Attractive forces between molecules of the same type are called cohesive forces. ... Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example.
Answer:
(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b) The force exerted by the water on the window is 36101.5 N
Explanation:
(a)
The absolute pressure is given by the formula
Where is the absolute pressure
is the atmospheric pressure
is the density
is the acceleration due to gravity (Take )
h is the height
From the question
h = 30.0 m
= 1.00 × 10³ kg/m³ = 1000 kg/m³
= 101.3 kPa = 101300 Pa
Using the formula
P = 101300 + (1000×9.8×30.0)
P = 101300 + 294000
P =395300 Pa
P = 395.3 kPa
Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b)
For the force exerted
From
P = F/A
Where P is the pressure
F is the force
and A is the area
Then, F = P × A
Here, The area will be area of the window of the underwater vehicle.
Diameter of the circular window = 34.1 cm = 0.341 m
From Area = πD²/4
Then, A = π×(0.341)²/4 = 0.0913269 m²
Now,
From F = P × A
F = 395300 × 0.0913269
F = 36101.5 N
Hence, the force exerted by the water on the window is 36101.5 N
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Scalar Quantity :-
→ These are the quantities with magnitude only . These quantities doesn't have to be mentioned with direction
eg.)=> Mass , Temprature .
Vector Quantity :-
→ These quantities are described with both Magnitude and Direction . These quantities follow special type of algebra called Vector algebra .
eg.)=> Force , Displacement
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Hope It Helps You. ☺