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3 years ago

Which of the following items can enter and exit an isolated system?

Physics

1 answer:

Mkey [24]3 years ago

7 0

Neither energy nor matter can enter and exit an isolated system.

Explanation:

There are three types of systems which refers to universe. They are

1. Open System : In an open system, both energy and matter have external interactions. Example:- boiling water in an open pan.

2. Closed system : In a closed system, only energy has interaction with surroundings. Example:- boiling water with a lid on the pan.

3. Isolated system : In an isolated system, neither energy-nor matter has external interactions with surroundings. Example : a thermos flask does not exchange energy and matter.

Hence, the correct option is (a) " neither energy nor matter ".

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An arrow is launched with an initial velocity of 7.6 m/s at an angle of 42 degrees above the horizontal. What is the horizontal

AleksAgata [21]

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

$v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}$

$\:\:\:\:\:=5.6\:\text{m/s}$

5 0

3 years ago

How is my Engineering Project?

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3 years ago

Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plate

vichka [17]

Answer:

CB = 4.45 x 10⁻⁹ F = 4.45 nF

Explanation:

The capacitance of a parallel plate capacitor is given by the following formula:

C = ε₀A/d

where,

C = Capacitance

ε₀ = Permeability of free space

A = Area of plates

d = Distance between plates

FOR CAPACITOR A:

C = CA = 17.8 nF = 17.8 x 10⁻⁹ F

A = A₁

d = d₁

Therefore,

CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1

FOR CAPACITOR B:

C = CB = ?

A = A₁/2

d = 2 d₁

Therefore,

CB = ε₀(A₁/2)/2d₁

CB = (1/4)(ε₀A₁/d₁)

using equation 1:

CB = (1/4)(17.8 X 10⁻⁹ F)

CB = 4.45 x 10⁻⁹ F = 4.45 nF

5 0

3 years ago

How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation

Mazyrski [523]

Answer:

175.96 g

Explanation:

Potential energy required for the man to climb 7.07 km = m g h.

= 64 x 9.8 x 7070

= 4.434 x 10⁶ J

= 4.434 X 10⁶ / 4.2 cals

= 1.0557 x 10⁶ cals

= 1.0557 x 10⁶ / 6000 g of butter

= 175.96 g of butter.

3 0

3 years ago

The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.

Sholpan [36]

The frequency of rotation of Mars is 0.0000113 Hertz.

Given the following data:

Period = 1 day and 37 minutes.

To find the frequency of rotation in Hertz:

First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.

Conversion:

1 day = 24 hours

24 hours to minutes = $60$ × $24$ = $1440$ minutes