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aleksklad [387]
2 years ago
11

question 18(multiple choice worth 1 points) (01.07 mc) the distance of planet jupiter from the sun is approximately 7.8 ⋅ 108 ki

lometers, and the distance of planet saturn from the sun is 1.5 ⋅ 109 kilometers. about how many more kilometers is the distance of saturn from the sun than the distance of jupiter from the sun? 6.3 ⋅ 101 kilometers 6.3 ⋅ 108 kilometers 7.2 ⋅ 108 kilometers 7.2 ⋅ 109 kilometers
Physics
1 answer:
Vlad [161]2 years ago
3 0

The distance of Saturn from the sun is more than the distance of Jupiter from the sun by:

C. 7.2 x 10⁸ kilometers

<h3>How is the difference in distance calculated?</h3>

The difference in distance is correctly obtained by subtracting the distance from Jupiter to the sun from the distance of Saturn from the sun.

Now, the distance of Saturn from the sun is:

1.5 x 10⁹ kilometers

Also, the distance of Jupiter from the sun is:

7.8 x 10⁸ kilometers

So, we subtract the smaller figure from the bigger figure to have:

1.5 x 10⁹ kilometers ₋ 7.8 x 10⁸ kilometers

= 15 x 10⁸ kilometers - 7.8 x 10⁸ kilometers

=  7.2 x 10⁸ kilometers

So, option C is right.

Learn more about the distance of the planets here:

brainly.com/question/11139157

#SPJ4

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Dafna1 [17]

Answer:

(a) Tangential velocity will be 38.648 m/sec

(b) Acceleration will be 133.617m/sec^2

Explanation:

We have given radius r = 11.2 m

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(a) We have to find the tangential velocity

We know that tangential velocity is given by  

v_t=\omega r=3.454\times 11.2=38.684m/sec

(b) We know that acceleration is given by

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8 0
3 years ago
Please Help!!! I WILL GIVE BRAINLIEST!!!! An electron is in motion at 4.0 × 10^6 m/s horizontally when it enters a region of spa
dybincka [34]

Answer:

a. 9.52 cm b. 4.34 × 10⁶ m/s

Explanation:

a. The horizontal distance traveled by the electron when it hits the plate.

The electric force F on the electron due to the electric field E of mass, m is

F = -eE = ma

a = -eE/m where a = acceleration of electron

The vertical distance moved by the electron is given by

Δy = ut +1/2at²

u = initial vertical velocity = 0. and take the top plate as y = 0 and bottom plate as y

So,

0 - y = 0 × t + 1/2at²

-y = 1/2at²

substituting a = -eE/m

-y = 1/2(-eE/m)t²

y = eEt²/2m

making t subject of the formula,

t = √(2ym/eE) where t is the time it takes to reach the bottom plate.

Since E = 4.0 × 10² N/C, y = distance between plates = 2.0 cm = 0.02 m, m = 9.109 × 10⁻³¹kg and e = 1.602 × 10⁻¹⁹ C

t = √[(2 × 0.02 m × 9.109 × 10⁻³¹kg)/(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C)]

t =  √[(0.36436 × 10⁻³¹kgm)/(6.408 × 10⁻¹⁷ N)]

t = √[(0.0569 × 10⁻¹⁴kgm/N)t

t = 0.238 × 10⁻⁷ s

t = 23.8 × 10⁻⁹ s

t = 23.8 ns

The horizontal distance moved when it hits the plates x = vt where v = initial horizontal velocity = 4.0 × 10⁶ m/s

x = 4.0 × 10⁶ m/s × 23.8 × 10⁻⁹ s

= 0.0952 m

= 9.52 cm

b. The velocity of the electron as it strikes the plate.

To find the velocity of the electron as it strikes the plates, we calculate its final vertical velocity V as it strikes the plate. This is gotten from

v' = u + at since u = 0,

v' = at

= -eEt/m

= -(1.602 × 10⁻¹⁹ C × 4.0 × 10² N/C × 0.238 × 10⁻⁷ s)/9.109 × 10⁻³¹kg

= -1.525 × 10⁻²⁴ Ns/9.109 × 10⁻³¹kg

= -0.167 × 10⁷ m/s

= -1.67 × 10⁶ m/s

So, the resultant velocity as it strikes the plate v = √(v'² + v²)

= √((-1.67 × 10⁶ m/s)² + (4 × 10⁶ m/s)²)

= √(2.7889  + 16) × 10⁶ m/s

= √18.7889 × 10⁶ m/s

= 4.335 × 10⁶ m/s

≅ 4.34 × 10⁶ m/s

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v = 0.72 x 1.70        

v = 1.224 m/s          

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