Ask question

Ask question

All categories

olga nikolaevna [1]

3 years ago

11

What is the wavelength of a light of frequency 7.21?

1 answer:

erik [133]3 years ago

4 0

MEMORIZED E=h*v h=6.626x10-34J*s INFORMED v=7.21x1014S-1CALCULATE E=h*v E=(6.626x10-34J*s)*(7.21x1014s-1) The "s" cancels out. s-1=1/s so you get s/s so you are left with Solution 4.78 10-19 J OR .478 aJ <span>Apex - 467 nm ^.^ hopefully thats the correct thing</span>

You might be interested in

Two clouds collide and form another, more massive cloud. One cloud is stationary, while the other is traveling at 1 m/s. After t

sweet [91]

Answer: 3

Explanation:

Given

One cloud is traveling at rate of $u_2=1\ m/s$

combined velocity of the two is $v=0.25\ m/s$

Suppose the masses of the clouds be $m_1,m_2$

Conserving momentum

$\Rightarrow m_1u_1+m_2u_2=\left(m_1+m_2\right)v\\\text{Divide whole equation by }m_2\\\Rightarrow \dfrac{m_1}{m_2}u_1+u_2=\left(\dfrac{m_1}{m_2}+1\right)v\\\\\Rightarrow 0+1=0.25\dfrac{m_1}{m_2}+0.25\\\\\Rightarrow \dfrac{m_1}{m_2}=\dfrac{0.75}{0.25}\\\\\Rightarrow \dfrac{m_1}{m_2}=3$

6 0

2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

2 years ago

A charge of 5.0 coulombs moves through a circuit in 0.50 second. What is the current in the circuit

user100 [1]

Answer:

10

Explanation:

i = 5/.5 = 10 Amps. Hope this helps :)

6 0

2 years ago

What mass has a rest energy of 100J?

alexandr1967 [171]

Answer:

option a is correct

Explanation:

<h2>I hope it's help you ❣️❣️</h2>

6 0

2 years ago

In an arcade game, a 0.126 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and relea

bearhunter [10]

Answer:

4.156 m/s

Explanation:

See attachment

4 0

2 years ago