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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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Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

ss7ja [257]

Answer:

The number of atoms are $1.86\times10^{8}$ .

Explanation:

Given that,

Diameter $D = 1.40\times10^{2}\ pm$

$D=1.40\times10^{2}\times10^{-12}\ m$

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

$Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}$

$Number\of\atoms =1.86\times10^{8}$

Hence, The number of atoms are $1.86\times10^{8}$ .

8 0

2 years ago

Does the mass of a pendulum affects the period of oscillation

pshichka [43]

Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.

8 0

2 years ago

A measurement has high<br> when it is very close to the<br> true value?

jarptica [38.1K]

Answer:

Accuracy

Explanation:

Accuracy means making measurements that are close to the value precision means making measurement that are close in value to eachother but not necessarily close to the true value.

I hope this helps! If not sorry.

7 0

2 years ago

The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate

ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

$\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}$ ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

$\theta=\frac{3.7m}{0.162m}=22.83rad$

$22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}$

to calculate the angular speed w we can use $\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}$