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LuckyWell [14K]
3 years ago
12

What is the importance of law of conservation of mass

Physics
1 answer:
EastWind [94]3 years ago
7 0
The principle of the Conservation of mass is so important, as it defines that in physics, nothing can be created or destroyed in an isolated system. It explains <span>that the number of molecules (parts) must be equal on both sides of the equation - BALANCED EQUATION</span>
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When friction occurs, which of the following types of energy is produced?
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heat energy

Explanation:

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Challenge Question (This is supposed to be hard, think critically and take it one step at a time):
aivan3 [116]

Answer:

You must travel at an average speed of 67.06 m/s to be on time to the lecture.

Explanation:

From the question, the total distance from your house to the science lecture is 75 miles.

Also, you get halfway there before you stop for a gas, that is, you have covered half of 75 miles, which 37.5 miles and you also have to cover 37.5 miles to get to the science lecture.

After filling up, you only have 15  minutes before the lecture starts,

To determine how fast you must drive to be on time to the lecture,

we will determine the average speed you need to travel.

From

Average speed = Distance / Time

Distance = 37.5 miles (Convert to meters)

(NOTE: 1 mile = 1609.344 meters)

Hence, 37.5 miles = 37.5 × 1609.344 miles = 60350.4 meters

∴ Distance = 60350.4 meters

Time = 15 minutes (Convert to seconds)

(NOTE: 1 minute = 60 seconds)

Hence, 15 minutes = 15 × 60 seconds = 900 seconds

Now, from

Average speed = Distance / Time

Average speed = 60350.4 m / 900 s

Average speed = 67.06 m/s

Hence, you must travel at an average speed of 67.06 m/s to be on time to the lecture.

8 0
4 years ago
A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera
iren [92.7K]

Answer:

(a)  a₁:  jogger  acceleration= 1.5 m/s²

(b)  a₂:  car  acceleration = 1.5 m/s²

(b)  d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

Explanation:

we apply uniformly accelerated motion formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁:  jogger displacement   

t₁ :  jogger time

v₀₁:  jogger initial speed

vf₁:  jogger  final speed

a₁:  jogger  acceleration

d₂: car displacement   

t₂ : car  time

v₀₂: car  initial speed

vf₂:  car  final speed

a₂:  car  acceleration

Data

v₀₁ = 0

vf₁ = 3 m/s

t₁ =2.0 s

v₀₂ = 38.0m/s

vf₂ = 41.0 m/s

t₂ = 2.0 s

Problem development

(a) Find the acceleration (magnitude only) of the jogger.

We apply the formula (1) for calculate acceleration :

vf₁= v₀₁+a₁*t₁

3 = 0 +(a₁)*(2)

a₁= (3)/(2)

a₁= 1.5 m/s²

(b) Determine the acceleration (magnitude only) of the car.

We apply the formula (1) for calculate acceleration :

vf₂= v₀₂+a₂*t₂

41 = 38 +(a₂)*(2)

a₂= (41 - 38)/(2)

a₂= 3 /2

a₂= 1.5 m/s²

(c) Does the car travel farther than the jogger during the 2.0 s? If so, how much farther?

We apply the formula (1) for calculate distance :

d₁= v₀₁*t₁+ (1/2)*a₁*t₁²= 0+ (1/2)*(1.5) *(2)² = 3 m

d₂= v₀₂*t₂+ (1/2)*a₂*t₂² =38*(2)+ (1/2)*(1.5) *(2)²= 79 m

d= 79 m-3 m

d= 76m : the car travels 76 meters longer than the jogger during the 2 seconds

3 0
3 years ago
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