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3 years ago

(SHOULDN'T BE HARD!!!!) How does gravity impact two objects in space?

1 answer:

8 0

Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies. It keeps planets in orbit.

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A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance

Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

$v=u+at$

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

t = 5 s is the time

So, the final velocity is

$v=0+(2)(5)=10 m/s$

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

$v^2-u^2=2as$

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

$a=2 m/s^2$ is the acceleration

v = 10 m/s is the final velocity

Solving for s,

$s=\frac{10^2-0^2}{2(2)}=25 m$

3 0

3 years ago

a ball is thrown horizontally from the roof of a building 45m tall and lands 24m from the base what is the initial speed?

Irina-Kira [14]

Its initial speed is 7.92 m/s

6 0

3 years ago

What is the final speed?

EastWind [94]

U=6.9ms-¹
a=0.62ms-²
t=3.4s
V=?
using
v=u+at
v=6.9+(0.62×3.4)
v= 6.9+2.108
v=9.008ms-¹

6 0

3 years ago

A pelican flying drops a fish from a height of 8.1 m. The fish travels 9.3 m horizontally before it hits the ground. The pelican

masya89 [10]

Using kinematics: 8.1 = 1/2(9.8)(t^2),
t = 1.2857 s.
Horizontal distance x travelled is x = vhorizontal * t, so 9.3 = v*1.2857, or v= 7.233 m/s horizontally.

7 0

4 years ago