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3 years ago

Two cars collide and come to a complete stop. where did all of their energy go?

1 answer:

3 0

Most of the energy will be absorbed by the materials that make up the cars, causing them to deform. The energy will also be converted into sound energy, causing a loud bang upon collision. Also, some energy will be converted to thermal energy, which will cause the cars to heat up slightly.

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The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi

Keith_Richards [23]

Answer:

Explanation:

a )

from lens makers formula

$\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})$

f is focal length , r₁ is radius of curvature of one face and r₂ is radius of curvature of second face

putting the values

$\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})$

1.462 = 2 - 1 / r₂

1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens f = 1.8 cm

image distance v = ?

lens formula

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

$\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}$

$\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}$

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

5 0

3 years ago

A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl

dolphi86 [110]

In the first case, the force acting on the spring is the weight of the mass:
$F=mg=(2.0 kg)(9.81 m/s^2)=19.6N$
This force causes a stretching of $x=6.4 cm=0.064 m$ on the spring, so we can use these data to find the spring constant:
$k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m$

In the second case, the first mass is replaced with a second mass, whose weight is
$F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N$
And since we know the spring constant, we can calculate the new elongation of the spring:
$x= \frac{F}{k}= \frac{24.5 N}{306.3 N/m}=0.080 m=8.0 cm$

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3 years ago

Recall from Chapter 1 that a watt is a unit of en- ergy per unit time, and one watt (W) is equal to one joule per second ( J·s–1

harkovskaia [24]

Answer:

Explanation:

The energy of a photon is given by the equation $E_p=h f$ , where h is the <em>Planck constant</em> and f the frequency of the photon. Thus, N photons of frequency f will give an energy of $E_N=N h f$ .

We also know that frequency and wavelength are related by $f=\frac{c}{\lambda}$ , so we have $E_N=\frac{N h c}{\lambda}$ , where c is the <em>speed of light</em>.

We will want the number of photons, so we can write

$N=\frac{\lambda E_N}{h c}$

We need to know then how much energy do we have to calculate N. The equation of power is $P=E/t$ , so for the power we have and considering 1 second we can calculate the total energy, and then only consider the 4% of it which will produce light, or better said, the N photons, which means it will be $E_N$ .

Putting this paragraph in equations:

$E_N=(\frac{4}{100})E=0.04Pt=(0.04)(100W)(1s)=4J$ .

And then we can substitute everything in our equation for number of photons, in S.I. and getting the values of constants from tables:

$N=\frac{\lambda E_N}{h c}=\frac{(520 \times10^{-9}m) (4J)}{(6.626\times10^{-34}Js) (299792458m/s)}=1.047 \times10^{19}$

3 0

3 years ago

5,6,1,9,21,37,68 ? find missing

maria [59]

Answer:

Missing numbers are 3,11,13, 15,17,19, 23,27,29 and wrong are 6,68

4 0

3 years ago

How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force

Learn more about gravitational force and electric force:

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5 0

3 years ago