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3 years ago

Which best describes the current atomic theory?

Physics

2 answers:

ivolga24 [154]3 years ago

6 0

Choice-C is a correct statement.

taurus [48]3 years ago

5 0

C. it is the most powerful explanation scientists have to offer at this time. I am 100% right, I just finish the test and got it right.

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A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the wate

Vilka [71]

Answer:

f = 5.3 Hz

Explanation:

To solve this problem, let's find the equation that describes the process, using Newton's second law

∑ F = ma

where the acceleration is

a = $\frac{d^2 y}{dt^2 }$

B- W = m \frac{d^2 y}{dt^2 }

To solve this problem we create a change in the reference system, we place the zero at the equilibrium point

B = W

In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains

B’= m $\frac{d^2 y'}{dt^2 }$

the thrust is given by the Archimedes relation

B = ρ_liquid g V_liquid

the volume is

V = π r² y'

we substitute

- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }

$\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0$

this differential equation has a solution of type

y = A cos (wt + Ф)

where

w² = ρ_liquid g π r² /m

angular velocity and frequency are related

w = 2π f

we substitute

4π² f² = ρ_liquid g π r² / m

f = $\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }$

calculate

f = $\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }$

f = 5.3 Hz

6 0

3 years ago

Student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.300 miles west, and finally dS = 0.200 miles south

melisa1 [442]

Taking the vertical component of the displacement
1.1 - 0.2 = 0.9 mile
The horizontal component of the displacement
-0.3 mile

The magnitude of the displacement is
√[ (0.9)² + (-0.3) ] = 0.95 mile

The direction is
θ = tan-1 (-0.3/0.9)
θ = 161.57 degrees.

8 0

3 years ago

Water vapor enters a turbine operating at steady state at 500°C, 40 bar, with a velocity of 200 m/s, and expands adiabatically t

faltersainse [42]

Answer:

W = 5701 KW

Explanation:

From the question let inlet be labelled as point 1 and exit as point 2, for the fluid steam, we can get the following;

Inlet (1): P1 = 40 bar ; T1 = 500°C and V1 = 200 m/s

Exit(2) : At saturated vapour; P2 = 0.8 bar and V2 = 150 m/s

Volumetric flow rate = 15 m^(3)/s

Now, to solve this question, we assume constant average values, steeady flow and adiabatic flow.

Specific volume for steam at P2 = 0.8 bar in the saturated vapour state can be gotten from saturated steam tables(find a sample of the table attached to this answer).

So from the table,

v2 = 2.087 m^(3)/kg

Now, mass flow rate (m) = (AV) /v

Where AV is the volumetric flow rate.

Thus, the mass flow rate at exit could be calculated as;

m = 15/(2.087) = 7.17 kg/s

We also know energy equation could be defined as;

Q-W = m[(h1 - h2) + {(V2(^2) - (V1(^2)} /2)} + g(Z2 - Z1)]

Since the flow is adiabatic, potential energy can be taken to be zero. Therefore, we get;

-W = m[(h2 - h1) + {(V2(^2) - (V1(^2)} /2)}

From, table 2, i attached , at P1 = 40 bar and T1 = 500°C; specific enthalpy was calculated to be h1 = 3445.3 KJ/Kg

Likewise, at P2 = 0.8 bar; from the table, we get specific enthalpy as;

h2 = 2665.8 KJ/Kg

So we now calculate power developed;

W = - 7.17 [(2665.8 - 3445.3) + {(150^(2) - 200^(2))/2000 = 5701KW

Since the sign is not negative but positive, it means that the power is developed from the system.

4 0

3 years ago

Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot

m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force $F_1=430 N$

other Pulls it with a rope of rope $F_2=360 N$

mass of crate $m=290 kg$

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

$f_r=F_1+F_2$

where $f_r$ is the friction force

$f_r=430+360$

$f_r=790 N$

$f_r=\mu N$

where $\mu$ is the coefficient of static friction

$N=mg$

$790=\mu 29\cdot 9.8$

$\mu =0.27$

6 0

3 years ago

Compare and contrast the general characteristics of a gas giant planet and one of the inner planets (such as Mercury, Earth, Ven

yKpoI14uk [10]

The gas planets usually have extremely high gravitational pulls, the surface isn't solid (since its a gas planet), and gas planets are larger than the inner planets.
<span>Similarities- These planets all have moons and they both revolve around the sun (obviously).

Hope this helps.</span>

6 0

3 years ago