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3 years ago

The orbital period of an object is 2 x 10^7 and its total radius is 4 x 10^4 m.

Physics

2 answers:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

12566

Explanation:

Use formula: vt=2pir/t

Ostrovityanka [42]3 years ago

7 0

Answer:

12,566 not 12560 because edgenuity doesnt want you to round by the tenths only by the ones

Explanation:

Because 2* pi * 4x10^10 = 251327412400 (Dont round the pi to 3.14)

and divide it by 2x10^7

you get 12,566.37 which its 12,566 (and dont add punctuation, stated on the quiz)

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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 $= 1.285 \times10^{-5} sec$ .

6 0

3 years ago

A motorcyclist drives around a bend with a 20 m radius, with a constant velocity of 3 m/s. The motorcyclist and the motorcycle h

Pavel [41]

Answer:

$a=0.45\ m/s^2$

Explanation:

Given that,

The radius of a bend, r = 20 m

Velocity of motorcyclist, v = 3 m/s

The combined mass of motorcyclist and the motorcycle is 50 kg

We need to find the motorcyclist’s centripetal acceleration. The formula used to find the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(3)^2}{20}\\\\a=0.45\ m/s^2$

So, the acceleration of the motorcyclist is $0.45\ m/s^2$ .

3 0

3 years ago

What is the net force required to accelerate a 16 kg box at a rate or 1.4 m/s^2

stich3 [128]

You can use Newton's Second Law which states:
$\sum \vec{F} = m\vec{a}$

Plug in given information:
$\sum \vec{F}=16(1.4) = 22.4N$

This is closest to option b which is your answer.

8 0

3 years ago

A 2kg mass is moving at 3m/s. What is its kinetic energy?

Illusion [34]

<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>

$\color{hotpink} \bold{9 \: J} \\$

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

$\underline{ \boxed{ \tt KE = \frac{1}{2} m{v}^{2} } }$

where,

v is the velocity in m/s
KE is the kinetic energy in J (joules)
m is the mass in kg

— — —

Based on the problem, the givens are:

KE (Kinetic energy) = ? (unknown)
m (mass) = 2 kg
v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

$\: \: \tt KE = \frac{1}{2} m{v}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt \: KE = \frac{1}{2} \times 2 \times {(3)}^{2} \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: KE = \frac{1}{2} \times(2\times 9) \\ \tt KE = \underline{ \boxed{ \blue{ \tt9 \: J}}}$

Therefore, the kinetic energy is 9 Joules.

3 0

2 years ago

In tenis does your knees have to be slightly flexed for both the forehand and backhand?

Elena L [17]

Answer:

yes

Explanation:

8 0

3 years ago