What distance is required for a train

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster tha

Daniel [21]

Answer:

Explanation:

If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .

So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .

v = u + a t

v₀ = 0 + a tmax

tmax = v₀ / a

The value of tmax is v₀ / a .

5 0

3 years ago

I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite

e-lub [12.9K]

Answer:

The range of powers is $- 5 \ D \le P \le - 2.667\ D$

Explanation:

From the question we are told that

The far point of the left eye is $n_f = 20 cm$

The near point of the left eye is $n = 15cm$

The near point with the glasses on is $n_g =25 \ cm$

From these parameter we can see that with the glass on that for near point the

Object distance would be $u = -25 \ cm$

Image distance would be $v = -15 \ cm$

To obtain the focal length we would apply the lens formula which is mathematically represented as

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

substituting values

$\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}$

$f = - \frac{75}{2} cm$

converting to meters

$f = - \frac{75}{2} * \frac{1}{100}$

$f = - \frac{75}{200} \ m$

Generally the power of the lens is mathematically represented as

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

Substituting values

$P = - \frac{200}{75} m$

$P = - 2.667 \ D$

From these parameter we can see that with the glass on that for far point the

Object distance would be $u_f = - \infty \ cm$

Image distance would be $v_f = -20 \ cm$

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

$\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}$

substituting values

$\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}$

$\frac{1}{f} = \frac{1}{-20} - 0$

$f_f = \frac{20}{1} \ cm$

converting to meters

$f_f = - \frac{20}{1} * \frac{1}{100}$

Generally the power of the lens is mathematically represented as

$P = \frac{1}{f_f}$

Substituting values

$P = - \frac{100}{20} m$

$P = - 5 \ D$

This implies that the range of powers of the lens in his glass is

$- 5 \ D \le P \le - 2.667\ D$

3 0

3 years ago

What is the transition from a gas to a liquid?

postnew [5]

The transition from gas to liquid is called condensation. An example would be water droplets forming on an ice cold glass placed in room temperature.

6 0

3 years ago

PLEASE HURRY

VLD [36.1K]

Answer:

Wilhelm Conrad Roentgen (1845-1923)

Antoine Henri Becquerel (1852-1908)

Pierre (1859-1906) and Marie (1867-1934) Curie

Explanation:

Wilhelm Conrad Roentgen (1845-1923)

Contribution: Discovery of x-rays in 1901.

Antoine Henri Becquerel (1852-1908)

Contribution: He discovered that radioactivity is the separation of x-rays and document and the difference between two.

Pierre (1859-1906) and Marie (1867-1934) Curie

Contribution: She discovered Polonium and Radium in1911

5 0

3 years ago

What type of pollution did the Clean Water Act succeed in limiting?

Ilya [14]

Sewage. If thats not it, then I need to see your choices. :)

4 0

3 years ago

Read 2 more answers