.57 MJ
Explanation:
ASSUMING it's fresh water with density of 1000 kg/m³
W = ΔPE = mgΔh = 14.0(1000)(9.81)(26.0) = 3,570,840 J
Salt water would require more.
The car travels at a speed of 25m/s.
<u>Explanation:</u>
Given-
Mass, m = 1500kg
Coefficient of friction, μk = 0.47
Distance, x = 68m
Speed, s = ?
We know,

and
F = μ X m X g
Therefore,
μ * m * g = m * a
μ * g = a
Let, g = 9.8m/s²
So,


We know,

where, v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
If the car comes to rest, the final velocity, v becomes 0.
So,

The car travels at a speed of 25m/s.
In zero order reactions the rate of reaction is independent of reactant concentrations. That is the rate of the reaction does not vary with increasing nor decreasing reactants concentrations. On the other hand, first order reaction are reactions in which the rate of reaction is directly proportional to the concentration of the reacting substance (reactants). In this case, i believe the rate of reaction will triple( increase by a factor of 3)
1) hypothesis
2) data
3) method
I think these are correct.