Ask question

Ask question

All categories

2 years ago

9

A new moon is discovered orbiting Neptune with an orbital speed of 9.3 x103 m/s. Neptune's mass is 1.0 x1026 kg. A) What is the

radius of the new moon's orbit? B) What is the orbital period? Assume that the orbit is circular. (G = 6.673 x10-11 N-m²/kg?

1 answer:

AysviL [449]2 years ago

6 0

The radius of the new moons orbit is R= 7.715 x 10^7 m
The orbital period of the moon is T= 14.48 hr

You might be interested in

8) The radius of the moon is approximately 1,350,000 m. What is the radius in centimeters? (100cm = 1m)

snow_tiger [21]

Answer:

a

Explanation:

a

6 0

3 years ago

An arteriole has a radius of 25 μm and it is 1000 μm long. The viscosity of blood is 3 x 10-3 Pa s and its density is 1.055 g cm

timama [110]

Answer: $1.955(10)^{13} \frac{Pa.s}{m^{3}}$

Explanation:

This can be solved by the Poiseuille’s law for a laminar flow:

$R=\frac{8 \eta L}{\pi r^{4}}$

Where:

$R$ is the resistance of the arteriole

$\eta=3(10)^{-3} Pa.s$ is the viscosity of blood

$L=1000 \mu m=1000(10)^{-6}m$ is the length of the arteriole

$r=25 \mu m=25(10)^{-6}m$ is the radius of the arteriole

$R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}$

$R=1.955(10)^{13} \frac{Pa.s}{m^{3}}$

3 0

3 years ago

A uniform electric field of magnitude 4.6 ✕ 104 N/C is perpendicular to a square sheet with sides 3.0 m long. What is the electr

rodikova [14]

Answer:

41.4* 10^4 N.m^2/C

Explanation:

given:

E= 4.6 * 10^4 N/C

electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field

then electric flux = ∫ E*n dA

= ∫ 4.6 * 10^4 * 3*3

= 41.4* 10^4 N.m^2/C

3 0

4 years ago

Read 2 more answers

The cylindrical tub of a dryer in a laundromat rotates counterclockwise about a horizontal axis at 41.5 rev/min as it dries the

frozen [14]

Answer:

$\theta = 49.81^0$

Explanation:

Given that:

$\omega = 41.5 \ rev/min\\\\\omega = 41.5 *\frac{1}{60}* 2 \pi\\\\\omega = 4.45 \ rad/s\\\\\\diameter = 0.748 m$

If we let the piece of the close lose contact at ∠θ;

Then ; from force balance;

we have:

$\\\\mg sin \theta = \frac{mv^2}{r}\\\\sin \theta = \frac{2v^2}{dg}\\\\\theta = sin^{-1} (\frac{2v^2}{dg})$

where;

$v = \frac{\omega d}{2}\\\\v = \frac{4.45 *0.748}{2}\\\\v = 1.6643\\\\v^2 = 2.77$

Again:

$\theta = sin^{-1}(\frac{2v^2}{dg})\\\\\theta = sin^{-1}( \frac{2*2.77}{0.74*9.8})\\\\\theta = 49.81^0$

6 0

3 years ago

Bicycle A with mass 40 Kg is traveling with a velocity of 4 m/s and Bicycle B with mass 20 Kg is traveling with a velocity of 2m

andrew-mc [135]

Answer:

bicycle A has a greater K.E.

Explanation:

K.E = 1/2mv²

bicycle A = 1/2 × 40 × 4² = 320J

bicycle B = 1/2 × 20 × 2² = 100J

bicycle A has a greater K.E. because it has bigger mass and moves with faster velocity

7 0

3 years ago