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3 years ago

Kepler found that the orbits of the planets were _____. circular amorphous oval elliptical

2 answers:

7 0

<span>Kepler found that the orbits of the planets were elliptical.
His work was so convincing that to this day, they still are. </span>

Dafna1 [17]3 years ago

3 0

Answer:

elliptical

Explanation:

Nicolaus Copernicus gave the heliocentric theory that sun is at the center if the solar system and all the planets and the moons revolved wound it. according to him, the planets revolved around the sun in circular paths.

But Kepler suggested a different shape of the orbit. According to the laws o planetary motion given by Kepler, the planets revolve around the sun in an elliptical orbit. The sun is at one of its focus.

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If eight water waves pass an ocean buoy each minute, and successive wave crests are 20 m apart, find the wave speed:____________

Llana [10]

Answer:

The wave speed is calculated below:

Explanation:

Given,

number of waves passed per minute = 8

time period = 1 minute = 60 s

distance between successive wave crests = 20 m

waves passing interval per second = $\frac{8}{60}$ $s^{-1}$

Now,

wave speed = 20 m × $\frac{8}{60}$ $s^{-1}$

= $\frac{8}3}$ m/s

= 2.67 m/s

Hence the wave speed is 2.67 m/s.

4 0

3 years ago

A proton, traveling with a velocity of 3.7 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 5.4

Vikki [24]

The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
$F=qvBsin\theta=qvB$
since $\theta = 90^{\circ}$ , because the velocity v and the force F in this problem are perpendicular, and so also the angle $\theta$ between the velocity and the magnetic field B should be $90^{\circ}$ .

Let's find the magnitude of the magnetic field; this is given by
$B= \frac{F}{qv}= \frac{5.4\cdot 10^{-14}N}{1.6\cdot 10^{-19}C \cdot 3.7\cdot 10^6 m/s}=0.091 T$

To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force

Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T

8 0

3 years ago

Alex has a 100 cm ruler pivoted at the centre. She ties a balloon filled with carbon doxide 16 cm from the pivot The total weigh

Leona [35]

If people send links report them

5 0

3 years ago

The gravitational force of the Earth exerts no torque on a satellite orbiting the Earth in an elliptical orbit. Compare the moti

stealth61 [152]

Answer:

3) None of the above

Explanation:

In orbital motion

by lepler law:

Planet sweeps out equal area in equal of time, so angular velocity is conserved.

Angular momentum is also conserved,

so option, 3) None of the above is right.

4 0

3 years ago

A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per m

Novosadov [1.4K]

Hi there!

Begin by using Gauss' Law to find the electric field.

$\oint {E \cdot} \, dA = \frac{Q_{encl}}{\epsilon_0}$

E = Electric field (N/C)
dA = differential area element

Q = enclosed charge (C)

ε₀ = Permittivity of free space (8.85 * 10⁻¹² C²/Nm²)

We can construct a large cylinder around the wire in order to determine the electric flux. The electric field lines will pass through the LATERAL surface area of the cylinder, so:
$A = 2\pi rL$

Where 'L' is the length of the cylinder and 'r' is the distance from the capacitor.

The enclosed charge is equivalent to the charge per meter length (λ) multiplied by the length, so:
$\oint {E \cdot} \, dA = \frac{\lambda L}{\epsilon_0}$

We can rewrite the dot product as EA (where cosθ = 1 since the normal vector points in the direction of the field).

A = the lateral surface area of a cylinder, so:

$E * 2\pi rL = \frac{\lambda L}{\epsilon_0}$

Rearrange to solve for 'E'.

$E = \frac{\lambda L }{2\pi r L \epsilon _0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

a)
Plug in the distance into 'r'.

$E = \frac{\lambda }{2\pi r \epsilon_0} \\\\E = \frac{0.0000072}{2\pi * 5.5 * (8.85 * 10^{-12})} = \boxed{23542.18 \frac{N}{C}}$

b)
Repeat:
$E = \frac{\lambda }{2\pi r \epsilon_0}\\\\E = \frac{0.0000072}{2\pi * 2.5 * (8.85 * 10^{-12})} = \boxed{51792.8 \frac{N}{C}}$

We can see that the distance from the wire is INVERSELY related to the electric field strength by a power of r⁻¹. The field strength DECREASES as the distance INCREASES.

4 0

2 years ago