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2 years ago

Calculate the amount of energy produced in a nuclear reaction in which the mass defect is 0.187456 amu.

2 answers:

8 0

For nuclear reactions, we determine the energy dissipated from the process from the Theory of relativity wherein energy is equal to the mass defect times the speed of light. We calculate as follows:

E = mc^2 = 0.187456 (3x10^8)^2 = 1.687x10^16 J

Hope this answers the question.

pishuonlain [190]2 years ago

4 0

Answer:

$E = 174.6 MeV$

Explanation:

As we know that energy corresponding to the 1 amu mass is given as

$E = mc^2$

if m = 1 amu

we have

$E = 931.5 MeV$

so we will have

mass = 0.187456 amu

$E = (0.187456)(931.5)$

$E = 174.6 MeV$

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charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel

IRISSAK [1]

Answer:

Explanation:

Electric field due to charge at origin

= k Q / r²

k is a constant , Q is charge and r is distance

= 9 x 10⁹ x 5 x 10⁻⁶ / .5²

= 180 x 10³ N /C

In vector form

E₁ = 180 x 10³ j

Electric field due to q₂ charge

= 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

= 30.33 x 10³ N / C

It will have negative slope θ with x axis

Tan θ = .5 / √.5² + .8²

= .5 / .94

θ = 28°

E₂ = 30.33 x 10³ cos 28 i - 30.33 x 10³ sin28j

= 26.78 x 10³ i - 14.24 x 10³ j

Total electric field

E = E₁ + E₂

= 180 x 10³ j +26.78 x 10³ i - 14.24 x 10³ j

= 26.78 x 10³ i + 165.76 X 10³ j

magnitude

= √(26.78² + 165.76² ) x 10³ N /C

= 167.8 x 10³ N / C .

3 0

2 years ago

Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a

Alex17521 [72]

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

$V =12.0 V$ , $E = 1.55*10^2 J$ , $f = 3600 Hz$

To determine the capacitance can use the equation

$U_c= \frac{1}{2}*C*V^2$

Solve to C'

$C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}$

$C=215.27 uF$

To find the inductance can use the frequency of the circuit

$f = \frac{1}{2\pi* \sqrt{C*L} }$

Solve to L'

$L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}$

$L = 9.078 uH$

6 0

1 year ago

When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a

Oxana [17]

Answer:

5 mg, $5\cdot 10^{-3}g$

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

$m=5\cdot 10^{-3}g$

If we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

$1 g : 1000 mg = 0.005 g : x$

and we find

$x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg$

4 0

2 years ago

Wood block 1 in (Figure 1), which has a mass of 1.0 kg, is at rest on a wood ramp. The angle of the ramp is 20º above horizontal

Sergeeva-Olga [200]

Answer:

Explanation:

For this problem we use the translational equilibrium condition. Our reference frame for block 1 is one axis parallel to the plane and the other perpendicular to the plane.

X axis

-Aₓ - f_e +T = 0 (1)

Y axis

N₁ - W_y = 0 ( 2)

let's use trigonometry for the weight components

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

We write the diagram for the second body.

Note that in the block the positive direction rd upwards, therefore for block 2 the positive direction must be downwards

W₂ -T = 0 (3)

we add the equations is 1 and 3

- W₁ sin θ - μ N₁ + W₂ = 0

from equation 2

N₁ = W₁ cos θ

we substitute

-W₁ sin θ - μ (W₁ cos θ) + W₂ = 0

W₂ = m₁ g (without ea - very expensive)

This is the smallest value that supports the equilibrium system

7 0

2 years ago

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m

3 0

2 years ago