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1 year ago

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Physics

2 answers:

Soloha48 [4]1 year ago

5 0

The momentum of the second ball was 15 kg.m/s.

<h3>What is inelastic collision?</h3>

In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.

In the given question,

Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.

Initial velocity of 1st ball of mass 3 kg is 1 m/s.

So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.

According to Principle of conservation of linear momentum for this inelastic collision,

Initial momentum of first ball + initial momentum of second ball = final momentum of the system

⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball

= 18 kg.m/s - 3 kg.m/s.

= 15 kg.m/s.

Hence, initial momentum of second ball = 15 kg.m/s.

Learn more about momentum here:

brainly.com/question/24030570

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liq [111]1 year ago

5 0

Answer:

the answer is 15kgm/s

Ptotal=18 kgm/

m1v1+m2v1=18

(3kg)(1)+m2v1=18

3+m2v1=18

subtract 3

m2v1=15

because p=mv

the momentum of the second ball is 15 kgm/

Explanation:

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Answer:

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2 years ago

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Answer:

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Explanation:

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6 0

2 years ago

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown st

kozerog [31]

Answer:

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

$g =9.8 \frac{m}{s^{2} }$

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

$y = 15*1 - ½ 9.8*1^{2}$ =15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

$y = 15*1.5 - ½ 9.8*1.5^{2}$ =22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

$y = 15*2 - ½ 9.8*2^{2}$ = 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

3 0

3 years ago

A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as

malfutka [58]

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2 $\pi \sqrt{0.5*9.8*0.3}$

T=1.2042s

The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

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6 0

11 months ago

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

oksian1 [2.3K]

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity $I$ = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity $I$ is proportional to 1/(distance)²

i.e

$I$ ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e $I$ ₂ = $I$ ₁/2

Hence,

$I$ ₂/ $I$ ₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

6 0

3 years ago