Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m
Answer:
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
Explanation:
Hi there!
The equation for the velocity of an object thrown upward is the following:
v = v0 + g · t
Where:
v = velocity of the ball.
v0 = initial velocity.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
t = time.
To find the velocity of the ball at t = 0.40 s, we have to replace "t" by 0.40 s in the equation:
v = v0 + g · t
v = 3.0 m/s - 9.8 m/s² · 0.40 s
v = -0.92 m/s
The velocity of the ball is 0.92 m/s in the downward direction (-0.92 m/s).
Answer:
Explanation:
Let the critical angle be C .
sinC = 1 / μ where μ is index of refraction .
sinC = 1 /1.2
= .833
C = 56°
Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )
sin i / sinr = 1.2 , i is angle of incidence
sini = 1.2 x sinr = 1.2 x sin 34 = .67
i = 42°.
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