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2 years ago

A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.

Physics

1 answer:

Pavel [41]2 years ago

7 0

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

$f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D$

So it would normally be written with a negative signs in front indicating a divergent lens.

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What two angles of elevation will enable a projectile to reach a target 15 km downrange on the same level as the gun if the pro

forsale [732]

Angle of projection: 28 degrees or 62 degrees

Explanation:

A projectile consists of two independent motions:

A uniform motion along the horizontal direction
A uniformly accelerated motion along the vertical direction

From the equations of motion, it is possible to derive an expression for the range of a projectile, which is:

$d=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

$g=9.8 m/s^2$ is the acceleration due to gravity

For the projectile in this problem, we have:

d = 15 km = 15,000 m is the range

$u=421 m/s$ is the initial speed

Solving the equation for $\theta$ , we find the possible angle:

$sin (2\theta) = \frac{dg}{u^2}=\frac{(15,000)(9.8)}{(421)^2}=0.829$

Which gives two values of the angle:

$2\theta = 56^{\circ} \rightarrow \theta=28^{\circ}\\2\theta=124^{\circ} \rightarrow \theta=62^{\circ}$

Learn more about projectile:

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4 0

2 years ago

Example 9.1

docker41 [41]

v1f = -0.16 ms

Explanation:

Use the conservation law of linear momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where

v1i = v2i = 0

m1 = 160 kg

m2 = 0.50 kg

v2f = 50m/s

v1f = ?

So we have

0 = (160 kg)v1f + (0.5 kg)(50 m/s)

v1f = -(25 kg-m/s)/(160 kg)

= -0.16 m/s

Note: the negative sign means that its direction is opposite that of the arrow.

6 0

2 years ago

A point charge of 6.8 C moves at 6.5 × 104 m/s at an angle of 15° to a magnetic field that has a field strength of 1.4 T. What i

Tamiku [17]

Force on a moving charge is given by

$F = q(v x B)$

Given that

q = 6.8 C

$v = 6.5 * 10^4 m/s$

B = 1.4 T

angle = 15 degree

$F = qvB sin\theta$

$F = 6.8*6.5 * 10^4 * 1.4 * sin15$

$F = 1.6 * 10^5 N$

6 0

2 years ago

Read 2 more answers

Electromagnetic radiation of 5.16Ă—1016 Hz frequency is applied on a metal surface and caused electron emission. Determine the w

IgorC [24]

Answer:

Work function of the metal, $W_o=3.38\times 10^{-17}\ J$

Explanation:

We are given that

Frequency of the electromagnetic radiation, $f=5.16\times 10^{16}$ Hz

The maximum kinetic energy of the emitted electron, $K=4.04\times 10^{-19}\ J$

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

$K=h\nu-w_o$

Where h=Plank's constant= $6.63\times 10^{-34} J.s$

$\nu$ =Frequency of light source

$w_o$ =Work function

Substitute the values in the given formula

Then, the work function of the metal is given by :

$W_o=h\nu -K$

$W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}$

$W_o=3.38\times 10^{-17}\ J$

So, the work function of the metal is $3.38\times 10^{-17}\ J$ . Hence, this is the required solution.

8 0

2 years ago

2.(01.01 LC)

babymother [125]

Answer:

Increases with increase in mass

Explanation:

gravity is proportional to mass and inversely proportional to the square of the distance between them

F = GMm/d²

3 0

2 years ago