All categories

3 years ago

How does the mass-energy equation e = mc2 relate to fission

2 answers:

3 0

It relates because the energy of an object to its mass and since fission requires energy to do so, e=mc^2 relates to it.

solniwko [45]3 years ago

3 0

in a fission reaction a large nucleus gets broken in to smaller nuclei

You might be interested in

A solenoid with 435 turns has a length of 7.50 cm and a cross-sectional area of 3.50 ✕ 10−9 m2. Find the solenoid's inductance a

OLga [1]

Answer:

Solenoid's inductance is 1.11 × 10^-8H

The average emf around the solenoid is 1.3 × 10^-5V

Explanation: Please see the attachments below

4 0

3 years ago

Reactants → products

Nat2105 [25]

The answer is A

Explanation: the conservation of matter means that the mass stays the same

4 0

2 years ago

Read 2 more answers

Find the voltage across the 15 Ω resistor. . 35 Ω 20 Ω 15 Ω 10 V

almond37 [142]

Answer:

35 Ω

Explanation:

3 0

2 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0

2 years ago

A block with m = 300 grams oscillates at the end of a linear spring with k = 6.5 N/m. (assume this is a top-down view and the bl

inna [77]

Answer:

a) T=1.35s

b) amplitude = 0.0923m

Explanation:

m=300 gr

k=6.5 N/m

first we need to get the angular frequency of the motion

so we have that

ω = √(k/m)

in this case motion is a simple harmonic so the period is defined by:

T= 2π / ω

T= 2π / √(k/m)

replacing the variables...

T= 2π / √(6.5/0.3)

T=1.35s (period of the block's motion)

and...

α max = | ω²r max |

2 = (2π/1.35)² * r max

r max= 0.0923m

6 0

3 years ago