What is the speed of a skater who travels a distance of 210 m in a time of 10 seconds?

Which are characteristics of electromagnetic waves?check all that apply.

emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, $c=3\cdot 10^8 m/s$ )

travel only through matter --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space --> FALSE, electromagnetic waves can also travel through matter

can bend around objects --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0

3 years ago

Read 2 more answers

1. A racing car with the driver weighs 1825 lb. Find the kinetic energy in ft*lb when traveling with a speed of 100 mi/hr.

Svetlanka [38]

Answer:

1. 610,000 lb ft

2. 490 J

Explanation:

1. First, convert mi/hr to ft/s:

100 mi/hr × (5280 ft / mi) × (1 hr / 3600 s) = 146.67 ft/s

Now find the kinetic energy:

KE = ½ mv²

KE = ½ (1825 lb / 32.2 ft/s²) (146.67 ft/s)²

KE = 610,000 lb ft

2. KE = ½ mv²

KE = ½ (5 kg) (14 m/s)²

KE = 490 J

6 0

3 years ago

A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the ground.

EastWind [94]

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6 0

2 years ago

Please help! I'm not sure what equation or the process to do this question.

lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy $U$ of the compressed spring is given by

$U = \dfrac{1}{2} kx^2$ ,

where $x$ is the length of compression and $k$ is the spring constant.

And the kinetic energy of the ball is

$K.E = \dfrac{1}{2}mv^2.$

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,$

solving for $v$ we get:

$v = x \sqrt{\dfrac{k}{m} }.$

And since momentum of the ball is $p=mv$ ,

$p =mx \sqrt{\dfrac{k}{m} }.$

Putting in numbers we get:

$p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.$

$\boxed{p=1.94kg\: m/s}$

5 0

2 years ago

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

Change in enthalpy is given by

$\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol$

Entropy is given by

$\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K$

Latent heat of fusion is given by

$L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol$

The latent heat of fusion is 481.76 J/mol

Melting point is given by

$T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K$

Melting occurs at 133.33 K

3 0

2 years ago