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3 years ago

state the principle of the Conservation of Linear Momentum and show how it follows from Newton's second law of motion

Physics

1 answer:

photoshop1234 [79]3 years ago

6 0

Explanation:

The linear momentum of a particle is defined as the product of the mass of the particle times the velocity of that particle. Conservation of momentum of a particle is a property exhibited by any particle where the total amount of momentum never changes. Linear momentum of a particle is a vector quantity and is denoted by →p

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A general contractor drives 5 miles to the hardware store in 10 minutes. He shops for needed tools for 30 minutes, then drives t

Alex17521 [72]

Given parameters:

Distance to hardware shop = 5 miles

Time to reach hardware shop = 10 minutes

Time spent at the shop = 30 minutes

Average speed to customer home = 45 mph

Time taken for the travel = 20 minutes

Unknown:

Average speed of the contractor = ?

Solution:

Average speed is the total distance covered divided by the total time taken.

Average speed = $\frac{total distance}{total time taken}$

total distance = distance to hardware shop + distance to customer's home

We do not know the distance to customer's home but we have been given the speed and time, so we can find the distance.

Distance = speed x time

Convert the time to hrs and solve;

60 minutes = 1 hr

20 minutes = $\frac{20}{60} hr$ = $\frac{1}{3} hr$

So, Distance = 45mph x $\frac{1}{3} hr$ = 15miles

Now;

Total distance = 5 + 15 = 20miles

Total time = time to reach hardware shop + time to reach customer's house

= 10 + 20

= 30 minutes

Convert the time from minutes to hrs;

60 minutes = 1hr

30 minutes = $\frac{30}{60}$ = 0.5hr

So;

Average speed = $\frac{20}{0.5}$ = 40mph

The average speed is 40mph

3 0

3 years ago

2. While standing near a bus stop, a student hears a distant horn beeping. The frequency emitted by the horn is 440 Hz. The bus

jarptica [38.1K]

Given Information:

Frequency of horn = f₀ = 440 Hz

Speed of sound = v = 330 m/s

Speed of bus = v₀ = 20 m/s

Answer:

Case 1. When the bus is crossing the student = 440 Hz

Case 2. When the bus is approaching the student = 414.9 Hz

Case 3. When the bus is moving away from the student = 468.4 Hz

Explanation:

There are 3 cases in this scenario:

Case 1. When the bus is crossing the student

Case 2. When the bus is approaching the student

Case 3. When the bus is moving away from the student

Let us explore each case:

Case 1. When the bus is crossing the student:

Student will hear the same frequency emitted by the horn that is 440 Hz.

f = 440 Hz

Case 2. When the bus is approaching the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330+20 )

f = 440 ( 330/ 350 )

f = 440 ( 0.943 )

f = 414.9 Hz

Case 3. When the bus is moving away from the student

f = f₀ ( v / v+v₀ )

f = 440 ( 330/ 330-20 )

f = 440 ( 330/ 310 )

f = 440 ( 1.0645 )

f = 468.4 Hz

6 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

A speeder is pulling directly away and increasing his distance from a police car that is moving at 24 m/s with respect to the gr

iogann1982 [59]

2,062,305 2,062,305 <span>2,062,305</span>

8 0

3 years ago

A 1.00 kg block of ice, at -25.0°C, is warmed by 35 kJ of energy. What is the final temperature of the ice?

ahrayia [7]

Answer:

-8.4°C

Explanation:

From the principle of heat capacity.

The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

6 0

3 years ago

state the principle of the Conservation of Linear Momentum and show how it follows from Newton's second law of motion​

state the principle of the Conservation of Linear Momentum and show how it follows from Newton's second law of motion