Match each media portrayal with the type of behavior the media portrayal is

For a huge luxury liner to move with constant velocity, its engines need to supply a forward thrust of 6.85 105 N. What is the m

Leviafan [203]

Answer:

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

Explanation:

By First and Second Newton's Law, the resistive force exerted by the water on the cruise ship has the same magnitude of forward thrust, with which it is antiparallel to. The equation of equilibrium for the luxury liner is:

$\Sigma F = F-R = 0$ (Eq. 1)

Where:

$F$ - Forward trust, measured in newtons.

$R$ - Resistive force exerted by the water, measured in newtons.

From (Eq. 1), we get that: ( $F = 6.85\times 10^{5}\,N$ )

$R = F$

$R = 6.85\times 10^{5}\,N$

The magnitude of the resistive force exerted by the water is $6.85\times 10^{5}$ newtons.

4 0

3 years ago

What is non-point source pollution, and why is this kind of pollution hard to monitor and control?

antoniya [11.8K]

The use of pesticides in farming land is example of non-point source pollution. The area of emitting of pollution is very large. That is why, it is difficult to monitor and control this type of pollution.

Americans get their water for everyday use through aquifer. Aquifers can be conserve by avoiding high amount of chemical uses and reducing household waste. Also the sustainable use of water cause conservation of natural resource.

Aquifer are basically underground water present in ground pores. The structure of aquifer just look like sponge. The water moves through the porous surfaces of the ground.

7 0

3 years ago

How is the surface area of an average person is 2m^2 in the chapter pressure

11Alexandr11 [23.1K]

Answer:

We have to show surface area $=2m^2$ ,with few conditions that is by considering Force $=200000\ N$ and Pressure $=100000\ Pa$ to be respectively.

Explanation:

The atmospheric pressure is $=10^{5}\ Pa$ on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is $=2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton$ .

Now to calculate surface area we can find it from $pressure=\frac{force}{area}$ and re-arranging it to.

$area=\frac{force}{pressure}$

So plugging the values,

Surface area $=\frac{20000}{10000}=2\ m^{2}$

Hence from the above calculations we can say that surface area is $2m^2$ .

So the surface area of an average person can be said to have $2m^2$ , using the concept of pressure and force.

3 0

2 years ago

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You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on

just olya [345]

Answer:

a) $v_{f}$ = 0.25 m / s b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

m₁ = 0.40 kg

v₁₀ = 9.0 m / s

m₂ = 14 kg

v₂₀ = 0

Initial

po = m₁ v₁₀

Final

$p_{f}$ = (m₁ + m₂) vf

po = pf

m₁ v₁₀ = (m₁ + m₂) $v_{f}$

$v_{f}$ = v₁₀ m₁ / (m₁ + m₂)

$v_{f}$ = 9.0 (0.40 / (0.40 +14)

$v_{f}$ = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

K₀ = ½ m₁ v₁₀² +0

K₀ = ½ 0.40 9 2

K₀ = 16.2 J

Final

$K_{f}$ = ½ (m₁ + m₂) $v_{f}$ 2

$K_{f}$ = ½ (0.4 +14) 0.25 2

$K_{f}$ = 0.45 J

ΔK = K₀ - $K_{f}$

ΔK = 16.2-0.445

ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

v₁₀’= v₁₀ -u

v₁₀’= 9 -.025

v₁₀‘= 8.75 m / s

v₂₀ ‘= v₂₀ -u

v₂₀‘= - 0.25 m / s

$v_{f}$ ‘= $v_{f}$ - u

$v_{f}$ = 0

Initial

K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

Ko = 15.31 + 0.4375

K o = 15.75 J

Final

$k_{f}$ = ½ (m₁ + m₂) vf’²

$k_{f}$ = 0

All initial kinetic energy is transformed into internal energy in this reference system

3 0

2 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

Lisa [10]

Answer:

1) v₀x = 13.76 m/s

2) v₀y = 19.66 m/s

3) ymax = 21.199 m

4) X = 55.1746 m

5) and 6) y = 18.4 m

Explanation:

1) v₀x = v₀*Cos α = 24 m/s* Cos 55° = 13.76 m/s

2) v₀y = v₀*Sin α = 24 m/s* Sin 55° = 19.66 m/s

3) ymax = y₀ + (v₀y²/(2g)) = 1.5 m + ((19.66 m/s)²/(2*9.81 m/s²)) = 21.199 m

4) We can use this equation

y = y₀ + (tan α)*x – (g / (2* v₀x²))*x²

where y = y₀ = 1.5 m

then

1.5 = 1.5 + tan (55°)*x - (9.81 / (2* (13.76)²))*x²

⇒ 0.02588 x² - 1.42815 x = 0

Solving this equation we get

x₁ = 0 and x₂ = 55.1746 m

The distance between the two girls is 55.1746 m

5) and 6) If v₀x = 15 m/s = vx and ymax = 24 m

y = ? when x = (xmax/2)

ymax = y₀ + (v₀y²/(2g)) ⇒ v₀y = √(2g*(ymax - y₀))

⇒ v₀y = √(2(9.81 m/s²)(24 m - 1.5 m)) = 21.01 m/s

then we get α' as follows

α' = tan⁻¹(v₀y/v₀x) = tan⁻¹(21.01 m/s/15 m/s) = 54.47°

v₀ = √(v₀x² + v₀y²) = √((15 m/s)² + (21.01 m/s)²) = 25.81 m/s

Now we can apply the equation of the path

y = ymax - ((gx²)/(2v₀²))

⇒ y = 24m - ((9.81)(55.1746/2)²/(2*25.81²))

⇒ y = 18.4 m

7 0

3 years ago

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