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creativ13 [48]
3 years ago
12

What does it mean if the moon is waxing and waning

Physics
1 answer:
konstantin123 [22]3 years ago
6 0
Both verbs come from Olde English. 
That's why everybody clearly understood their meaning until
a hundred years ago, but nobody understands them now.


"Waxing"  =  growing

For two weeks after the New Moon, it's growing toward Full.
First it's a waxing crescent for a week, then it's waxing gibbous.


"Waning"  =  shrinking

For two weeks after the Full Moon, it's shrinking toward New.
First it's waning gibbous for a week, then it's a waning crescent.
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Block A is also connected to a horizontally-mounted spring with a spring constant of 281 J/m2. What is the angular frequency (in
Svetradugi [14.3K]

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, the formula to be used here is

ω = 2π/T

Where ω is the angular frequency (in rad/s)

T is the period - the time taken for Block A to complete one oscillation and return to it's original position.

To solve for this period T, the formula below should be used

T = 2π√m/k

where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)

5 0
3 years ago
Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more
umka2103 [35]

Answer:

None.

Explanation:

  • Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
  • As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

       \Delta U = U_{f} -U_{0} = m*g*h - 0  = m*g*h

  • As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.
6 0
3 years ago
An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.
Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }

B = \boxed{7.07 *10^{-10} T}

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

\boxed{B = 0 T}

3 0
2 years ago
A .2kg Basketball is pitched with a velocity or 40 m/s and then bed and into the picture with a velocity of 60 m/s. What is the
Tamiku [17]

Answer:

40kgm

Explanation:

∆p = m(v - u)

= 2(60 - 40)

= 2 × 20

= 40kgm/s

4 0
3 years ago
Newton's first law of motion states that an object will keep a constant speed and direction unless acted upon by an unbalanced f
trasher [3.6K]

Answer:

The unbalanced force that caused the ball to stop was friction

Explanation:

As Newton's second law states, the acceleration of an object is proportional to the net force applied on the object:

F=ma

therefore, in order to move at constant speed, an object should have a net force of zero (balanced forces) acting on it.

In this case, the ball slows down and eventually comes to a stop: it means that the ball is decelerating, so there are unbalanced forces (net force different from zero) acting on it. The unbalanced force acting on the ball is the friction: friction is a force against the motion of the object, which is due to the contact between the surface of the ball and the surface of the street, and this force is responsible for slowing down the ball.

3 0
3 years ago
Read 2 more answers
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