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vova2212 [387]
3 years ago
5

A. Water has many unique properties. Identify a property of water and explain why this property makes water vital to living orga

nism.
b. Write one or two sentences that summarize Pascal's principle.

c. A force of 14,00 N is exerted on a piston that has an area of 0.020 m^2. What is the force exerted by a second piston that has an area of 0.040 m^2.

d. An object's weight is greater than its buoyant force in water. What would you expect to happen when the object is placed in water? Explain your answer.

e. A cannon is recovered from a shipwreck. Why does the buoyant force on the cannon stay the same as long as it is fully under water? Explain your reasoning.

f. An object sinks in both water and soil, so it displaces the same volume of each liquid. Oi is less dense than water. Why is the buoyant force acting on the object greater in water than it is in oil?

g. Use the flow rate equation to help explain how the motion of a whitewater raft changes as the river channel becomes narrower.
Physics
1 answer:
rosijanka [135]3 years ago
8 0

a) It can dissolve other substances and has high specific heat capacity.

b) Pressure is transferred equally in all directions

c) F=14*0.04/0.02=28 N

d) I'd expect that it will sink cause weight will pull it to the bottom

e) Because the buoyant force is stipulated by difference between pressures in the top and bottom surfaces of a body.

f) Because this force increases in denser liquids: F_b=\rho gV, where rho is density.

g) Q=Av, where v - speed, A - area. With the same flow rate v=Q/A, so larger area decreases speed and vice versa.

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Use the worked example above to help you solve this problem. An Alaskan rescue plane drops a package of emergency rations to str
Vlad [161]

Answer:

(a) The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact is 77.19°

Explanation:

The parameters given are;

Velocity of the plane, vₓ = 39.0 m/s

Height of the plane above the ground, h = 1.50 × 10² m = 1,500 m

(a) The time, t, before the package hits the ground when dropped from the plane is given by the relation;

h = u·t + 1/2×g×t²

Where:

g = Acceleration due to gravity = 9.81 m/s²

u = Initial vertical velocity = 0 m/s

Hence;

1500 = 0×t + 1/2 × 9.81 × t² = 4.905·t²

∴ t = √(1500/4.905) = 17.49 s

The horizontal distance the package travels before landing = 17.49 × 39 ≈ 682 m

The package lands 682 meters horizontally ahead from the point the package was dropped from the plane

(b) The vertical velocity, v_y, of the package just before landing is given by the relation;

v_y² = u² + 2·g·h

u = 0 m/s

∴ v_y² = 0 + 2×9.81×1500 = 29430 m²/s²

v_y = √29430  = 171.55 m/s

Hence the horizontal component = 39.0 m/s

The vertical component = 171.55 m/s

(c) The angle of impact, θ, is given as follows;

tan \theta = \dfrac{v_y}{v_x}  = \dfrac{171.55}{39.0 } = 4.4

∴ θ = tan⁻¹(4.4) = 77.19°.

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3 years ago
An electric generator has an 18-cmcm-diameter, 120-turn coil that rotates at 60 HzHz in a uniform magnetic field that is perpend
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3 years ago
9- Under what circumstances would a vector have components that are equal in
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Explanation:

c. if the vector is oriented at 0° from the X -axis.

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2 years ago
There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

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3 years ago
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An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
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