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motikmotik
3 years ago
15

List one example of contact and non contact force

Physics
1 answer:
Savatey [412]3 years ago
5 0

Answer Contact Force =Muscular force ,frictional force ,spring force, normal force, tension force are few examples.

Non-Contact Force= Gravitational force ,electrostatic force, magnetic force and electrochemical force are few examples

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A nonzero net force acts on a particle and does work. Which one of the following statements is true?
arlik [135]

Answer:

option (D) is correct.

Explanation:

According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.

the kinetic energy of a body is directly proportional to the square of the speed of the body.

As the kinetic energy change, the speed of the body also change.

Option (D) is correct.

5 0
3 years ago
How do I solve for this?
vova2212 [387]
I can’t see anything
7 0
3 years ago
A rubber ducky is placed 20 cm from a thin convex lens with a focal length of 15 cm. Which statement correctly describes the nat
blondinia [14]
I believe the answer is B, a real and inverted image is formed on the side of the lens opposite the rubber ducky. The focal length is 15 cm and therefore the center of curvature (2F) will be 30 cm. When the object is placed between F and 2F (in this case 20 cm) in front of a convex lens, an inverted, real image is formed on the other side of the lens.
6 0
3 years ago
Read 2 more answers
A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?
charle [14.2K]
Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
5 0
3 years ago
A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ
FromTheMoon [43]

Answer:

6.27*10^{23}kg

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

F=\frac{GMm}{r^{2} }  and a=rω^2

where ω=\frac{2\pi }{T} and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\  M=\frac{r^{3}}{G}  (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg

6 0
3 years ago
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