What is the accletation of a 1500kg with a net force of 7500 N
1 answer:
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When light travels from a medium with higher refractive index to a medium with lower refractive index, there is a critical angle after which all the light is reflected (so, there is no refraction).
The value of this critical angle can be derived by Snell's law, and it is equal to
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In our problem, n1=1.47 and n2=1.33, so the critical angle is
The value of this critical angle can be derived by Snell's law, and it is equal to
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In our problem, n1=1.47 and n2=1.33, so the critical angle is
Answer:
v₁ = 37.5 cm / s
Explanation:
For this exercise we can use that angular and linear velocity are related
v = w r
in the case of the spool the angular velocity for the whole system is constant,
They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,
w = v₀ /r₀
for the outside of the spool r₁ = 1.5 cm
w = v₁ / r₁1
since the angular velocity is the same we set the two expressions equal
v1 =
let's calculate
v₁ =
v₁ = 37.5 cm / s