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Gnesinka [82]
2 years ago
12

Is radiation the transfer of energy in the form of particles

Physics
1 answer:
katen-ka-za [31]2 years ago
7 0
Radiation is the emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles that cause ionization. 
Or, Radiation is the energy transmitted by heat, light, electricity, etc. However, your question term is convection.
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The Doppler effect causes sounds moving toward an observer to sound _____.
vladimir1956 [14]

Answer: softer

The sound waves become weaker, so to an observer this wave is lighter sound.

8 0
2 years ago
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What is the role of equations in this course?
Karolina [17]
That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.
4 0
3 years ago
CAN SOMEONE HEELLLLLLLLLP MEEEEEEE
Soloha48 [4]
Its B , the definition of cardiovascular endurance is <span>the ability of the heart, lungs and blood vessels to deliver oxygen to your body tissues.</span>
7 0
3 years ago
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A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p
saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, x, of an SHM is given by

x = A\cos(\omega t)

A is the amplitude and \omega is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or \frac{\pi}{4} radian.

From trigonometry, \sin A =\cos B if A and B are complementary.

At t = 0, x = 3.5

3.5 = A\cos(\omega \times0)

A =3.5

So

x = 3.5\cos(\omega t)

At t = 0.12, x = 1.5

1.5 = 3.5\cos(0.12\omega)

\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286

0.12\omega =\cos^{-1}0.4286

0.12\omega = 1.13

\omega = 9.4

The period, T, is related to \omega by

T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67

5 0
2 years ago
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
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