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3 years ago

Why does the momentum before a collision equal the momentum after a collision?

Physics

2 answers:

Art [367]3 years ago

8 0

Answer: Correct option is C) because momentum is conserved

Explanation:

According to law of conservation of linear momentum if net external force on a system is zero then the total momentum of the system will be conserved .

<u>For a collision occurring between object 1 and object 2 in an isolated system</u>

If we consider the two colliding object as a single system then while collision the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

Thus momentum before collision equal the momentum after a collision because momentum is conserved .

kotykmax [81]3 years ago

3 0

The answer is C. Due to conservation of momentum.

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Charges can be separated in an isolated system

Vesnalui [34]

Answer:

yes. can always be separated

4 0

3 years ago

An object has a mass of 10grams and a volume of 20m3. what is the density of the object

Pavlova-9 [17]

Explanation:

density = mass/ volume

in the question the mass is given in grams so will convert it into kg,

10 g = 0.01 kg

density= 0.01/ 20 = 1 / 2000 = 0.005 kg/m^3

7 0

3 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

3 years ago

Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc

IgorC [24]

Answer:

The answers to the questions are;

a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.

b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.

c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.

Explanation:

To solve the question, we write out the known variables as follows

1 g of fat = 9.00kcal

Number of steps the student climbs = 95 steps

Height of each step = 0.150 m

Time it takes for the student to reach the top of the stairs = 57.5 s.

Efficiency of human muscles = 20 %

Mass of student, m = 65 kg

a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.

The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..

That is work done, W, = P. E. = m·g·h

Where:

h = The total height climbed by the student

g = Acceleration due to gravity = 9.81 m/s²

Therefore;

h = Height of each step × Number of steps the student climbs =

= 0.150 m/(step) × 95 steps = 14.25 m

Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²

= 9086.5125 J

We remember that the efficiency of the muscle is 20 %

The formula for efficiency is

Efficiency = $\frac{Ene rgy Out put}{Energ y In put} \times 100 %$

The work produced by the muscle = Energy Output = 9086.5125 J

Energy input is given by

$\frac{Out put} {Effici ency}$ = 9086.5125 J/ (0.2) = 45432.5625 J

= 45.432 kJ

From the question, 1 g of fat = 9.00 kcal and

1 kcal = 4186 J

Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J

Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ

To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs $\frac{37674 kJ}{45.432 kJ}$ times to make up the 37674 kJ energy contained in 1 kg of fat

That is $\frac{37674 kJ}{45.432 kJ}$ = 829.23 times

b. Power is the rate of doing work

That is Power output = $\frac{ WorkO utput }{Time}$ = $\frac{9086.5125 J}{57.5 s}$ = 158.026 watts

c. No as the activity student will have to spend a total time of

829.23 × 57.5 s = 47680.67 s climbing up the stairs alone and

47680.67 s = ‪13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× ‪13.24 Hours

= 26.49 hrs = 1.104 days exercising which is not humanly possible.

3 0

3 years ago

What is a watt equivalent to in terms of kilograms, meters and seco

Ierofanga [76]

1 watt = 1 joule per second

1 joule = 1 newton-meter

1 newton = 1 kg-meter per second²

Put it all together and you have . . .

1 watt = 1 <em>kg-meter² / second³

</em>

5 0

3 years ago