All categories

3 years ago

What is true of systems in terms of their size and boundaries?

Physics

2 answers:

Nataly [62]3 years ago

5 0

A system is contained by its boundary; therefore, the size of a system is limited by its boundary.

dusya [7]3 years ago

5 0

All systems differ in size from subatomic to the extent of the universe. All systems have limits and many systems have matter and power that runs within them. A boundary is an ended surface enclosing a system through which energy and mass may enroll or omit the system.

You might be interested in

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

Mila [183]

Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.

5 0

2 years ago

Light hits a mirror at a 45Â° angle. It will be reflected at an angle _____.

Eddi Din [679]

It will be equal to 45Â°

6 0

2 years ago

suppose both the dog and the girl run at a speed of 2 m/s.Calculate both of there kinetic energies. kinetic energy of dog =

Andre45 [30]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

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

Very large forces are produced in joints when a person jumps from some height to the ground. Calculate the force produced if an

krok68 [10]

Answer:

A) 31 kJ

B) 1.92 KJ

C) 40 , 2.48

Explanation:

weight of person ( m ) = 79 kg

height of jump ( h ) = 0.510 m

Compression of joint material ( d ) = 1.30 cm ≈ 0.013 m

A) calculate the force

Fd = mgh

F = mgh / d

W = mg

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.013) )

= 774.99 ( 40.231 ) ≈ 31 KJ

B) calculate the force when the stopping distance = 0.345 m

d = 0.345 m

Fd = mgh hence F = mgh / d

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

= 79 * 9.81 ( 1 + (0.51 / 0.345) )

= 774.99 ( 2.478 ) = 1.92 KJ

C) Ratio of force in part a with weight of person

= 31000 / ( 79 * 9.81 ) = 31000 / 774.99 = 40

Ratio of force in part b with weight of person

= 1920 / 774.99 = 2.48

4 0

3 years ago