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kozerog [31]
3 years ago
9

Which is always true in a closed system?

Physics
1 answer:
Tanya [424]3 years ago
4 0

Answer:

The correct option is momentum is conserved

Explanation:

A closed system is a system that is independent/free of external factors/force and does not exchange matter with its surrounding. Since a close system is free of external factors/force; <em>acceleration is constant in it, mass is conserved in it and there will be changes in velocities of objects in the closed system</em>.

This question actually seeks to test our knowledge of the law of momentum. The law of conservation of momentum states that the momentum of a closed system is conserved.

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A race car travels 40 m/s around a banked (45° with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of
OLEGan [10]

Answer:

c)F_{net} = 0.640 kN

Explanation:

As we know that resultant force is the net force that is acting on the system

As per Newton's II law we know that net force is product of mass and acceleration

so we will have

F_{net} = ma

here we know

m = 80 kg

for circular motion acceleration is given as

a_c = \frac{v^2}{R}

a_c = \frac{40^2}{200} = 8 m/s^2

now we have

F_{net} = 80 \times 8

F_{net} = 640 N

F_{net} = 0.640 kN

7 0
3 years ago
If the force applied to an object is not greater than the starting friction, what will happen to the object? Which answer is cor
AleksandrR [38]

Answer:

The object will move in the opposite direction of the force applied. - 2.

8 0
3 years ago
The minimum number of vectors of unequal magnitude required to
EleoNora [17]

Answer:

d

Explanation:

4 0
3 years ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
Why is it not suitable to determine the volume of an irregular charcoal using displacement method ​
AleksandrR [38]

Answer:

Displacement method of volume measurement is no suitable

Explanation:

Displacement method of volume measurement is no suitable for the objects that do not get immersed into the water completely because of the hindrance in accuracy of the measurement.

4 0
3 years ago
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