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3 years ago

Part B.

Physics

1 answer:

Reptile [31]3 years ago

8 0

Answer:

for Visible light, changing the frequency will change the color

Explanation:

edmentum

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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

3 years ago

A positively charged particle moves through an electric field. As part of a complicated trajectory, the particle passes through

kow [346]

Answer:

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

8 0

3 years ago

How to represent milligram in kilogram by standard formula?

Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take $\frac{1}{1,000,000}$ which equals 0.000001 kg