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3 years ago

6. A car moving in a straight line at constant speed:

Physics

1 answer:

ddd [48]3 years ago

8 0

Answer:

D. has no overall force acting on it.

Explanation:

Why?

Because in a straight line at the constant speed means the car moving in the same velocity, which is not acceleration neither deceleration, and it cannot be on a downhill slope. So the correct answer is

<h3>→ D. has no overall force acting on it.</h3>

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A car traveling at 21 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

diamong [38]

The formula for acceleration is the velocity times the inverse of time so it would be 21 times 1/13. So roughly 0.0769... is the acceleration(m/s^2).

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3 years ago

Que propiedad de la Luz se produce cuando ves tu cara reflejada en la cuchara?

kramer

Answer:

refracción

Explanation:

I don't speak spanish but that's the answer

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3 years ago

When flying the LNAV approach, the missed approach point (MAP) would be indicated by reaching __________. A. an altitude of 3100

frutty [35]

Answer:

When flying the LNAV Approach, the missed approach point (MAP) would be indicated by reaching:

C. the RW30 waypoint.

Explanation:

In Aviation, LNAV stands for Lateral Navigation. The option a is incorrect as an altitude of 3100 feet refers to the decision altitude not the missed approach point.
The option b is incorrect as a distance of 1.5 NM to RW30 referring to the Visual descent point (VDP) is 1.5 nautical miles for the Runway (RW) 30 from threshold.
The option c is correct as Missed approach point is designed to coincide with the runway threshold. The RW 30 way point is referring to the way point to the threshold for the Runway 30.

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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.

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

One disadvantage of using proprietary licensed software is that

Vaselesa [24]

Answer:

its b

Explanation:

5 0

3 years ago

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