What replaces a cold current that sinks to the ocean floor?

Which is a characteristic of projectile motion?

<h2>Right answer: It follows a curved path </h2>

The movement of a projectile is a movement in two dimensions (forming a curved path: a parabola shape) with constant acceleration.

A projectile is any body or object that is thrown or projected by means of some force and continues in motion by its own inertia. This means the only force that acts on it while in motion is the acceleration of gravity (in this case we are on Earth, so the gravity value is $9.8\frac{m}{s^{2}}$ ).

Where gravity influences the vertical movement of the projectile, while the horizontal movement of the projectile is the result of the tendency of any object to remain in motion at a constant speed (according to Newton's 1st law of motion sometimes called Law of Inertia).

The other options are incorrect because are false:

-The forward motion negates air resistance: There is always at least a small percent of air resistance, as long as that movement is done on Earth.

-It has variable acceleration: In projectile motion acceleration is constant (gravity acceleration) .

-It is unaffected by gravity: The only force that acts on the projectile is due gravity.

5 0

3 years ago

Read 2 more answers

Suppose you increase your walking speed from 6 m/s to 13 m/s in a period of 1 s. What is your acceleration? m/s2

ziro4ka [17]

Average acceleration is
Change in Velocity/change in time
So you could then do Vf-Vi/Tf-Ti
Which would look like 13m/s-6m/s / 1s-0s
Which then is 7m/s/1s which means the acceleration is 7m/s^2

5 0

3 years ago

A 20 g bullet is shot from a 10 kg gun with a velocity of 400 m/s. What

Nady [450]

Answer: -.80m/s

Explanation:

3 0

3 years ago

How long will it take a ball to roll 10 meters along the floor at a speed of 0.5m/s

Julli [10]

Answer:

5s

Explanation:

because m/s=s/t so,

0.5m/s=10m/t

t=10m/0.5m/s

t=5sec

5 0

1 year 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