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marishachu [46]

2 years ago

5

You travel 23 meters north in 16 seconds, 5 meters south in 4 seconds, and 16 meters north in 18 seconds. Calculate your total d

istance traveled and your displacement.

1 answer:

dolphi86 [110]2 years ago

4 0

Answer:

d: 44m

Δd:34

Explanation:

d: 23+5+16= 44m

Δd: 23-5+16= 34m

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If you compare an electrical circuit to a system that carries water, what would the switch represent?

ehidna [41]

Try B. The handle on the faucet. It stops the energy and lets it flow like how the switch lets the energy flow but then stops it also

8 0

2 years ago

A light wave travels through water (n=1.33) at an angle of 35º. What angle

san4es73 [151]

Answer:

$49.7^{\circ}$

Explanation:

We can solve the problem by using Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 = 1.33 is the index of refraction of the first medium (water)

n2 = 1.00 is the index of refraction of the second medium (air)

$\theta_1 = 35^{\circ}$ is the angle of incidence of the wave in water

$\theta_2$ is the angle of refraction of the wave in air

Solving for the angle of refraction,

$sin \theta_2 = \frac{n_1 sin \theta_1}{n_2}=\frac{(1.33) sin 35^{\circ}}{1.00}=0.763\\\theta_2 = sin^{-1} (0.763)=49.7^{\circ}$

6 0

2 years ago

Read 2 more answers

A puck moves 2.35 m/s in a -22 degree direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50

Whitepunk [10]

Answer:

Displacement in Y direction is 0.434 m

Explanation:

initial velocity of the puck is 2.35 m/s at -22 degree

so here it is given as

$v_i = 2.35 cos22 \hat i - 2.35 sin22 \hat j$

$v_i = 2.18 \hat i - 0.88 \hat j$

final velocity is given as 6.42 m/s at 50 degree

so we have

$v_f = 6.42 cos50 \hat i + 6.42 sin50\hat j$

$v_f = 4.13 \hat i + 4.92\hat j$

now displacement in Y direction is given as

$\Delta y = \frac{v_f + v_i}{2}t$

$\Delta y = \frac{-0.88 + 4.92}{2} (0.215)$

$\Delta y = 0.434 m$

5 0

3 years ago

What is the energy of a photon that has the same wavelength as an electron having a kinetic energy of 15 ev?

serg [7]

Answer: $6.268(10)^{-16}J$

Explanation:

The kinetic energy of an electron $K_{e}$ is given by the following equation:

$K_{e}=\frac{(p_{e})^{2} }{2m_{e}}$ (1)

Where:

$K_{e}=15eV=2.403^{-18}J=2.403^{-18}\frac{kgm^{2}}{s^{2}}$

$p_{e}$ is the momentum of the electron

$m_{e}=9.11(10)^{-31}kg$ is the mass of the electron

From (1) we can find $p_{e}$ :

$p_{e}=\sqrt{2K_{e}m_{e}}$ (2)

$p_{e}=\sqrt{2(2.403^{-18}J)(9.11(10)^{-31}kg)}$

$p_{e}=2.091(10)^{-24}\frac{kgm}{s}$ (3)

Now, in order to find the wavelength of the electron $\lambda_{e}$ with this given kinetic energy (hence momentum), we will use the De Broglie wavelength equation:

$\lambda_{e}=\frac{h}{p_{e}}$ (4)

Where:

$h=6.626(10)^{-34}J.s=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

So, we will use the value of $p_{e}$ found in (3) for equation (4):

$\lambda_{e}=\frac{6.626(10)^{-34}J.s}{2.091(10)^{-24}\frac{kgm}{s}}$

$\lambda_{e}=3.168(10)^{-10}m$ (5)

We are told the wavelength of the photon $\lambda_{p}$ is the same as the wavelength of the electron:

$\lambda_{e}=\lambda_{p}=3.168(10)^{-10}m$ (6)

Therefore we will use this wavelength to find the energy of the photon $E_{p}$ using the following equation:

$E_{p}=\frac{hc}{lambda_{p}}$ (7)

Where $c=3(10)^{8}m/s$ is the spped of light in vacuum

$E_{p}=\frac{(6.626(10)^{-34}J.s)(3(10)^{8}m/s)}{3.168(10)^{-10}m}$

Finally:

$E_{p}=6.268(10)^{-16}J$

4 0

3 years ago

A person 1.8m tall stands 0.75m from a reflecting globe in a garden.

Maru [420]

Answer:

1. The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. The person's image is 3.38 m tall.

Explanation:

From the given question, object distance, u = 0.75 m, object height = 1.8 m, radius of curvature of the reflecting globe, r = 8 cm = 0.08 m.

f = $\frac{r}{2}$ = $\frac{0.08}{2}$ = 0.04 m

1. The image distance, v, can be determined by applying mirror formula:

$\frac{1}{f}$ = $\frac{1}{u}$ + $\frac{1}{v}$

$\frac{1}{0.04}$ = $\frac{1}{0.75}$ + $\frac{1}{v}$

$\frac{4}{100}$ - $\frac{75}{100}$ = $\frac{1}{v}$

$\frac{1}{v}$ = $\frac{4 - 75}{100}$

= - $\frac{71}{100}$

⇒ v = - $\frac{100}{71}$

= - 1.41 m

The image of the person is 1.41 m, virtual and formed at the back of the surface of the globe.

2. $\frac{image distance}{object distance}$ = $\frac{image height}{object height}$

$\frac{1.41}{0.75}$ = $\frac{v}{1.8}$

v = $\frac{2.538}{0.75}$

= 3.384

v = 3.38 m

The person's image is 3.38 m tall.

6 0

2 years ago