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3 years ago

How does the amount of energy in light change as the wavelength increases

Physics

1 answer:

Phoenix [80]3 years ago

5 0

Answer

we know,

$E= \dfrac{hc}{\lambda}$

where

E = energy

h = planks constant

λ = wavelength

from the above equation, we can see that energy is inversely proportional to wavelength.

$E\ \alpha\ \dfrac{1}{\lambda}$

that is why

Violet color light has low wavelength but the energy of Violet light is very high whereas red light has low energy and high wavelength.

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A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun

Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100.0}{300}$

$t=0.33\ sec$

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s=0+\dfrac{1}{2}\times9.8\times(0.33)^2$

$s=0.53\ m$

Hence, The distance is 0.53 m.

5 0

3 years ago

It takes 500 J of work to compress a spring 10 cm. What is the force constant of the spring? (b) When a 3.0 kg block is pushed a

NARA [144]

Answer:

Explanation:

a) Energy stored in spring = 1/2 k x² = .5 x k 0.1²

500 = 5 x 10⁻³ k ,

k = (500/5) x 10³ = 10⁵ N/m

b )

k = 4.5 x 10¹ = 45 N/m

Stored energy = 1/2 k x² = .5 x 45 x 8² x 10⁻⁴ =1440 x 10⁻⁴ J

This energy gets dissipated by friction .

work done by friction = μ mg d

d is the distance traveled under friction

so 1440 x 10⁻⁴ = μ x 3 x 9.8 x 2

μ = 245 x 10⁻⁴ or 0.00245 which appears to be very small. .

8 0

3 years ago

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

How is the Earth's magnetic field like a bar magnet's magnetic field?

valina [46]

In the outer layers of earths atmosphere gases are in ionized state primarily on account of cosmic rays . as earth rotates , strong electric current are set up due to movement of ions . these currents form earth magnetic field . and thus two equal and opposite poles of earth formed

8 0

3 years ago

natta225 [31]

Answer:

maybe it is the first one

7 0

3 years ago