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3 years ago

Please help me with these calculating velocity questions im so stressed out

Physics

1 answer:

Darina [25.2K]3 years ago

3 0

Answer: -10m/s

Explanation:

Velocity is solved using the equation displacement over time. Part C shows a displacement or change in x of 30 and a change in time of 3 secs (7-10) so the equation would be 30m/3s= -10m/s it is negative because look at the slope. Its decreasing right? So that means the objects is slowing down which means it can't have a positive velocity

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Which of the following is an example of acceleration? I. A car speeds up. II. A car slows down. III. A car travels in a straight

svetlana [45]

I., II., and IV. are examples of acceleration. III. isn't.

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4 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

NeTakaya

Answer:

Explanation:

General equation of the electromagnetic wave:

$E(x, t)= E_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

where

$\phi =$ Phase angle, 0

c = speed of the electromagnetic wave, 3 × 10⁸

$\lambda =$ wavelength of electromagnetic wave, 698 × 10⁻⁹m

E₀ = 3.5V/m

Electric field equation

$E(x, t)= 3.5sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\E(x, t)= 3.5sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

Magnetic field Equation

$B(x, t)= B_0sin[\frac{2\pi}{\lambda}(x-ct)+\phi ]$

Where B₀= E₀/c

$B_0 = \frac{E_0}{c} = \frac{3.5}{3\times10^8}=1.2 \times 10^{-8}T$

$B(x, t)= 1.2\times10^{-8}sin[\frac{2\pi}{6.98\times10^{-7}}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6}(x-3\times 10^8t)]\\\\B(x, t)= 1.2\times10^{-8}sin[{9 \times 10^6x-2.7\times 10^{15}t)]$

6 0

3 years ago

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma

tresset_1 [31]

Answer:

Explanation:

a. Given that:

$m-$ mass of first & second piece, $3m-$ mass of 3rd piece, $\bar v_1$ -velocity of first piece( $-23\dot i \ m/s$ ) and $\bar v_2$ as velocity of 2nd piece ( $-23\dot j \ m/s$ )

Let $\bar v_3$ be velocity of 3rd piece=?

#Vessel is at rest before explosion. Considering conservation of linear momentum:

$m\bar v_1 +m\bar v_2 +3m \bar v_3=0$ #Dividing both sides by $m, m\neq 0$

$\bar v_1+ \bar v_2 +3\bar v_3=0\\3\bar v_3=-(\bar v_1 +\bar v_2)\\\bar v_3=-\frac{1}{3}(\bar v_1 +\bar v_2)$

#Plug the $\bar v_1 ,\bar v_2$ values:

$\bar v_3=-\frac{1}{3}(-23\dot i -23\dot j)=\frac{1}{3}(23\dot i +23\dot j)$

#So the magnitude of the third piece is:

$|\bar v_3|=\sqrt{(23/3)^2+(23/3)^2}\\=10.84m/s$

Magnitude of the 3rd piece is 10.84 m/s

b. To find direction of the magnitude (as an angle relative to the $x$ -axis), we find $\angle \theta$ . The angle is obtained by getting the tan inverse as: