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Gre4nikov [31]
2 years ago
5

A magnifier has a magnification of 8 x. How far from the lens should an object be placed so that it’s virtual image is at the ne

ar-point distance of 25 cm?
Physics
1 answer:
larisa86 [58]2 years ago
8 0

The distance of the object from the lens is 3.12 cm.

<h3>What is magnification?</h3>

Magnification is the process of of enlarging an apparent size of an object.

<h3>Object distance</h3>

The object distance is calculated using the following lens formula;

M = \frac{V}{U}

where;

  • M is the magnification
  • V is the image distance
  • U is object distance

- 8 = \frac{-25}{U} \\\\U = \frac{-25}{-8} \\\\U = 3.12 \ cm

Thus, the distance of the object from the lens is 3.12 cm.

Learn more about object distance here: brainly.com/question/24894435

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Se apunta un rifle horizontalmente con mira a un blanco pequeño que está a 200m en el suelo. La velocidad inicial de la bala es
vfiekz [6]

Answer:

Lo importante a tener en cuenta sobre esta pregunta es que la velocidad horizontal de la bala no hace ninguna diferencia en cuanto al tiempo que tarda en caer al suelo.

Debido a que el arma no ha aplicado ninguna fuerza vertical a la bala, la única fuerza que afecta la bala es la gravedad. Esto significa que la bala tarda tanto en caer al suelo como lo haría si se cayera, a pesar de que ahora viaja una gran distancia horizontal en la duración.

Para encontrar el tiempo de viaje antes de tocar el suelo, tenemos 3 valores:

-El desplazamiento desde el suelo que la bala debe viajar, s = 1.5m

-La aceleración que experimenta la bala. Como la gravedad está acelerando la bala hacia abajo, a = g = ~ 9.81m / s ^ 2

-La velocidad inicial de la bala verticalmente. Como la bala es estacionaria verticalmente (solo viaja horizontalmente al inicio), u = 0m

Examinamos nuestras ecuaciones de movimiento, comúnmente conocidas como ecuaciones SUVAT. Es posible que necesite aprender estos para su examen, pero algunas tablas de examen los proporcionan.

Debido a que tenemos s, u y a, y estamos buscando el tiempo t, la ecuación relevante es

s = ut + 0.5 (en ^ 2)

Completando nuestros valores tenemos:

1.5 = 0t + 0.5 (9.81 x t ^ 2)

1.5 = 4.905 x t ^ 2

Divide 1.5 entre 4.905 para encontrar t ^ 2

t ^ 2 = 0.3058 ...

Simplemente encontramos la raíz cuadrada de t ^ 2 para encontrar t, el tiempo que tarda la bala en llegar al suelo:

t = 0.553s (3 cifras significativas)

Para encontrar la distancia horizontal, d, que la bala ha viajado antes de tocar el suelo, podemos usar la ecuación que vincula el desplazamiento s con cierta velocidad v durante un tiempo t:

s = vt

La velocidad horizontal de la bala, v = 430

El tiempo antes de que la bala toque el suelo, t = 0.553

Entonces d = vt = 430 * 0.553 = 238m (3 cifras significativas)

3 0
3 years ago
Two objects experienced the same displacement in the same amount of time. The two objects must have the same -
nikdorinn [45]

If two object experiences the same displacement at the same time, the two objects must have the same average velocity.

The average velocity of an object is defined as the change in displacement per change in time of motion. This can be written as follows;

avg. \ velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\V = \frac{\Delta \ x}{\Delta t}

If two object experiences the same displacement at the same time, it means that the change in displacement with time is constant.

\frac{\Delta x_1}{\Delta t_1} = \frac{\Delta x_2}{\Delta t_2} = V_{avg}

Thus, we can conclude that if two object experiences the same displacement at the same time, the two objects must have the same average velocity.

Learn more here: brainly.com/question/23856383

6 0
2 years ago
"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and
spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0
3 years ago
In Act III, Scene 1, who hears these words from Titania?
k0ka [10]
OMGG I  LOVE LOVE LOVE the midnight summer's dream
Either A or D
Bottom's the donkey right?
I'm sure that it's D or A since I know C , Pease is a ... fairy?
Need to refresh my memory!
3 0
3 years ago
"when you double the vehicle's weight, you will __________ the vehicle's stopping distance. "
Stolb23 [73]

With same braking power you will be stopping faster on the original weight therefore the answer to fill the blank is increase. The stopping distance will increase as there'll be higher energy to dissipate than lighter cars applied with the braking force similar with that of the lighter car. Also the skid and drag will add to the distance as well as the inertia of the moving heavier vehicle would be greater as well.

5 0
2 years ago
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