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4 years ago

5

Using a simple machine, a student is able to lift a 500N weight by applying only 100N. What is the mechanical advantage of the s

imple machine? A) 0.2 B) 1 C) 5 D) 50000

1 answer:

Vanyuwa [196]4 years ago

4 0

Answer:

The answer is 5

Explanation:

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In a nuclear fusion reaction two 2H atoms are combined to produce one 4He.

Mrrafil [7]

Answer:

a)= 0.025602u

b) = 23.848MeV

c) N = 1.546 × 10¹³

Explanation:

The reaction is

²₁H + ²₁H ⇄ ⁴₂H + Q

a) The mass difference is

Δm = 2m(²₁H) - m (⁴₂H)

= 2(2.014102u) - 4.002602u

= 0.025602u

b) Use the Einstein mass energy relation ship

The enegy release is the mass difference times 931.5MeV/U

E = (0.025602) (931.5)

= 23.848MeV

c)

the number of reaction need per seconds is

N = Q/E

= 59W/ 23.848MeV

$= \frac{59}{(23.848 \times 10^6 )(1.6 \times 10^1^9) } \\\\= 1.546 \times 10^1^3$

N = 1.546 × 10¹³

5 0

3 years ago

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a

sergij07 [2.7K]

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

= $4.59\times 10^{-19} \ J$

or,

= $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

= $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒ $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒ $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒ $=1.4375 \ eV$

5 0

3 years ago

Which statement describes the effects of forces on an object

lukranit [14]

Answer:

Sorry this isn’t going to be any help. You don’t have any statement that I’m able to see.

Explanation:

8 0

3 years ago

If we want to describe work , we must have what ?

ivanzaharov [21]

An external force that is being applied in the direction of the displacement

4 0

3 years ago

A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu

yarga [219]

Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

Temperature = 296 K

If the initial volume is V then the final volume will be V/3.

We need to calculate the work done

Using formula of work done

$W=nRT\ ln(\dfrac{V_{f}}{V_{i}})$

Put the value into the formula

$W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})$

$W=1.90\times8.314\times296\ ln(\dfrac{1}{3})$

$W=−5136.88\ J$

The Work done on the system.

Hence, The work done is 5136.88 J.

5 0

3 years ago