What is the momentum of an object at rest, in m/s?

A billiard ball with a mass of 1.5kg is moving at 25m/s and strikes a second ball with a mass of 2.3 kg that is motionless. Find

aliya0001 [1]

Answer:

The velocity of the second ball after the collision will be 16.3 m/s.

Step-by-Step Explanation:

Law of conservation of momentum states that the total momentum of an isolated remains the same before and after the collision.

Let: mass of the first ball = m1 = 1.5 kg

mass of the second ball = m2 = 2.3 kg

Momentum before the collision:

velocity of the first ball = v1 = 25 m/s

velocity of the second ball = v2 = 0 m/s

[tex] Initial momentum = m1v1 + m2v2 = 1.5*25 + 2.3*0 = 37.5 kgm/s [\tex]

Momentum after the collision:

velocity of the first ball = v1 = 0 m/s

velocity of the second ball = v2 = x

[tex]momentum after the collision= m1v1 + m2v2 = 1.5*0 + 2.3*x = 2.3*x[\tex]

According to law of conservation of momentum:

Initial momentum = Momentum after collision

37.5 = 2.3*x

⇒ x = 37.5/2.3

x = 16.3 m/s

4 0

3 years ago

5. A construction worker on a high-rise building is on a platform suspended between two cables as illustrated below. The constru

Natalka [10]

Answer:

a) Tc = 750 [N] ;b) See the explanation below.

Explanation:

To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.

b)

The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.

a)

In this way we can find by means of a sum of forces on the y axis equal to zero:

- 850 - 450 + 550 + Tc = 0

Tc = 750 [N]

4 0

3 years ago

A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 475°C. The aluminum plate has

Anuta_ua [19.1K]

Answer:

$T_0 = 338.916 Degree\ celcius$

Explanation:

Given data:

Thickness of aluminium sheet 10 cm

initial temperature = 25 degree celcius

Assumption

Thermal properties remain constant, transfer of heat by radiation is negligible.

from the information given in the question we have

T_S ≈T_∞ , it implies we have h → ∞

from table 4.2 Biot number → ∞ the value of

$\lambda_1 = 1.5708 and A_1= 1.2732$

The fourier number is

$t = \frac{\alpha t}{l^2} = \frac{97.1\times 10^{-6} \times 15}{0.05^2} = 0.5826$

Temperature at center after 15 second of heating

$\theta _{0 wall} = \frac{T_0 - T_{\infity}}{T_i -T_{\infity}} = A_i e^{\lambda_1^2 t}$

$T_0 = T_i -T_{\infity} \times A_i e^{\lambda_1^2 t}$

$T_0 = (25 - 475) 1.2732 e^{-1.5708^2 \times 0.5826} + 475 = 356 degree celcius$

$T_0 = 338.916 Degree\ celcius$

8 0

3 years ago

(^=˃ᆺ˂) hiya! I need some help please (^=˃ᆺ˂)

IrinaK [193]

Hello.

Some eartquakes are more daminging becasue of the high frequency. Some earthquakes have higher then others. And the location places with less people have smaller earthqakes then places with a lot. The earthquake in i<span>ndian Ocean</span> hurt over a miilon people but the one in Hati only left things in poor condation.

Have a nice day

7 0

3 years ago

Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The

Brut [27]

Answer:

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .