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3 years ago

Which statement about subatomic particles is true?

Physics

1 answer:

navik [9.2K]3 years ago

4 0

A. An electron has far less mass than either a proton or neutron.

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In an electrostatic field, path 1 between points A and B is twice as long as path 2. The electrostatic work done on a negatively

Elanso [62]

Answer:

W2 = W1

Explanation:

work is independent of the path taken between the points.

8 0

2 years ago

Help me please!!!

lilavasa [31]

Both planets are similar in shape and have a rocky surface. Not sure about the phases though

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2 years ago

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A force of 60 N is applied to a skier to pull him along a horizontal surface so that his speed remains constant. If the coeffici

matrenka [14]

Answer:

$1200\ \text{N}$

Explanation:

F = Force on the skier = 60 N

$\mu$ = Coefficient of friction = 0.05

w = Weight of skier

Force is given by

$F=\mu w$

$\Rightarrow w=\dfrac{F}{\mu}$

$\Rightarrow w=\dfrac{60}{0.05}=\dfrac{6000}{5}$

$\Rightarrow w=1200\ \text{N}$

Weight of the skier on which the force is being applied is $1200\ \text{N}$ .

7 0

3 years ago

You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart

hichkok12 [17]

Answer:

$v = 3.5 \times 10^5 m/s$

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

$\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}$

now we have

$\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}$

$\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}$

so we will have

$r = 3.46 \times 10^8 m$

Now by energy conservation

$-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}$

$-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5$

$v = 3.5 \times 10^5 m/s$

7 0

2 years ago

A satellite is launched to orbit the Earth at an altitude of 3.25 107 m for use in the Global Positioning System (GPS). Take the

Korolek [52]

Answer:Orbital period =21.22hrs

Explanation:

given that

mass of earth M = 5.97 x 10^24 kg

radius of a satellite's orbit, R= earth's radius + height of the satellite

6.38X 10^6 + 3.25 X10^7 m =3.89 X 10^7m

Speed of satellite, v= $\sqrt GM/R$

where G = 6.673 x 10-11 N m2/kg2

V= \sqrt (6.673x10^-11 x 5.97x10^ 24)/(3.89 X 10^ 7m)

V =10,241082.2

v= 3,200.2m/s

a) Orbital period

$\sqrt GM/R$ = $\frac{2\pi r}{T}$

V= $\frac{2\pi r}{T}$

T= 2 $\pi$ r/ V

= 2 X 3.142 X 3.89 X 10^7m/ 3,200.2m/s

=76,385.1 s

60 sec= 1min

60mins = 1hr

76,385.1s =hr

76,385.1/3600=21.22hrs

3 0

3 years ago