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3 years ago

A wave travelling along the positive x-axis side with a

Physics

1 answer:

andrew11 [14]3 years ago

5 0

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

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Compared to its weight on Earth, a 10-kg object on the moon will weigh Question 9 options: less. the same amount. more.

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it will weigh less on the moon.

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3 years ago

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A block of aluminum had a volume of 17.0 mL and a mass of 45.9g. What is it’s density?

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The density of aluminum is 2.7 g/ml.

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3 years ago

Which change would create light?

hichkok12 [17]

Answer:

Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.

Explanation:

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3 years ago

An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea

german

Answer:

Q1) Time taking by particle to travel the 40 km wrt. earth = $1.34\times10^{-6}$ sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = $1.285\times10^{-5}$ sec.

Explanation:

Given :

Speed of particle wrt. earth $v=0.99537c$

Distance between where particle is created and earth surface = 40 km

we know that,

⇒ $v = \frac{x}{t}$

Where $x = 40\times10^{3} m$ , $v = 0.99537c$ , we know speed of light $c = 3 \times10^{8}$

∴ $t = \frac{x}{v}$

$= \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }$

$t = 1.34\times10^{-6} sec$

∴ Thus, time taking by particle to travel the 40 km wrt. earth $t = 1.34\times10^{-6} sec$

According to the lorentz transformation,

⇒ $l = l_{o} \sqrt{1-\frac{v^2}{c^2} }$

Where $l =$ improper length, $l_{o} =$ proper length (distance measured wrt. rest frame) = 40 km

$l = 40 \sqrt{1-\frac{v^2}{c^2 }$

$l = 40 \times 0.096$

$l = 3.84$ km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

$\Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }$

Where $\Delta t =$ improper time (wrt. earth frame time) $=1.34\times10^{-6} sec$ , $\Delta t _{o} =$ proper time (wrt. particle frame).

$1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}$

$\Delta t_{o} = 1.285 \times10^{-5} sec$

Thus, the time taking by particle to travel from where it is created to the surface of the earth $= 1.285 \times10^{-5} sec$ .

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3 years ago

A grasshopper leaps into the air at a 62° angle above the horizontal, and follows a parabolic arc in free fall after it leaves t

harina [27]

Answer:

d. None of the above.

Explanation:

In a parabolic motion, you have that in the complete trajectory the component velocity is constant and the vertical component changes in time. Then, the total velocity vector is not zero.

In the complete trajectory the gravitational acceleration is always present. Then, the grasshopper's acceleration vector is not zero.

At the top of the arc the grasshopper is not at equilibrium because the gravitational force is constantly acting on the grasshopper.

Then, the correct answer is:

d. None of the above.

5 0

3 years ago