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Tcecarenko [31]
2 years ago
6

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collisi

on? Explain why or why not.
Physics
2 answers:
mart [117]2 years ago
6 0
Conservation of momentum<span> is a fundamental law of physics. This law states that the </span>momentum<span> of a system is constant if there are </span>no external forces acting on the system. In a situation in which two balls, each with a mass of 0.5 kg, collide on a pool table<span> the law of conservation of momentum is not satisfied because there are external forces that moved the balls. </span>
swat322 years ago
4 0
Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s, and the total momentum after the collision is 
<span>1.5 kg • m/s. The momentum stays the same after the collision.</span>
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A small child gives a plastic frog a big push at the bottom of a slippery 2.0 meter long, 1.0 meter high ramp, starting it with
valentinak56 [21]
Refer to the diagram shown below.

Because the ramp is slippery, ignore dynamic friction.
Let m =  the mass of the frog.
g = 9.8 m/s²

The KE (kinetic energy) at the bottom of the ramp is
KE₁ = (1/2)*(m kg)*(5 m/s)² = 12.5 m J

Let v =  the velocity at the top of the ramp.
The KE at the top of the ramp is
KE₂ = (1/2)*m*v²= 0.5 mv² J
The PE (potential energy) at the top of the ramp relative to the bottom is
PE₂ = (m kg)*(9.8 m/s²)*(1 m) = 9.8m J

Conservation of energy requires that
KE₁ = KE₂ + PE₂
12.5m = 0.5mv² + 9.8m
0.5v² = 2.7
v = 2.324 m/s

Answer: 2.324 m/s

7 0
2 years ago
A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu
bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let \omega is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

\alpha =\dfrac{\omega_f-\omega_i}{t}

\alpha =\dfrac{\omega-0}{t}

\alpha =\dfrac{\omega}{t}

t=\dfrac{\omega}{\alpha }............(1)

Using second equation of kinematics as :

\theta=\omega_i t+\dfrac{1}{2}\alpha t^2

\theta=\dfrac{1}{2}\alpha t^2

Using equation (1) in above equation

\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }

In one revolution, \theta=4\pi (in 2 revolutions)

4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }

\omega=\sqrt{8\pi \alpha}

\omega=2\sqrt{2\pi \alpha}

Hence, this is the required solution.

6 0
2 years ago
Read 2 more answers
The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.
Kipish [7]

The missing stresses in the question:

Engineering Engineering strain

Stress (MpA)

235 0.194

250 0.296

Answer:

σ = 245MPa

Explanation:

For this problem, we first need to convert engineering stresses and strains to true stresses and strains so that the constants K and n may be determined. Since σT = σ(1 + ε)

So,

σT1 = (235 MPa)(1 + 0.94) = 280 MPa

σT2 = (250 MPa)(1 + 0.296) = 324MPa

Similarly for strains, since;

εT = ln(1 + ε)

Thus;

ε T1 = ln (1 + 0.194) = 0.177

εT2 = ln (1 + 0.296) = 0.259

We know that true stress;

σε =(σT/K)^(1/n)

So taking log of both sides, to get;

log σT = log K + n log εT

which allows us to set up two simultaneous equations for the above pairs of true stresses and true strains, with K and

n as unknowns.

Thus ,

log (280) = log K + n log (0.177)

and

log (324) = log K + n log (0.259)

Solving both log equations simultaneously, we get;

K = 543 MPa and n = 0.383

Now converting, ε to true strain;

ε T= ln (1 + 0.25) = 0.223

From the true stress equation earlier quoted, we'll make σT the subject and get;

σT = (εT^n) x k

σT = 543 x (0.223^(0.383)) = 306 MPa

Now converting this value of σT

to an engineering stress using the first equation quoted in this answer, gives;

σ = σT/(1 +ε) = 306/(1+0.25) = 245MPa

7 0
2 years ago
B. Complete the table to show the effect of each change on each electric quantity. (1 point)
notka56 [123]

Answer:

Effect on electric force

Multiply one charge by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is directly proportional to the charge.

Hence if one charge is doubled, then the electric force is doubled.

Multiply distance by 2

The electric force is given by F=kq1q2/r2

From the equation, the force is inversely proportional to the square of the distance of separation.

If the distance is doubled, F is decreased by 22. This means that the force is multiplied by 1/4.

Effect on electric potential energy

Multiply one charge by 2.

The electric potential energy is given by U=kq1q2/r

From the equation, the electric potential energy is directly proportional to the charge q.

If one charge is doubled, the electric potential energy is doubled.

Multiply distance by 2

The electric force is given by U=kq1q2/r

From the equation, the electric potential energy is inversely proportional to the distance of separation r.

If the distance is doubled, U is divided by 2. This means that the electric potential energy is multiplied by 1/2.

Effect on potential difference

Potential difference is defined as the change in electric potential energy.

Increase in the charge causes an increase in the potential difference and an increase in the distance of separation decreases the potential difference.

4 0
2 years ago
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