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3 years ago

Which statement correctly lists the four forces from the weakest to the strongest?

2 answers:

4 0

The strongest of the natural forces is Strong Nuclear, followed by Electromagnetic, then Weak Nuclear and lastly Gravitational.
So from weakest to strongest: gravitational --> weak nuclear --> electromagnetic --> strong nuclear
Answer C)

mash [69]3 years ago

3 0

The correct answer of this question is : C i.e Gravitational < Weak nuclear force < Electromagnetic force < Strong nuclear force.

EXPLANATION:

There are four fundamental forces in nature which are known as gravitational force, weak nuclear force, electromagnetic force and strong nuclear force.

Out of these four fundamental forces, the gravitational force is the weakest forces in nature,but has infinite range.

Weak nuclear force is produced due to W/Z bosons. Hence, it is stronger than gravitational force but weaker as compared to electromagnetic and strong nuclear force. It is a short range force.

Electromagnetic force is the third one which is weaker as compared to strong nuclear force but has infinite range.

Strong nuclear force is a short range, spin dependent strong attractive force that arises between nucleons due to exchange of mesons.

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An infinite line of charge produces a field of magnitude 4.90 ✕ 104 n/c at a distance of 1.9 m. calculate the linear charge dens

bekas [8.4K]

From Gauss law we have
E = 2kx/r where x=linear charge density, r=distance from the wire=1.7 m
k =9 x 10^9 SI units.....we have been given E = 3.8 x 10^4 N/C ..... so plugging numbers we get

x = 3.59 x 10^ -6 C/ M (coulomb per meter) =linear charge density

6 0

3 years ago

Sunlight has its maximum intensity at a wavelength of 4.83 x 10'm; wha energy does this correspond to in e?

Lera25 [3.4K]

Answer:

E = 2,575 eV

Explanation:

For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength

E = h f

c = λ f

Where the Planck constant has a value of 6.63 10⁻³⁴ J s

Let's replace

E = h c / λ

Let's calculate for wavelengths

λ = 4.83 10-7 m (blue)

E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷

E = 4.12 10-19 J

The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J

E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

E = 2,575 eV

5 0

3 years ago

Did I do this right??

Nikitich [7]

Answer:

yes

Explanation:

good job

7 0

2 years ago

A rigid container holds 0.30g of hydrogen gas.

Strike441 [17]

Answer:

Part A: $\mathbf{Q =94 \ J}$ to two significant figures

Part B: $\mathbf{Q =160 \ J}$ to two significant figures

Part C: $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures

6 0

3 years ago

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = $\frac{\textup{V}}{\textup{R}}$

I₀ = $\frac{\textup{6.53}}{\textup{594}}$

I₀ = 0.0109 A

also,

I = $I_0[1-e^{-\frac{t}{\tau}}]$

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = $0.0109[1-e^{-\frac{\tau}{\tau}}]$

I = $0.0109[1-2.717^{-1}]$

I = 0.00688 A

I = 6.88 mA

5 0

3 years ago