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3 years ago

Precipitation is greater over what and less over what? A. Land, sea B. Sea, land

1 answer:

3 0

I would say letter A to this because we need rain water for crops plants. Also we need it so that we can drink it. Its less great over sea because as it merge with the salt in the ocean it becomes less drinkable and less usable.

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Move the red arrow onto the picture to show how you think warm air would move over this area of land and water on a dark cold ni

Bess [88]

There’s no pictures?

7 0

2 years ago

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The object moves with BLANK from A to B. it BLANK from B to C. it moves with BLANK from C to D.

Ganezh [65]

Note that this is a position vs time graph.

From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.

From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.

4 0

2 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

Three quarterbacks challenged each other to a throwing competition. They want to see who can throw the ball the furthest. What S

bazaltina [42]

Answer:

Meter

Explanation:

The competition between the three quarterbacks is with respect to how far the ball would be thrown by each person, which is the distance covered by the ball. The thrown ball is an example of projectile, which would move over a certain distance.

With respect to the measure to be used in the competition, the appropriate SI unit is meter. This is the measure of length or distance covered.

5 0

3 years ago

Which change indicates that the universe is expanding?

Aleonysh [2.5K]

Red shift of distant galaxies

5 0

3 years ago

Read 2 more answers