When an unbalanced force acts on an object

If the person can't accommodate and the glasses is + 2.50d, at which distance will the person see clearly

bezimeni [28]

Power of the glasses required is given as +2.5 D

this shows that person is having Farsightedness in which he can objects placed far clearly but he is not able to see clearly to the near objects

For normal vision near point of eye is 25 cm

Let say person can see the objects clearly at distance "d" from the eye

now from the formula

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here given that

$\frac{1}{f} = 2.5$

$f = 40 cm$

also we know that

$d_o = 25 cm$

from above equation now

$\frac{1}{d} + \frac{1}{25} = \frac{1}{40}$

$d = 66.6 cm$

So the distance of object must be 66.6 cm

8 0

3 years ago

Read 2 more answers

A pendulum consists of a 2.0-kg block hanging on a 1.5-m length string. A 10-g bullet moving with a horizontal velocity of 900 m

pav-90 [236]

Answer:

The maximum height above its initial position is:

$h_{max}=1.53\: m$

Explanation:

Using momentum conservation:

$m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb}$ (1)

Where:

m(b) is the mass of the bullet
m(B) is the mass of the block
v(ib) is the initial velocity of the bullet
v(fb) is the final velocity of the bullet
v(fB) is the final velocity of the block

Let's find v(fb) using equation (1)

$m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}$

$v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}$

$v_{fB}=\frac{0.1(900-300)}{2}$

$v_{fB}=30\: m/s$

We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.

$\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}$

$h_{max}=\frac{1}{2g}v_{fB}$

$h_{max}=\frac{1}{2(9.81)}30$

$h_{max}=1.53\: m$

I hope it helps you!

8 0

3 years ago

1. What part of the Earth do “faults” appear?

alekssr [168]

Answer:

Faults are found in collisions zones, and tectonic plates push up against mountain ranges for example the Himalayas or the Rocky Mountains.

Explanation:

6 0

2 years ago

A quantity of gas is contained in a sealed container of fixed volume. The temperature of the

Alchen [17]

Answer:

If the Kelvin temperature of a gas is increased, the volume of the gas increases. This can be understood by imagining the particles of gas in the container moving with a greater energy when the temperature is increased.

Explanation:

If you heat a gas you give the molecules more energy so they move faster. This means more impacts on the walls of the container and an increase in the pressure. Conversely if you cool the molecules down they will slow and the pressure will be decreased.

To calculate a change in pressure or temperature using Gay Lussac's Law.

6 0

2 years ago

A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3

den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

$h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m$

$h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m$

$m = 148g*\frac{1kg}{1000g} = 0.148kg$

$g = 9.8 \frac{m}{s^2}$

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

$U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J$

The increase in potential energy of the ball is 115.82 J

5 0

3 years ago