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2 years ago

What is a likely consequence of continued human population growth?

Physics

1 answer:

yan [13]2 years ago

7 0

Answer:

The answer is C. Abundant natural resources

Explanation:

A is wrong because humans cause pollution, B is wrong because we use natural resources and their supply would deteriorate, and D is wrong because we already break down rainforests and they are getting smaller.

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Big Ben, a large artifact in England, has a mass of 1x10^8 kilograms and the Empire State Building 1x10^9 kilograms. The distanc

TiliK225 [7]

Answer:

The force, exerted by Big Ben on the Empire State Building is 2.66972 × 10⁻⁷ N

Explanation:

The question relates to the force of gravity experienced between two bodies

The given parameters are;

The mass of Big Ben, M₁ = 1 × 10⁸ kg

The mass of the Empire State Building, M₂ = 1 × 10⁹ kg

The distance between the two Big Ben and the Empire State Building, r = 5,000,000 meters

By Newton's Law of gravitation, we have;

$F=G \times \dfrac{M_{1} \times M_{2}}{r^{2}}$

Where;

F = The force exerted by Big Ben on the Empire State Building and vice versa

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M₁, M₂, and r are the given parameters

By plugging in the values of the parameters and the constant into the equation for Newton's Law of gravitation, we have;

$F=6.67430 \times 10^{-11} \times \dfrac{1 \times 10^8 \times 1 \times 10^9}{(5,000,000)^{2}} = 2.66972 \times 10^{-7}$

The force, 'F', exerted by Big Ben on the Empire State Building is F = 2.66972 × 10⁻⁷ N.

3 0

2 years ago

Why did the following change occur? the snow that covered the ground all winter began to melt, temperatures rose, and flowers st

FromTheMoon [43]

Answer:C

Explanation:

8 0

2 years ago

Read 2 more answers

An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

7 0

2 years ago

Newton's Cradle.A Newton's cradle is composed of row metal balls, each hung from a string attached to a horizontal rod, so that

denis-greek [22]

Answer:

In Newton's Cradle there are N number of balls are suspended from the thread such that they all touch each other.

Now when we pull one ball apart and released then after the collision one ball on the other side moves up

Then in next time if we pull two balls from one side then after collision it will pull upwards exactly two balls on the other side after collision.

So here the number of balls on the other side is exactly same every time we pull the balls.

This is due to the conservation of momentum principle

As we know that all the balls are identical here then when we pull any number of balls on one side and then release them then after collision the momentum of the balls is transferred to same amount of the balls on the other side

Since there is no external force on this system so we can say that that this momentum conservation will exist for all cases and all number.

4 0

3 years ago

A piston–cylinder device initially contains 0.6 m3 of saturated water vapor at 250 kPa. At this state, the piston is resting on

AysviL [449]

Answer:

T2 = 662°c Q(in) = 911kj W(out) = 180kj

Explanation:

1. Firstly, we find the mass from the initial volume and specific volume from steam table A-5 for given pressure and state.

M = V(1)/a(1) = 0.6/0.7183(kg)

Where a(1) = specific volume

M = 0.835kg

The work is determined from the final pressure and the volume change that occurs when it is reached.

W = P2(V2 - V1)

W = P2V2

= 300×0.6kg = 180kg

Therefore, to get the final temperature and final pressure, the final temperature is determined fro steam table A-6 using interpolation.

T2 = T1* + T2*-T1*/a2*-a1* (a2-a1*)

T2 = 600 + 700-600/1.4958-1.34139 (1.43746-1.34139)

T2 = 662°c

2. The final internal energy can be determined in the same way just like the initial ones were done. From A-5 on steam table given the pressure

The heat Q(in) = ∆u + W(out)

= m(U2-U1) + W(out)

= 0.835(3412.3-2536.8)kj

+ 180

Q(in) = 911kj

6 0

3 years ago