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2 years ago

12

The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s)

of the orbit and shaded segments are the same?

1 answer:

strojnjashka [21]2 years ago

4 0

Answer: not sure

Explanation:

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A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,

Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

$v_f^2=v_o^2+2a*d$ Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

$v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}$

$v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}$

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

$15.4^2 = 30.8^2+2(-0.5)*d$

$2(0.5)*d = 30.8^2 - 15.4^2$

$d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m$

$d = 717.6 m\frac{1km}{1000m} = 0.7176Km$

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2 years ago

Make the following conversion. 275 mm = _____ cm

scZoUnD [109]

10mm=1cm So.......... 275 mm=27.5 cm

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2 years ago

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What does Einstein's equation E = mc2 explain? A. That a decrease of mass in the nucleus is what provides the energy to hold the

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D. How Much Energy Comes From A Bond Breaking.

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3 years ago

The relationship among mass force and acceleration is explained by

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Newton's second law of motion. F = m a .

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2 years ago

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A purse at radius 2.10 m and a wallet at radius 3.15 m travel in uniform circular motion on the floor of a merry-go-round as the

vlada-n [284]

Complete question is:

A purse at radius 2.10 m and a wallet at radius 3.15 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is $1.7 m/s^2 \hat i+3.20m/s^2 \hat j$ . At that instant and in unit-vector notation, what is the acceleration of the wallet?

Answer:

$a'= 2.55 m/s^2\hat i + 4.8 m/s^2 \hat j$

Explanation:

Given: Purse is at radius, $R=2.10 m$

Wallet is at radius, $R'=3.15 m$

acceleration of the purse, $a =$ $1.7 m/s^2 \hat i+3.20m/s^2 \hat j$

$a=r\omega^2$

Both the purse and wallet would have same angular velocity $\omega$ .

$\sqrt \frac{a}{R}=\sqrt \frac{a'}{R'}$

$a'=\frac{a}{R}\times R \\ \Rightarrow a' = \frac{1.7 m/s^2 \hat i+3.20m/s^2 \hat j}{2.10 m}\times 3.15 m \\ a'= 2.55 m/s^2\hat i + 4.8 m/s^2 \hat j$

5 0

2 years ago