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2 years ago

Janet was pulling a box that weighed 20N across 5m. How much work did she exert?

Physics

1 answer:

Galina-37 [17]2 years ago

6 0

Answer:

100nm

Explanation:

Work=Force×distance

=20×5

=100nm

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How are theories and laws connected

Anni [7]

Answer:

Laws are statements about something that's been observed and stated while a theory is an explanation of what's been observed. This connection between them forms a main idea that many people regulate as "what's normal."

Explanation:

4 0

3 years ago

Read 2 more answers

Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What

Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

$v2=\frac {p1v1}{p2}$

Substituting the given values then

$v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l$

Therefore, the volume is approximately 0.426 L

6 0

3 years ago

Which one of the following statements concerning the momentum of a system when the net force acting on the system is equal to ze

Alex

Answer:

Explanation:

According to newton's 2nd law rate of change of momentum is directly proportional to the force applied on the body. Since, net Force is zero this means momentum did not change or momentum of the body remained constant.

Hence, the system have constant value of momentum. Therefore, option D is correct.

5 0

3 years ago

An airplane refuels in Bakersfield and continues on to Fresno. It travels an average speed of 610 km/h. IF the trip takes 2.75 h

rusak2 [61]

2.75x610km

1220+7.5x61

1220+427 ... just over

1647

8 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