Shelby travels 6 meters East, 2 meters North, 3 meters West, and then 2 meters South. What is their displacement?

What is a circumpolar star? A)a star that is close to the north celestial pole a star that is close to the south celestial pole

Vedmedyk [2.9K]

Answer:

A star that always remains above your horizon and appears to rotate around the celestial pole.

Explanation:

A) a star that is close to the north celestial pole: a circumpolar star could be close to the north celestial pole, but this answer is omitting the south celestial pole.

B) a star that is close to the south celestial pole: a circumpolar star could be close to the south celestial pole, but this answer is omitting the north celestial pole.

C) a star that always remains above your horizon and appears to rotate around the celestial pole: this is the definition of a circumpolar star.

D) a star that makes a daily circle around the celestial sphere: every star does this.

E) a star that is visible from the Arctic or Antarctic circles : there are many starts visible from there that are not circumpolar.

6 0

3 years ago

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w

REY [17]

Answer:

0.7515875 eV

$4\times 10^{14}\leq f$

Explanation:

f = Maximum frequency = $7.9\times 10^{14}\ Hz$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

W = Work function = 2.52 eV

Converting to Joules

$W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J$

Maximum photon energy is given by

$E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J$

Maximum Kinetic energy is given by

$K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J$

Converting to eV

$1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV$

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

$W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz$

The range of frequencies for which no electrons are ejected is

$4\times 10^{14}\leq f$

5 0

3 years ago

How to find angular velocity from linear velocity and radius.

Lilit [14]

Answer:

ω=v/r.

Explanation:

angular velocity= linear velocity/radius

6 0

2 years ago

A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget

avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

7 0

3 years ago

An object moving 20 m/s

yan [13]

Answer:

We are given:

initial velocity (u) = 20m/s

acceleration (a) = 4 m/s²

time (t) = 8 seconds

displacement (s) = s m

Solving for Displacement:

From the seconds equation of motion:

s = ut + 1/2 * at²

replacing the variables

s = 20(8) + 1/2 * (4)*(8)*(8)

s = 160 + 128

s = 288 m

7 0

3 years ago