true or false. A small pick-up truck traveling at 55 km/h has more momentum than a full-sized bus traveling at the same speed

The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to th

pickupchik [31]

Answer:

a

The focal length of the lens in water is $f_{water} = 262.68 cm$

b

The focal length of the mirror in water is $f =79.0cm$

Explanation:

From the question we are told that

The index of refraction of the lens material = $n_2$

The index of refraction of the medium surrounding the lens = $n_1$

The lens maker's formula is mathematically represented as

$\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

Where $f$ is the focal length

$n$ is the index of refraction

$R_1 and R_2$ are the radius of curvature of sphere 1 and 2 of the lens

From the question When the lens in air we have

$\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]$

When immersed in liquid the formula becomes

$\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]$

The ratio of the focal length of the the two medium is mathematically evaluated as

$\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }$

From the question

$f_{air }$ = 79.0 cm

$n_2 = 1.55$

and the refractive index of water(material surrounding the lens) has a constant value of $n_1 = 1.33$

$\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}$

$f_{water} = 262.68 cm$

b

The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.

7 0

3 years ago

What is the mathematical relationship between wavelength and energy transmission?.

WITCHER [35]

The mathematical relationship between wavelength and energy transmission E = hv.

<h3>What is Wavelength and Energy transmission?</h3>

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is often defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm). The wavelength is most frequently described in nanometers (nm), which are units of 10⁻⁹ m, or angstroms (Å), which are units of 10⁻¹⁰ m, for infrared (IR), visible light (UV), and gamma radiation (γ).

The most fundamental aspect of global energy integration is energy transmission. With the flow of electricity produced from coal as well as from hydro, nuclear, wind, and solar energy all being transported through power networks, electric energy transmission is a significant source of energy transport.

Wavelength and frequency are connected to energy in the same way as they are to light. Greater energy is correlated with shorter wavelengths and higher frequencies. Therefore, lower energy is produced by longer wavelengths and lower frequencies. E = hv is the energy equation.

to learn more about wavelength and frequency go to - brainly.com/question/2174631

#SPJ4

6 0

2 years ago

A car of weight 6,400 N has four wheels. Each wheel has an area of

Natasha2012 [34]

Answer:

80N/cm^2

Explanation:

Given data

A car of weight 6,400N

Area= 80cm^2

The weight is distributed evenly on the 4 wheels

hence 6400/4

=1600 N

We know that

Pressure = force/Area

Pressure= 1600/80

Pressure= 20 N/cm^2

For the four wheels, the pressure is

=20*4

=80N/cm^2

6 0

3 years ago

A bubble, located 0.200 m beneath the surface in a glass of beer, rises to the top. The air pressure at the top is 1.01 ???? 10

SIZIF [17.4K]

Answer:

the ratio of the bubble’s volume at the top to its volume at the bottom is 1.019

Explanation:

given information

h = 0.2 m

$P_{0}$ = 1.01 x $10^{5}$ Pa

$P_{1} V_{1} = P_{2} V_{2}$

$\frac{V_{2} }{V_{1}} = \frac{P_{1} }{P_{2}}$

$P_{1}$ = $P_{0}$ + ρgh, ρ = 1000 kg/ $m^{3}$

$P_{1}$ = 1.01 x $10^{5}$ Pa + (1000 x 9.8 x 0.2) = 1,0296 x $10^{5}$ Pa

$P_{2}$ = $P_{0}$ = $10^{5}$ Pa

thus,

$\frac{V_{2} }{V_{1}} = 1,0296 x [tex]10^{5}$ / $10^{5}$ = 1.019

8 0

3 years ago

Riders on the Tower of Doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.5

umka2103 [35]

Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:

A) The apparent weight is 3548 N directed upwards

B) The factor of change of weights is 5

Given parameters

Free fall time t1 = 2.0 s

The time tot stops at t2 = 0.5 s

Mass m = 65 kg

To find

A) Apparent weight

B) Factor that exceeds the actual weight

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

This exercise we will do them in two parts

1st part. Freefall

We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero

v = v₀ - g t

Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time

v = -gt

v = -9.8 2.0

v = -19.6 m / s

The negative sign indicates that the velocity is directed downwards

2nd part. Braking movement

We look for the acceleration to stop the body, in this case the final velocity is zero

$v_f$ = v + a t

0 = v + at

a = $- \frac{v}{t}$ - v / t

a = $-\ \frac{-19.6}{0.5}$

a = + 39.2 m / s²

The positive sign indicates that the acceleration is directed upward.

A) Newton's second law states that the force is proportional to the mass and the acceleration of the body

∑ F = m a

F -W = m a

F = m (a + g)

F = 65 (39.2 + 9.8)

F = 3185 N

Body weight is

W = mg

W = 65 9.8

W = 637 N

therefore the apparent weight is

ΔW = F -W

ΔW = 3185 - 637

Δw = 2548 N

B) the weight change factor

Factor = $\frac{F}{W}$ F / W

Factor = $\frac{3185}{937}$

factor = 5

In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:

A) The apparent weight is 3548 N directed upwards

B) The factor of change of weights is 5

Learn more about Newton's second law and kinematics here:

brainly.com/question/15378807

5 0

3 years ago