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3 years ago

All of the outer planets are much larger than the inner planets true or false

1 answer:

4 0

That statement is false because yes Jupiter and Saturn are large, however, planets like Uranus and Neptune are quite small if not smaller than some of the inner planets.

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Question 6. B) and c)

maks197457 [2]

The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds

4 0

3 years ago

Now, consider the resultant electric field e⃗ net at p. with reference to the coordinate system shown in the previous part, whic

Rzqust [24]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices:
only the x component
only the y component
both the x and y components
neither the x nor the y component

The answer is neither the x nor the y component

5 0

3 years ago

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

nlexa [21]

Answer:

Explanation:

5 0

2 years ago

All objects____ energy. Eliminate Conduct Radiate Insulate

Trava [24]

the answer is c. radiate

7 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

Given that the escape velocity for Earth, $v_e_E$ ≈ 11,186 m/s, we have;

The escape velocity on the planet = $v_e$ ≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.

3 0

3 years ago