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2 years ago

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Rock can be weathered and eroded in the rock cycle. What might happen with the resulting rock particles? A. New rock is formed f

rom other rock by great heat and pressure. B. Igneous rock is formed. C. Rock is melted. D. Sediments can be lithified to form new rock.

2 answers:

Nostrana [21]2 years ago

4 0

Answer: A
After rocks are eroded the particles turn into a sedimentary rock.

IceJOKER [234]2 years ago

3 0

Really the right answer is D:sediments can be lithified to new rock that the right answer

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A simple watermelon launcher is designed as a spring with a light platform for the watermelon. When an 8.00 kg watermelon is put

olga_2 [115]

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

$F_s = F_g$

$kx = mg$

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

$k = \frac{mg}{x}$

$k = \frac{8*9.8}{0.1}$

$k = 784N/m$

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

$PE = \frac{1}{2} kx^2$

$PE =\frac{1}{2} 784*0.4^2$

$PE = 63.72J$

Therefore the energy stored in the spring is 63.72J

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2 years ago

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu

Anit [1.1K]

A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N</span><span> ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N

Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2

</span>

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2 years ago

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A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he

Viefleur [7K]

Answer:

cold air is more dense than warm water so it sinks to the bottom of the pool

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2 years ago

_____ discovered that light also showed properties commonly found in waves.

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Through the work of Max Planck<span>, Einstein, </span>Louis de Broglie<span>, </span>Arthur Compton<span>, </span>Niels Bohr<span>, current scientific theory holds that all particles also have a wave nature (and vice versa).</span>

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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago