Horizontal velocity: 81.9 km/h
Vertical velocity: 57.4 km/h
Explanation:
We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.
The horizontal velocity of the stunt bike is given by:
where
v = 100 km/h is the magnitude of the velocity
is the angle of projection
Substituting, we find
The vertical velocity instead is given by
where
Substituting,
Learn more about vector components:
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For only through exploration of the unknown we have come as far as we are now.
Discoveries are what laid the foundation for the technology and all advancement of the society we have today. Similar to how we continue to voyage into the unknowns of this world to find resources to supply for our needs, studies through new materials, species we encounter will all benefit the most common problems we have such as energy resource, medical needs and cure for the incurable diseases. These explanetary exploration may be the next key which holds potential to solve these problems.
Answer:
velocity of the object
Explanation:
For an object moving at a constant acceleration, we would expect to see a position graph with a curved shape and a velocity graph with a straight shape.
Answer:
v_f = 6.92 x 10^(4) m/s
Explanation:
From conservation of energy,
E = (1/2)mv² - GmM/r
Where M is mass of sun
Thus,
E_i = E_f will give;
(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)
m will cancel out to give ;
(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)
Let's make v_f the subject;
v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]
G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²
Mass of sun is 1.9891 x 10^(30) kg
v_i = 2.1×10⁴ m/s
r_i = 2.5 × 10^(11) m
r_f = 4.9 × 10^(10) m
Plugging in all these values, we have;
v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12
v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]
v_f = √[(441000000) + (435.38 x 10^(7))
v_f = 6.92 x 10^(4) m/s
When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.
