Ask question

Ask question

All categories

LuckyWell [14K]

3 years ago

12

Begin any simulation, and turn on Gravity Force in the central menu on the right. The gravity force arrow shows the direction an

d strength of the gravitational force each body feels. How do the gravity force arrows change throughout the orbit?

1 answer:

Katyanochek1 [597]3 years ago

3 0

Answer:

The gravity arrow for each body rotates, always pointing toward the other body. Both arrows grow longer when the bodies come closer to one another and shorter when they move farther apart. This change shows that the gravitational force is stronger the closer together the bodies are.

Explanation:

You might be interested in

current is passed through two parallel conductors in the same direction. If the conductors are placed near each other,they will

svlad2 [7]

If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will attract each other.

<h3>What is electric current?</h3>

Electric current can be defined as the flow of electrons.

Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.

Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.

On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.

The flow of current is known as electricity.

Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.

In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.

Learn more about parallel conductors at: brainly.com/question/17148082

#SPJ1

6 0

2 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago

Which "spheres" are interacting when water evaporates from plants

Novay_Z [31]

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

8 0

3 years ago

Read 2 more answers

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the

SOVA2 [1]

Answer:

Johnny created an electromagnet out of a solenoid (a coil of wires with 20 loops), an iron core (of 1 nail), and a single 9 V battery. When Johnny does this, he creates a small magnetic field that allows him to pick up 2 paper clips. Using a CER format, explain to Johnny three things he could change that would increase the strength of his magnetic field and why each change increases the magnetic field. You may want to write three paragraphs to make this easier for the reader to understand

5 0

3 years ago