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4 years ago

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How much energy is required to remove a proton from 157N? The masses of the atoms 157N, 146C and 11H are 15.000109 u, 14.003242

u and 1.007825 u, respectively. How much energy is required to remove a neutron? The mass of a neutron is 1.008665 u.

1 answer:

vlada-n [284]4 years ago

8 0

Answer:

10.207377 MeV

10.833345 MeV

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

The reaction to remove a proton is as follows

$^{15}_7N=^{14}_6C+^{1}_1H$

From the mass energy equivalence

$E=(14.003242+1.007825-15.000109)u\times (c)^2\\\Rightarrow E=0.010958uc^2\\\Rightarrow E=0.010958\times 931.5\\\Rightarrow E=10.207377\ MeV$

The energy is required to remove a proton is 10.207377 MeV

The reaction to remove a proton is as follows

$^{15}_7N=^{14}_7N+n$

From the mass energy equivalence

$E=(14.003074+1.008665-15.000109)u\times (c)^2\\\Rightarrow E=0.01136uc^2\\\Rightarrow E=0.01163\times 931.5\\\Rightarrow E=10.833345\ MeV$

Mass of $^{14}_7N=14.003074u$

The energy is required to remove a neutron is 10.833345 MeV

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Answer:

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A leaky 10-kg bucket is lifted from the ground to a height of 11 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

nalin [4]

Answer:

the work done to lift the bucket = 3491 Joules

Explanation:

Given:

Mass of bucket = 10kg

distance the bucket is lifted = height = 11m

Weight of rope= 0.9kg/m

g= 9.8m/s²

initial mass of water = 33kg

x = height in meters above the ground

Let W = work

Using riemann sum:

the work done to lift the bucket =∑(W done by bucket, W done by rope and W done by water)

= $\int\limits^a_b {(Mass of Bucket + Mass of Rope + Mass of water)*g*d} \, dx$

Work done in lifting the bucket (W) = force × distance

Force (F) = mass × acceleration due to gravity

Force = 9.8 * 10 = 98N

W done by bucket = 98×11 = 1078 Joules

Work done to lift the rope:

At Height of x meters (0≤x≤11)

Mass of rope = weight of rope × change in distance

= 0.8kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*0.8(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 7.84 (11x-x²/2)upper limit 11 to lower limit 0

W done = 7.84 [(11×11-(11²/2)) - (11×0-(0²/2))]

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Work done in lifting the water

At Height of x meters (0≤x≤11)

Rate of water leakage = 36kg ÷ 11m = $\frac{36}{11}$ kg/m

Mass of rope = weight of rope × change in distance

= $\frac{36}{11}$ kg/m × (11-x)m = 3.27kg/m × (11-x)m

W done = integral of (mass×g ×distance) with upper 11 and lower limit 0

W done = $\int\limits^1 _0 {9.8*3.27(11-x)} \, dx$

Note : upper limit is 11 not 1, problem with math editor

W done = 32.046 (11x-x²/2)upper limit 11 to lower limit 0

W done = 32.046 [(11×11-(11²/2)) - (11×0-(0²/2))]

= 32.046(60.5 -0) = 1938.783 Joules

the work done to lift the bucket =W done by bucket+ W done by rope +W done by water)

the work done to lift the bucket = 1078 +474.32+1938.783 = 3491.103

the work done to lift the bucket = 3491 Joules

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4 years ago

The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the

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Answer:

The activation energy is $Q = 328.31 \ K J/mol$

Explanation:

From the question we are told that

The rate constant is k

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The value of k is $k_1 = 1.05 *10^{-8} \ kg /m^4 \cdot s$

at temperature $T_2 = 400 ^oC = 400 + 273 = 673 \ K$

The value of k is $k_2 = 2.95 *10^{-4} \ kg /m^4 \cdot s$

The rate constant is mathematically represented as

$k = Ce^{- \frac{Q}{RT} }$

Where Q is the activation energy

R is the ideal gas constant with a value of $R = 8.314 \ J /mol \cdot K$

C is a constant

T is the temperature

For the first rate constant

$k_1 = Ce ^{-\frac{Q}{RT_1} }$

For the second rate constant

$k_2 = Ce ^{-\frac{Q}{RT_2} }$

Now the ratio between the two given rate constant is

$\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}$

=> $ln [\frac{k_1}{k_2} ] = \frac{Q}{R} * [\frac{1}{\frac{T_2 -1}{T_1} } ]$

substituting values

$ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] = \frac{Q}{8.314} * [\frac{1}{\frac{673 -1}{573} } ]$

=> $Q = 328.31 \ K J/mol$

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3 years ago

Mekhanik [1.2K]

Answer:

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2. 4.8m

3. 0.075 litre

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5. 5600g

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7. 63mm

Explanation:

1. 1000m in 1 km so divide by 1000

2. 100cm in m so divide by 100

3. 1000ml in 1l so divide by 1000

4. multiply the mass value by 1000

5.multiply the mass value by 1000

6. divide the length value by 100

7. multiply the length value by 10

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3 years ago

An electron is located on a pinpoint having a diameter of 3.52 µm. What is the minimum uncertainty in the speed of the electron?

Komok [63]

Explanation:

It is given that,

An electron is located on a pinpoint having a diameter of 3.52 µm, $\Delta x=3.52\times 10^{-6}\ m$

We need to find the minimum uncertainty in the speed of the electron. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x \geq \dfrac{h}{4\pi}$

$\Delta p \geq \dfrac{h}{4\pi \Delta x}$

Since, p = m v

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$\Delta v \geq \dfrac{h}{4\pi \Delta x m}$

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3 years ago