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ratelena [41]
3 years ago
11

During a new moon, is the moon A) in between Earth and the sun B) in front of the sun, in back of Earth C) in front of Earth, in

front of the sun D) alongside Earth in front of the sun
Physics
2 answers:
Ira Lisetskai [31]3 years ago
5 0
B. in front of the sun, in back of earth
Lady_Fox [76]3 years ago
4 0

During a new moon, the moon is in between the earth and the sun

Answer: Option A

<u>Explanation:</u>

New moon day also known as solar eclipse that occurs when the moon is located  in between the sun and the earth. In that phase, the sun, moon and earth are in alignment.

Usually they cannot be seen from earth with our naked eyes. Here the sun and the earth will be facing the opposite sides. Hence option B.C and D are irrelevant.

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Communications satellites are placed in orbits so that they always remain above the same point of the earth's surface.A Part com
masya89 [10]

Answer:

24 hours

7\times 10^{-5}\ rad/s

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

T=24\times 60\times 60\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s

The angular velocity of the satellite is 7\times 10^{-5}\ rad/s

3 0
3 years ago
Why is a noble gas different from other elements
fredd [130]
Because noble gases has full of 8 electron in her external cell......
7 0
3 years ago
Read 2 more answers
A yoyo with a mass of m = 150 g is released from rest as shown in the figure.
avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The  weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

  • I is moment of inertia
  • α is angular acceleration
  • T is tension in the rope
  • r is inner radius
  • R is outer radius
  • f is frictional force

rT - Rf = Iα  ----- (1)

T - f = Ma  -------- (2)

a = Rα

where;

  • a is the linear acceleration of the yoyo

Torque equation for frictional force;

f = (\frac{r}{R} T) - (\frac{I}{R^2} )a

solve (1) and (2)

a = \frac{TR(R - r)}{I + MR^2}

since the yoyo is pulled in vertical direction, T = mg a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

3 0
2 years ago
Five lamp, each labbled "6V,3W" are operated at normal brightness. What is the total energy supplied to the lamps in five second
REY [17]

Answer:

E = 75 J

Explanation:

First, we will calculate the total power consumed by the five lamps:

Total\ Power = P = (5)(Power\ of\ one\ lamp)\\P = (5)(3\ W)\\P = 15\ W

Now, the energy supply can be calculated as follows:

E = Pt

where,

E = Energy = ?

t = time = 5 s

Therefore,

E = (15 W)(5 s)

<u>E = 75 J</u>

8 0
2 years ago
A body of mass 20kg initially at rest is subjected to a force of 40m for 15sec. Calculate the change in k.e
Ede4ka [16]

Explanation:

mass(m)=20kg

velocity(v)=d/t=2.67

k.e=?

now,

k.e=1/2mv^2

=1/2*20*(2.67)^2

=71J

3 0
2 years ago
Read 2 more answers
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