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inessss [21]
3 years ago
13

Which of these are part of our solar system? select all that apply

Physics
2 answers:
Len [333]3 years ago
6 0
A)planets
b)the sun
c)moons
e)comets
f)asteroids
Papessa [141]3 years ago
4 0
Everything on this list is part of the solar system EXCEPT Polaris.
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A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
jeyben [28]

Answer:

-4*10⁴ units.

Explanation:

As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.

5 0
3 years ago
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons
Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

5 0
3 years ago
A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation. (a) Calculate the
Illusion [34]

angular vel to tangential vel

v=r omega

v = 56 x 100/60 x 2 pi

v = 56x5/3x6

v=560m/s as estimate

100 revs, 5.00m

5 0
3 years ago
Read 2 more answers
Convert 325 cm to yards I need the answer ASAP
Ira Lisetskai [31]
325 /91 = 3.571 yards
7 0
3 years ago
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