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2 years ago

Which kinds of objects emit visible light in the electromagnetic spectrum?

Physics

1 answer:

xeze [42]2 years ago

6 0

answer :

Visible: Our eyes detect visible light. Fireflies, light bulbs, and stars all emit visible light. Ultraviolet: Ultraviolet radiation is emitted by the Sun and are the reason skin tans and burns. "Hot" objects in space emit UV radiation as well.

Which objects emit electromagnetic waves?

All objects emit electromagnetic radiation according to their temperature. Colder objects emit waves with very low frequency (such as radio or microwaves), while hot objects emit visible light or even ultraviolet and higher frequencies.

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A projectile is launched at ground level with an initial speed of 54.5 m/s at an angle of 35.0° above the horizontal. It strikes

Alchen [17]

<h2>Answer: x=125m, y=48.308m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=54.5m/s$ is the projectile's initial speed

$\theta=35\°$ is the angle

$t=2.80s$ is the time since the projectile is launched until it strikes the target

$x$ is the final horizontal position of the projectile (the value we want to find)

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0$ is the initial height of the projectile (we are told it was launched at ground level)

$y$ is the final height of the projectile (the value we want to find)

$g=9.8m/s^{2}$ is the acceleration due gravity

Having this clear, let's begin with x (1):

$x=(54.5m/s)cos(35\°)(2.8s)$ (3)

$x=125m$ (4) This is the horizontal final position of the projectile

For y (2):

$y=0+(54.5m/s)sin(35\°)(2.8s)-\frac{(9.8m/s^{2})(2.8s)^{2}}{2}$ (5)

$y=48.308m$ (6) This is the vertical final position of the projectile

4 0

3 years ago

Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0

2 years ago

During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.

Allushta [10]

Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0

2 years ago

Continuous sinusoidal perturbation Assume that the string is at rest and perfectly horizontal again, and we will restart the clo

Elena-2011 [213]

a) $3.14 \cdot 10^{-4} s$

b) See plot attached

c) 10.0 m

d) 0.500 cm

Explanation:

The position of the tip of the lever at time t is described by the equation:

$y(t)=(0.500 cm) sin[(2.00\cdot 10^4 s^{-1})t]$ (1)

The generic equation that describes a wave is

$y(t)=A sin (\frac{2\pi}{T} t)$ (2)

where

A is the amplitude of the wave

T is the period of the wave

t is the time

By comparing (1) and (2), we see that for the wave in this problem we have

$\frac{2\pi}{T}=2.00\cdot 10^4 s^{-1}$

Therefore, the period is

$T=\frac{2\pi}{2.00\cdot 10^4}=3.14 \cdot 10^{-4} s$

The sketch of the profile of the wave until t = 4T is shown in attachment.

A wave is described by a sinusoidal function: in this problem, the wave is described by a sine, therefore at t = 0 the displacement is zero, y = 0.

The wave than periodically repeats itself every period. In this sketch, we draw the wave over 4 periods, so until t = 4T.

The maximum displacement of the wave is given by the value of y when $sin(...)=1$ , and from eq(1), we see that this is equal to

y = 0.500 cm

So, this is the maximum displacement represented in the sketch.

When standing waves are produced in a string, the ends of the string act as they are nodes (points with zero displacement): therefore, the wavelength of a wave in a string is equal to twice the length of the string itself:

$\lambda=2L$