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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS AND I NEED ALL CORRECT ANSWERS ONLY!!!

Physics

2 answers:

Mrrafil [7]3 years ago

8 0

This statement Doorbells, scrap yards, and rock concerts are all examples of every day uses of electromagnets, is correct or true.

Hope this helps chu

Have a great day ♡♡

Lemur [1.5K]3 years ago

3 0

I believe the answer is true. Hope this helped!

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What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth

Tanya [424]

Answer:

$r = 3.84 \times 10^7 m$

Explanation:

Let say the distance at which gravitational force due to both Earth and moon is zero is given by force force balance

so it is given as

$F_{moon} = F_{Earth}$

$\frac{GM_{moon}}{r^2} = \frac{GM_{earth}}{(d - r)^2}$

so we have

$\frac{M_{moon}}{r^2} = \frac{M_{earth}}{(d - r)^2}$

$\frac{7.35 \times 10^{22}}{r^2} = \frac{5.97 \times 10^{24}}{(3.84 \times 10^8 - r)^2}$

now we have

$\frac{2.71}{r} = \frac{24.4}{3.84\times 10^8 - r}$

$10.4 \times 10^8 - 2.71 r = 24.4 r$

now we have

$r = 3.84 \times 10^7 m$

5 0

3 years ago

A 225-kg object and a 525-kg object are separated by 3.80 m. (a) Find the magnitude of the net gravitational force exerted by th

pav-90 [236]

Answer: $F_{net}=3.383\times 10^{-7}\ N$

Explanation:

Given

Mass of first object $m_1=225\ kg$

Mass of second object $m_2=525\ kg$

Distance between them $d=3.8\ m$

$m_3=61\ kg$ object is placed between them

So force exerted by $m_1$ on $m_3$

$F_{13}=\frac{Gm_1m_3}{1.9^2}$

$F_{13}=\frac{6.674\times 10^{-11}(225\times 61)}{1.9^2}$

$F_{13}=2.5374141274×10^{−7}\ N$

Force exerted by $m_2\ on\ m_3$

$F_{23}=\frac{Gm_2m_3}{1.9^2}$

$F_{23}=\frac{6.674\times 10^{-11}(525\times 61)}{1.9^2}$

$F_{23}=5.920632964\times 10^{-7}\ N$

So net force on $m_3$ is

$F_{net}=F_{23}-F_{13}$

$F_{net}=5.920632964\times 10^{-7}-2.5374141274\times 10^{-7}$

$F_{net}=3.383\times 10^{-7}\ N$

i.e. net force is towards $m_2$

(b)For net force to be zero on $m_3$ , suppose

So force exerted by $m_1$ and $m_2$ must be equal

$F_{13}=F_{23}$

$\Rightarrow \frac{Gm_1m_3}{x^2}=\frac{Gm_2m_3}{(3.8-x)^2}$

$\Rightarrow \frac{m_1}{x^2}=\frac{m_2}{(3.8-x)^2}$

$\Rightarrow (\frac{3.8-x}{x})^2=\frac{m_2}{m_1}$

$\Rightarrow \frac{3.8-x}{x}=\sqrt{\frac{525}{225}}$

$\Rightarrow 3.8-x=1.52752x$

$\Rightarrow 3.8=2.52x$

$\Rightarrow x=1.507\ m$

4 0

3 years ago

Does the sign of the charge (positive or negative) affect how that charge is dissipated? Explain.

labwork [276]

This is an add proceeded by

5 0

3 years ago

What acceleration results from exerting a 125N force on a 0.65kg ball?

labwork [276]

first you do your pyramid f is on top and ma is
on bottom were m=mass and a=acceleration
were in this case you do f÷m=force÷mass so again in this case 125÷0.65=192.3 to be more accurate 192.3076923077

5 0

4 years ago

I need help ASAP please :)

rusak2 [61]

Density offers a convenient means of obtaining the mass of a body from its volume or vice versa; the mass is equal to the volume multiplied by the density (M = Vd), while the volume is equal to the mass divided by the density (V = M/d).

M = V d

M = 1.4 * 2 = 2.8 kg

7 0

2 years ago