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3 years ago

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The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

egligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. The index of refraction of air is nair = 1.000 and that of water is nwater = 1.333. (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall.

1 answer:

monitta3 years ago

5 0

Answer:

i know the questin but i got to try and find it

Explanation:

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To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

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I think the answer is A.

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3 years ago

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Una pelota se desliza hacia arriba por una pendiente se halla inicialmente a 6m de la parte más baja de dicha pendiente y tiene

jolli1 [7]

Answer:

a) vprom = - 0.6 [m/s] b) ver explicacion c) vf = 10.8 [m/s]

Explanation:

Para poder solucionar este problema debemos completar el enunciado del problema así como la pregunta, realizando una búsqueda en Internet encontramos el enunciado completo, así como la pregunta.

"Una pelota que se desliza hacia arriba por una pendiente se halla inicialmente a 6 m de la parte más baja de dicha pendiente y tiene una velocidad de 4 m/s. Cinco segundos después se encuentra a 3 m de la parte más baja. Si suponemos una aceleración constante, "

¿cual fue la velocidad media?

¿Cuál es el significado de una velocidad media negativa?

¿Cuáles son la aceleración media y la velocidad final?

Para facilitar esta solución debemos realizar un esquema del movimiento de la pelota en el plano inclinado. En la imagen adjunta se puede ver un esquema del movimiento de la pelota sobre el plano inclinado en los diferentes tiempos mencionados.

<u>¿cual fue la velocidad media?</u>

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La velocidad media se debe calcular utilizando la expresión de velocidad = espacio / tiempo.

$v_{prom}=x/t\\v_{prom}=(3-6)/5\\v_{prom}=-0.6 [m/s]$

<u>¿Cuál es el significado de una velocidad media negativa? </u>

La velocidad media negativa significa que se dirige hacia abajo en sentido contrario

<u>¿Cuáles son la aceleración media y la velocidad final?</u>

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$x=v_{o}*t+0.5*a*t^{2} \\-3=(4)*(5)+0.5*a*(5)^2\\a=1.36[m/s^2]\\\\Velocity\\v_{f}=v_{o}+a*t = 4+(1.36*5)\\v_{f}=10.8[m/s]$

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3 years ago

11) Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a sp

sleet_krkn [62]

Answer:

100 m

Explanation:

Arthur and Betty should be walking the same amount of time if they start walking at the same time and stop when they meet = t

Speed of Arthur = 3 m/s

Speed of Betty = 2 m/s

Distance = Speed × time

Distance covered by Arthur = 3t

Distance covered by Betty = 2t

The distance covered by both of them will be 100 m

$3t+2t=100\\\Rightarrow 5t=100\\\Rightarrow t=\frac{100}{5}\\\Rightarrow t=20\ s$

The speed of dog is 5 m/s

Spot is running back and forth at 5 m/s for 20 seconds

In 20 seconds the distance covered by the dog is

$\mathbf{5\times 20}=\mathbf{100\ m}$

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4 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

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3 years ago

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Lelechka [254]

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