Answer:
Explanation:
We shall write the velocities given in vector form to make the solution easy.
The velocity of water with respect to earth that is waV(e) makes 30 degree with north or 60 degree with east so in vector form
waV(e) = 2.2 cos 60 i + 2.2 sin 60 j
waV(e) = 1.1 i + 1.9 j
Similarly , velocity of wind with respect to earth that is wiV(e) , is making 50 degree with west or - ve of x axes so we cal write it in vector form as follows
wiV(e) = - 4.5 cos 50 i - 4.5 sin 50 j
wiV(e) = - 2.89 i - 3.45 j
Now we have to calculate velocity of wind with respect to water that is
wiVwa
wiV( wa) = wiV ( e)+ eV(wa)
= wiV( e)- waV(e)
- 2.89 i - 3.45 j - 1.1 i - 1.9 j
= - 3.99 i - 5.35 j
Magnitude of this relative velocity
D² = 3.99² + 5.35²
d = 6.67 m /s
Answer:
Maximum force, F = 1809.55 N
Explanation:
Given that,
Diameter of the anterior cruciate ligament, d = 4.8 mm
Radius, r = 2.4 mm
The tensile strength of the anterior cruciate ligament,
We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,
So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N
To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.
By definition the relationship between pressure and temperature is given by
Here
P = Pressure
T = Temperature
The ratio of specific heats. For air normally is 1.4.
Our values are given as,
Therefore replacing we have,
Solving for
Therefore the maximum theoretical pressure at the exit is
Answer:
0.2 m/s
Explanation:
given,
mass of astronaut, M = 85 Kg
mass of hammer, m = 1 Kg
velocity of hammer , v =17 m/s
speed of astronaut, v' = ?
initial speed of the astronaut and the hammer be equal to zero = ?
Using conservation of momentum
(M + m) V = M v' + m v
(M + m) x 0 = 85 x v' + 1 x 17
85 v' = -17
v' = -0.2 m/s
negative sign represent the astronaut is moving in opposite direction of hammer.
Hence, the speed of the astronaut is equal to 0.2 m/s
We have that the speed over the ground of a mosquito flying 2 m/s relative to the air against a 2 m/s headwind is
X=0m/s
From the question we are told that
mosquito flying 2 m/s
against a 2 m/s headwind
Generally
The speed over the ground is the Flight Speed minus resistance speed
Generally the equation for the speed over the ground is mathematically given as
X=Flight Speed-resistance speed
Therefore
X=2-2
X=0m/s
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