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3 years ago

6

Gina's teacher says that Jupiter has a much greater gravitational force than Earth, while the Moon has a much smaller force. How

would Gina's weight and mass change on these surfaces?

1 answer:

belka [17]3 years ago

8 0

On jupiter there more of a gravitational pull, pulling you closer to the "center" making you weight more.

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Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomical o

Vlada [557]

Answer:

The answer is "telescopes".

Explanation:

Throughout ancient times, astronomical observatories have indeed been available, and so many historical locations were reserved for astronomical observations. All contemporary astronomers lacked within those older telescopes were lenses until 1610. A telescope is indeed an instrument used to view far-off objects. Telescopes often are being used to look at planets and stars.

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3 years ago

A group of atoms with aligned magnetic poles are known as which of the following?

cestrela7 [59]

A magnetic domain is a group of atoms aligns with magnetic poles. Domains are usually <span>light and dark stripes visible within each grain.</span>

4 0

3 years ago

Read 2 more answers

In an arcade game, a 0.126 kg disk is shot across a frictionless horizontal surface by compressing it against a spring and relea

bearhunter [10]

Answer:

4.156 m/s

Explanation:

See attachment

4 0

3 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

3 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

7 0

2 years ago