Which particle is commonly used to initiate a fission chain reaction

If the wavelength of a wave is quadrupied then what happens to the frequency of the wqve(Assume speed is constant)

dimulka [17.4K]

Answer:

What happens to the wavelength of a wave if you double the frequency?

If the frequency of a wave is increased, what happens to its wavelength? As the frequency increases, the wavelength decreases. 2. If the frequency is doubled, the wavelength is only half as long.

Explanation:

4 0

2 years ago

The moon’s period of revolution is 27 1/3 days, and its period of rotation about its axis is _____.

mestny [16]

It lasts 29 1/2 days.
-the sidereal month- is the true period of the moon's revolution around Earth. It lasts 27 1/3 days.
-the difference of 2 days between the synodic and sidereal cycles is due to the Earth- moon system also moving in an orbit around the sun.

5 0

2 years ago

Read 2 more answers

Three small masses are positioned as follows: 2.0 kg at (0.0 m, 0.0 m), 2.0 kg at (2.0 m, 0.0 m), and 4.0 kg at (2.0 m, 1.0 m).

melamori03 [73]

Refer to the diagram shown below.

The given data is

mass, kg Coordinates. m
------------- -----------------
2 (0, 0)
2 (2, 0)
4 (2, 1)

Total mass, M = 2+2+4 = 8kg
Let (x,y) be the coordinates of M.

Then, taking moments about the origin, we obtain
8x = 2*0 + 2*2 + 4*2 = 12
x = 1.5 m

8y = 2*0 + 2*0 + 4*1 = 4
y = 0.5 m

Answer: (1.5, 0.5) m

6 0

2 years ago

Two small nonconducting spheres have a total charge of 93.0 μC . Part A

Travka [436]

Answer:

charge on each

Q1 = 2.06 × $10^{-5}$ C

Q2 = 7.23 × $10^{-5}$ C

when force were attractive

Q1 = 1.07 × $10^{-4}$ C

Q2 = -1.39 × $10^{-5}$ C

Explanation:

given data

total charge = 93.0 μC

apart distance r = 1.14 m

force exerted F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC = 93 × $10^{-6}$ C

and

According to Coulomb's law force between two sphere is

Force F = $\frac{K Q1 Q2 }{r^2}$ .........1

Q1Q2 = $\frac{F*r^2}{k}$

here F is force and r is apart distance and k is 9 × $10^{9}$ N-m²/C² put all value we get

Q1Q2 = $\frac{ 10.3*1.14^2}{9*10^9}$

Q1Q2 = 1.49 × $10^{-9}$ C²

and

we have Q2 = 93 × $10^{-6}$ C - Q1

put here value

Q1² - 93 × $10^{-6}$ Q1 + 1.49 × $10^{-9}$ = 0

solve we get

Q1 = 2.06 × $10^{-5}$ C

and

Q1Q2 = 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = 1.49 × $10^{-9}$

Q2 = 7.23 × $10^{-5}$ C

and

if force is attractive we get here

Q1Q2 = - 1.49 × $10^{-9}$ C²

then

Q1² - 93 × $10^{-6}$ Q1 - 1.49 × $10^{-9}$ = 0

we get here

Q1 = 1.07 × $10^{-4}$ C

and

Q1Q2 = - 1.49 × $10^{-9}$

2.06 × $10^{-5}$ Q2 = - 1.49 × $10^{-9}$

Q2 = -1.39 × $10^{-5}$ C

5 0

2 years ago

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$