You hold your physics textbook in your hand. (Assume that no other objects are in contact with the book.)
1 answer:
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Answer:
B) d = √ ( 4 h₂ ( H - 2h₂))
Explanation:
A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance
2) If the height is very small downwards, it touches the earth a little and the horizon is small,
B) to find an equation for horizontal distance (d)
We must maximize the speed at the bottom of the slide let's use energy
Starting point Higher
Em₀ = U = m g h₁
Final point. Lower (slide bottom)
Emf = K + U = ½ m v² + m gh₂
As there is no friction the energy is conserved
mgh₁ = ½ m v² + mgh₂
v² = 2 g (h₁-h₂)
This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)
The distance traveled when leaving the table can be searched with kinematics, projectile launch
x = v₀ₓ t
y = t - ½ g t²
The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero
t = √ 2h₂ / g
We substitute in the other equation
d = √ (2g (h₁-h₂)) √ 2h₂ / g
d = √ (4 h₂ (h₁-h₂))
H = h₁ + h₂
h₁ = H -h₂
d = √ ( 4 h₂ ( H - 2h₂))
Complete question:
Find the pressure exerted by a waterbed with dimensions of 2 m x 2 m which is 30 cm thick. (hint: use 1000 kg/m³ as density of water)
Answer:
The pressure exerted by the waterbed is 2940 N/m²
Explanation:
Given;
Length of waterbed, L = 2 m
Width of waterbed, W = 2 m
Height of waterbed, H = 30 cm = 0.3 m
density of water, ρ = 1000 kg/m³
Hydrostatic pressure derivation:
The hydrostatic pressure exerted by the waterbed is directly proportional to the height of the waterbed. Thus, the hydrostatic pressure increases with increase in height of the waterbed.
Hydrostatic Pressure exerted by the waterbed:
P = ρgH
P = 1000 x 9.8 x 0.3
P = 2940 N/m²
Therefore, the pressure exerted by the waterbed is 2940 N/m²