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rewona [7]
3 years ago
8

What is a true statement about the rotation period of the moon?

Physics
1 answer:
mrs_skeptik [129]3 years ago
5 0
The time it takes for the Moon to rotate once around its axis is equal to the time it takes for the Moon to orbit once around Earth
You might be interested in
Which two elements have similar properties and 8 electrons in their<br> outermost shells?
statuscvo [17]

Answer:

A. Neon and argon

General Formulas and Concepts:

  • Reading the Periodic Table of Elements
  • Valence Shell configurations

Explanation:

Elements that have a full valence shell (the outermost shell), are in Group 18 of the Periodic Table.

Because their outermost shell is full, they are inert, meaning that they do not like to bond or react with other elements.

Therefore, these gases with a stable full valence shell are known as Noble Gases.

  1. Helium
  2. Neon
  3. Argon
  4. Krypton
  5. Xenon
  6. Radon

A. would be the correct answer.

5 0
3 years ago
Find the magnitude of the electric field at a point midway between two charges +13.6 x 10-9 C and +61.0 x 109 C separated by a d
Ostrovityanka [42]

Answer:

Ep =  3797.05  N/C  in the direction leaving the charge q₂ towards point P

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Data

k=  8.99*10⁹ N*m²/C²

q₁ = +13.6 x 10⁻⁹C

q₂ = +61.0*10⁻⁹C

d₁ =d₂= 33.5 cm = 0.335 m

Look at the attached graphic:

E₁: Electric Field at point  P due to charge q₁. As the charge  q₁ is positive   (q₁+) ,the field leaves the charge

E₂: Electric Field at point  P due to charge q₂. As the charge q₂ is positive (q₂+) ,the field leaves the charge

E₁ = k*q₁/d₁² = 8.99*10⁹ *13.6 *10⁻⁹/(0.335)² = 1089.45 N/C

E₂ = k*q₂/d₂²=- 8.99*10⁹ *61*10⁻⁹/(0.335)² = - 4886.5   N/C

Magnitude of the electric field at a point midway between q₁ and q₂

Ep= - 4886.5+ 1089.45 = -3797.05  N/C

Ep =  3797.5  N/C  in the direction leaving the charge q₂ towards point P

6 0
3 years ago
Find the magnitude of the gravitational force a 68.4 kg person would experience while standing on the surface of earth with a ma
luda_lava [24]
Given: Mass of a person M = 68.4 Kg.

           Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

           G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: F = ?

Formula: F = GMeM/r²

               F = (6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴)(68.4 Kg)/(6.37 x 10⁶ m)²

               F = 672.41 N




6 0
4 years ago
Trafegando por uma avenida com velocidade constante de 54km/h, num dado instante, o motorista percebe o sinal vermelho à sua fre
Furkat [3]

Answer:

The acceleration is - 3 m/s2.

Explanation:

initial speed, u = 54 km/h = 15 m/s

final speed, v = 0

time, t = 5 s

Let the acceleration is a.

use first equation of motion

v = u + at

0 = 15 + a x 5

a = - 3 m/s2

5 0
3 years ago
Can someone help me convert these?
fomenos

Assuming you are supposed to write each conversion in scientific notation:

(2) 1 m = 100 cm, so

(67 cm) × (1/100 m/cm) = 67/100 m = 0.67 m = 6.7 × 10 ⁻¹ m

(3) 1 km = 1,000 m, so

(1.2 km) × (1000 m/km) = 1200 m = 1.2 × 10³ m

(4) 1 m = 1,000 mm = 10³ mm, so

(6.2 × 10 ⁻³ m) × (10³ mm/m) = 6.2 mm

(5) 1 m = 1,000,000,000 nm = 10⁹ nm, so

(4.05 × 10³ nm) × (1/10⁹ m/nm) = 4.05 × 10 ⁻⁶ m

(6) 1 g = 1,000,000 µg = 10⁶ µg, so

(3200 µg) × (1/10⁶ g/µg) = 3200 × 10 ⁻⁶ g = 3.2 × 10 ⁻³ g

4 0
3 years ago
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