Answer:
weigh is 2353.13 N
Explanation:
Given data
density = 7890.00 kg/m3
lighter = 299 N
to find out
the volume of the anchor and weigh in air
solution
from question we can say that
apparent weight = actual weight - buoyant force
we know weight = mg and buoyant force = water density × g
so volume of anchor is = actual weight - apparent weight / buoyant force
volume of anchor is = 299 / 1000 × 9.81
volume of anchor is = 0.0304791 m³
and
weight of anchor is mg
here mass m = density Fe g
density Fe = 7870 from table 14-1
so weight = 7870 × 0.0304791 × 9.81
weigh is 2353.13 N
Answer:
vf = 11.2 m/s
Explanation:
m = 10 Kg
F = 2*10² N
x = 4.00 m
μ = 0.44
vi = 0 m/s
vf = ?
We can apply Newton's 2nd Law
∑ Fx = m*a (→)
F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m
⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²
then , we use the equation
vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)
⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s
Answer:
The remaining lights would shine with the same brightness.
Explanation:
Answer:
Part of the question is missing but here is the equation for the function;
Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.
Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3
Explanation:
What is applied is the principle of dimensional homogenuity
From the equation V = (1/3)zxt2.
- V has a dimension of [L/T]
- t has a dimension of [T]
- from the equation, make z the subject of the relation
- z = v/xt2 where 1/3 is treated as a constant
- Substituting into the equation for z
- z = L/T / L x T2
- the dimension for z = 1/T3 i.e 1/ T - raised to power 3
You are who you are because of your enviorment. It depends if your in a healthy environment or a toxic one which changes your act. (Hope this helps)
