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OleMash [197]
3 years ago
4

An alpha particle (charge 2e) is sent at high speed toward an atomic nucleus. The electric force acting on the alpha particle is

79.8 N when it is 2.29 x 10(-14) m away from the nucleus. What is the charge on the atomic nucleus? (e = 1.60 x 10(-19) C, Kc = 9 x 10(9) N•m(2)/c(2)
Physics
1 answer:
babunello [35]3 years ago
3 0

Answer:

1.45×10⁻¹⁶ C

Explanation:

Applying,

F = kqq'/r²................ Equation 1

Where F = force acting on the alpha particle, k = coulomb's constant, q = charge on the alpha particle, q' = charge on the atomic nucleus, r = distance of seperation between the alpha particle and the atomic nucleus.

make q' the subject of the equation

q' = Fr²/(kq)................. Equation 2

From the question,

Given: F = 79.8 N, r = 2.29×10⁻¹⁴ m, q = 2e = (2×1.6×10⁻¹⁹) = 3.2×10⁻¹⁹ C, k = 9×10⁹ Nm²/C²

Substitute these values into equation 2

q' = [79.8×(2.29×10⁻¹⁴)²]/(3.2×10⁻¹⁹×9×10⁹)

q' = (4.18×10⁻²⁶)/(2.88×10⁻¹⁰)

q' = 1.45×10⁻¹⁶ C

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A car is on cruise control at v=30m/s. If the wind resistance is equal to. F=0.42v² what energy will the car expend to drive 1km
weeeeeb [17]

Answer:

D)378kJ

Explanation:

Applying,

E = F×d................ Equation 1

Where E = Energy expanded by the car, F = Wind resistance, d = distance

From the question,

F = 0.42v²............ Equation 2

Susbtitute equation 2 into equation 1

E = 0.42v²d.............. Equation 3

Given: v = 30 m/s, d = 1 km = 1000 m

Substitute these values into equation 3

E = 0.42(30²)(1000)

E = 378000

E = 378 kJ

Hence the right option is D)378kJ

6 0
3 years ago
What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

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An object has a mass of 120 kg on the moon. What is the force of gravity acting on the object on the moon?
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3 years ago
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kirill115 [55]

Answer:

t_{1/2}=6 h

Explanation:

Let's use the decay equation.

A=A_{0}e^{-\lambda t}

Where:

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  • A₀ is the initial activity
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We know that \lambda=\frac{ln(2)}{t_{1/2}}

So we have:

\lambda=\frac{ln(A/A_{0})}{t}

\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}

t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}

t_{1/2}=6 h

Therefore, the half-life of the source is 6 hours.

I hope it helps you!

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