What power lens is needed to correct for farsightednesswherethe uncorrrected near point is 75 cm?

Answers

Answer 1

Answer:

To solve this problem we will apply the concept related to the lens power with which farsightedness can be corrected. Mathematically this value is given by the relationship,

$P = (1)/(f)$

Here,

f =focal length

In turn, said expression can be exposed in terms of the distance of the object and the image as:

$P = (1)/(p)+(1)/(q)$

Here,

p = Object Distance ( By convention is 25cm)

q = Image distance

Replacing we have,

$P = (1)/(0.25)+(1)/(-0.75)$

$P = +2.67D$

Therefore the power lens that is needed to correct for farsightedness is +2.67D

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Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?

Answers

Hi, thank you for posting your question here at Brainly.

To solve this problem, we use Coulomb's Law:

F = kQ1Q2/d^2, where k = 9x10^9

Q1 = 3.5 uC
Q2 = -3.5 uC
Q3 = 4.0 uC

But first, we find the distance between Q1 and Q3 and between Q2 and Q3.

d between Q1 and Q2:
d = sqrt[(0-0.4)^2+(0.3-0)^2]
d = 0.5 m

d between Q1 and Q3:
d = sqrt[(0-0.4)^2+(-0.3-0)^2]
d = 0.5 m

Through force balance, F between Q2 and Q3 - F between Q1 and Q3:

$F_(net) = ((9x 10^(9))(-3.5)(4) )/( 0.5^(2) ) -((9x 10^(9))(3.5)(4) )/( 0.5^(2) )=-1.008* 10^(12)$

Thus, the net force is -1 x 10^-12 C

Final answer:

The total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC is zero. This is because the forces due to each of these charges on the third charge are equal in magnitude but opposite in direction, hence they cancel each other completely.

Explanation:

The question asks for the magnitude and direction of the total electric force exerted by point charges -3.5 μC and 3.5 μC on a point charge 4.0 μC. This is related to Coulomb's Law, which describes the force between charged objects. Specifically, Coulomb's Law states that the force (F) between two point charges is directly proportional to the product of their charges (q1*q2) and inversely proportional to the square of the distance (r) between them. It also depends on the permittivity of free space (ε₀).

First, you would determine the force between each of the point charges and the third charge separately, and then superpose these forces to find the total force. The force in each case can be calculated using the equation F = k*|q1*q2|/r², where k is Coulomb's constant (8.99 * 10^9 N.m²/C²). You would need to make sure you take into account the signs of the charges when deciding the directions of the forces and when superposing the separate forces.

Assume upwards to be the positive direction. The 3.5 uC charge forces and -3.5 uC charge forces on the 4 uC charge would be opposite in direction (one downwards and one upwards) and identical in magnitude. Therefore, they will cancel each other out, and hence, the total electric force on the third charge (4 uC) will be zero.

Learn more about Total Electric Force here:

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A proton moves perpendicular to a uniform magnetic field B with arrow at a speed of 2.20 107 m/s and experiences an acceleration of 1.90 1013 m/s2 in the positive x-direction when its velocity is in the positive z-direction. Determine the magnitude and direction of the field.

Answers

Answer:

The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

Explanation:

Given that,

Speed $v = 2.20*10^7\ m/s$

Acceleration $a=1.90*10^(13)\ m/s^2$

We need to calculate the magnetic field

Using formula of magnetic field

$F=qvB$ ....(I)

Using newton's second law

$F= ma$ ....(II)

From equation (I) and (II)

$ma=qvB$

Put the value into the formula

$1.90*10^(13)*1.67*10^(-27)=1.6*10^(-19)*2.20*10^(7)*B$

$3.173*10^(-14)=1.6*10^(-19)*2.20*10^(7)*B$

$B=(3.173*10^(-14))/(1.6*10^(-19)*2.20*10^(7))$

$B=0.009014\ T$

We need to calculate the direction of the field

Using the right hand rule, point the right hand fingers along the velocity which is in the positive z direction.

Now, if we curl the fingers along the direction of magnetic field that is in the negative y direction, then the thumb will point in the positive x direction.

Hence, The magnitude and direction of the magnetic field is 0.009014 T in the negative y direction.

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

The formula for the rotational kinetic energy is

$KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)$

$KE_(rot) =18,671.31 \ J$

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> $573*2\pi *(1)/(60) =60$ rad/s

I= $(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537$ kgm^2

K=1/2(number of rods)(I)(ω)= $(1)/(2) *(5)(0.537)(60)^2=4833$ J

I know it's very late, but hope this helps anyone else trying to find the answer.

Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?

Answers

Final answer:

Given the same initial linear speed, a solid ball, solid disk, and hoop will expend energy on both rotation and translation. The solid ball, having the lowest moment of inertia, uses the most energy for translation and, therefore, will travel the highest up an incline.

Explanation:

In the context of this problem related to physics, the deciding factor is the distribution of mass, which influences each object's moment of inertia. Objects set to roll tend to use energy in two ways: translation (moving along the incline) and rotation (spinning about their center). Moment of inertia essentially measures how much of the object's energy goes towards rotation.

For a solid ball, solid disk, and hoop with the same mass and radius, the hoop has the highest moment of inertia with all of its mass at the maximum distance from the center. Followed by the solid disk, with its mass spread evenly from the center to its edge. Lastly, the solid ball has the lowest moment of inertia as its mass is concentrated towards the center.

This means that, given the same initial linear speed, the hoop will expend most of its energy on rotation rather than moving up the incline (translation). The solid disk will have a more balanced distribution between translation and rotation, and finally, the solid ball will use the least amount of energy on rotation and the most on translation. As such, the solid ball will go the farthest up the incline.

Learn more about the Physics of Rotating Objects here:

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An eighteen gauge copper wire has a nominal diameter of 1.02mm. This wire carries a constant current of 1.67A to a 200w lamp. The density of free electrons is 8.5 x 1028 electrons per cubic metre. Find the magnitude of:i. The current density ii. The drift velocity

Answers

Answer:

The current density is $J = 2.04 * 10^(6) A /m^2$

The drift velocity is $v_d = 1.5 * 10^(-4) m/s$

Explanation:

From the question we are told that

The nominal diameter of the wire is $d = 1.02 mm= (1.02)/(1000) = 0.00102 \ m$

The current carried by the wire is $I = 1.67 A$

The power rating of the lamp is $P = 200 W$

The density of electron is $n = 8.5 * 10^(28) \ e/m^3$

The current density is mathematically represented as

$J = (I)/(A)$

Where A is the area which is mathematically evaluated as

$A = \pi (d^2)/(4)$

Substituting values

$A = 3.142 * ((1.02 * 10^(-3))^2 )/(4)$

$A = 8.0*10^(-4)m^2$

$J = (1.67)/(8.0*10^(-4))$

$J = 2.04 * 10^(6) A /m^2$

The drift velocity is mathematically represented as

$v_d = (J)/(ne)$

Where e is the charge on one electron which has a value $e = 1.602 *10^(-19) C$

$v_d =(2.04 * 10^6 )/(8.5 *10^(28) * 1.6 * 10^(-19))$

$v_d = 1.5 * 10^(-4) m/s$

A beam of light travels from a medium with an index of refraction of 1.27 to a medium with an index of refraction of 1.46. If the incoming beam makes an angle of 14.0° with the normal, at what angle from the normal will it refract?