A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is
supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
Calculate the tension in the cable.
1 answer:
The tension in the cable is mathematically given as
T =266.659 N
<h3>What is the tension in the cable.?</h3>Generally, the equation for the angle of the boom with horizontal is mathematically given as
A = tan-1(5 /10)
The angle of cable with horizontal
B = tan-1(4 / 10)
B= 21.80 degrees
Hence
175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1
T =266.659 N
In conclusion, the tension in the cable.
T =266.659 N
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Answer:
i) FALSE, ii) TRUE, iii) FALSE, iv) FALSE
Explanation:
When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts
n = c / v
where v is the speed of light in the material medium.
The direction of the ray can be determined by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
Let's apply these equations to our case where
nₓ <n_y
i) The expression of the refractive index
nₓ < n_y
v_y <vₓ
therefore the expression is FALSE
ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray
so the expression is TRUE
iii) The speed of light is constant in all material media
the statement is FALSE
iv) light approaches normal
Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.
the statement of FALSE