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Nadusha1986
2 months ago
10

Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc

us on objects which are as close as 17.2 cm from your eyes, but no closer. The purpose of glasses would be to make objects that are far away appear to be at your far point. What would be the refractive power (in diopters - do not enter units) for your prescription?If your glasses make distant objects appear to be at your far point, then any object that is closer to you will appear even closer than it really is. What is the closest distance from your face at which you could hold a book and still be able to see it clearly while wearing your glasses?
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
5 0
La imagen del objeto distante se formará en el punto lejano o a 52.5 cm, así que la distancia del objeto u = infinito, la distancia de la imagen v = -52.5 cm, la longitud focal requerida = f. Usando la fórmula de la lente 1 / v - 1 / u = 1 / f, obtenemos f = -52.5 cm = -0.525 m. La potencia P = 1 / f = -1 / 0.525 = -1.90. Ahora, para el ojo con gafas, debemos encontrar el nuevo punto cercano.
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Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart
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Answer:F=\frac{(M+m)g}{\mu _s}

Explanation:

Provided:

The trolley, with mass M, is allowed to roll freely without friction.

The coefficient of friction between the trolley and mass m is \mu _s.

A force F is applied to mass m.

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
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Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

Projectile Motion

When an object is projected near the surface of the Earth at an angle \theta to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

where y_o represents the initial height from ground level, v_{oy} is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

v_y=v_{oy}-gt

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of v_o.

The peak height is attained when v_y=0, which allows us to compute the time to reach that height.

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum height reached is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is equal to 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Continuing with the calculations for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Substituting known values yields

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

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