An inclined plane of length 12m is used to pull a load of 3000N to height of 3m by applying an effort of 1000N. Calculate its MA
1 answer:
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Answer
given,
mass of the ball = 3 kg
swing in vertical circle with radius = 2 m
work done by the gravity = ?
work done by the tension = ?
Work done by the gravity = - m g Δh
Δ h = 2 + 2 = 4 m
Work done by the gravity =
= -117.6 J
work done by gravity is equal to -117.6 J
Work done by tension will be equal to zero.
Zero because tension is always perpendicular to velocity
work done by tension is equal to 0 J
Answer:
Explanation:
Given:
Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s
Area of cross-section = 0.50 cm²
Length of channel =0.25 cm
Now for the new channel
Area of cross-section = 0.30 cm²
Length of channel =0.10 cm
let the Solute Diffusion rate of new channel = s
now equating the diffusion rate per unit volume for both the channels
thus,
Answer:
The density of the woman is 950.8 kg/m³
Explanation:
Given;
fraction of the woman's volume above the surface = 4.92%
then, fraction of the woman's volume below the surface = 100 - 4.92% = 95.08%
the specific gravity of the woman
The density of the woman is calculate as;
Density of fresh water = 1000 kg/m³
Density of the woman = 0.9508 x 1000 kg/m³
Density of the woman = 950.8 kg/m³
Therefore, the density of the woman is 950.8 kg/m³