Compare the mass on Neptune (68.11) to the weight on Neptune (905.863). What is the difference? Why is there a difference?

I have a problem in this questions?

photoshop1234 [79]

Answer:

8.46E+1

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = 39 C

Charge 2 (q₂) = –53 C

Force (F) of attraction = 26×10⁸ N

Electrical constant K) = 9×10⁹ Nm²/C²

Distance apart (r) =?

The distance between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

26×10⁸ = 9×10⁹ × 39 × 53 / r²

26×10⁸ = 1.8603×10¹³ / r²

Cross multiply

26×10⁸ × r² = 1.8603×10¹³

Divide both side by 26×10⁸

r² = 1.8603×10¹³ / 26×10⁸

r² = 7155

Take the square root of both side

r = √7155

r = 84.6 m

r = 8.46E+1 m

7 0

3 years ago

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

7 0

3 years ago

Which three factors are used to calculate gravitational potential energy?

NeX [460]

Answer:

Height, mass, acceleration.

Explanation:

I hope it helps u dear! ^_^

7 0

2 years ago

(ASAP) would it be 125 m/s2 to calculate for her speeding up?

serg [7]

Answer:

$0\:\mathrm{ m/s^2}$

Explanation:

Recall the formula for acceleration:

$\displaystyle\\a=\frac{v_f-v_i}{\Delta t}$ , where $v_f$ is final velocity, $v_i$ is initial velocity, and $\Delta t$ is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

$\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}$

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0

2 years ago

A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity