A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the grou
1 answer:
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
= 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
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Answer:
The weight of measuring stick is 9.8 N
Explanation:
given information:
the mass of the rock, = 1 kg
measuring stick, x =1 m
d = 0.25 m
to find the weight of measuring stick, we can use the following equation:
τ = Fd
τ = 0
-
= 0
F_{r} = the force of the rock
F_{s} = the force of measuring stick
= m g
= 1 kg x 9.8 m/s
= 9.8 N
thus, the weight of measuring stick is 9.8 N
Answer:
Explanation:
<u>Net Forces and Acceleration</u>
The second Newton's Law relates the net force acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.
According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.
The magnitude of the resulting force is computed as the hypotenuse of a right triangle
The acceleration can be obtained from the formula
Note we are using only magnitudes here
The work done on the sail is 600 J
Explanation:
The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:
where
m is the mass of the sail
g is the acceleration of gravity
(mg) is the weight of the sail
is the change in height of the sail
In this problem we have
mg = 150 N (weight)
Substituting, we find the work done:
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