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FrozenT [24]
3 years ago
13

(a) An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kin

etic energy? (b) An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum?
Physics
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

A) The smaller object; B) The larger object

Explanation:

A) Lets say the small object is 2 kg and the large one is 6 kg. Lets say they also have 30 kg*m/s of momentum each. The small object would have 15 m/s velocity and the large would have 5 m/s.

Now for kinetic energy(.5*m*v²), the small object is .5*2*15², which is 225 J

The large object is .5*6*5², which is 75 J, so the smaller object has more Kinetic energy. Since velocity is squared, it is more important than how large mass is.

B) Same masses as part A. Lets say the kinetic energy is 45 J for both of them. For the small object, 45=.5*2*v²

.5*2 is 1, so 45/1 is 45. Take the square root and we get v= 6.71 m/s

For the large object, 45=.5*6*v²

.5*6 is 3, so 45/3 is 15. Take the square root and we get v=3.87 m/s

Now we plug the velocities into p=mv

For the small object, p=2*6.71, which gives us p=13.42 kg*m/s

For the large object, p=6*3.87, which gives us p=23.22 kg*m/s

The larger object has the larger momentum.

Hope this helps

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Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f
vampirchik [111]

Answer:

\large \boxed{\text{30 rev/s}}

Explanation:

This question is based on the Law of Conservation of Angular Momentum.

Angular momentum (L) equals the moment of inertia (I) times the angular speed (ω).

L = Iω

If momentum is conserved,

I₁ω₁ = I₂ω₂

Data:

 I₁ = 3.5    kg·m²s⁻¹

ω₁ = 6.0    rev·s⁻¹

 I₂ = 0.70 kg·m²s⁻¹

Calculation:

\begin{array}{rcl}I_{1}\omega_{1} &= &I_{2}\omega_{2}\\\text{3.5 kg$\cdot$m$^{2}$}\times \text{6.0 rev/s} &= &\text{0.70 kg$\cdot$m$^{2}$}\times\omega_{2}\\\text{21 rev/s} &= &0.70\omega_{2}\\\omega_{2} & = & \dfrac{\text{21 rev/s}}{0.70}\\\\&=&\textbf{30 rev/s}\\\end{array}\\\text{The skater's final rotational speed is $\large \boxed{\textbf{30 rev/s}}$}

8 0
3 years ago
The center-seeking change in velocity of an object moving in a circle is:
NeTakaya

The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.

So, by Newton's laws, we know that an object moving with a given velocity will remain in constant motion with a constant velocity until we apply an acceleration.

So we define acceleration as the rate of change of the velocity, also remember that velocity is a vector (has magnitude and direction), so, if there is a change the direction of the velocity, we have an acceleration that causes that.

In circular motion, the velocity vector is always perpendicular to the radius of the circle, and it can only be possible if the velocity direction is changing constantly. This will happen because of something called centripetal acceleration.

This acceleration points radially inwards (to the center of the circle) so is also perpendicular to the velocity of the moving object, and this is what causes the constant change in the direction of the velocity of the moving object.

Just to give an example, if you have a string with a mass on one end, and with your hand, you rotate the mass (from the string), the tension of the string would be the centripetal acceleration.

If you want to learn more about circular motion, you can read:

brainly.com/question/2285236

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2 years ago
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g You want to make simultaneous measurements of the position and momentum of an electron and a proton that are moving along a st
Maslowich

Answer:

1832

Explanation:

From;

Δp Δx = h/4π

Δp = uncertainty in momentum

Δx = uncertainty in position

h= Plank's constant

But p =mv hence, Δp= Δmv

m= mass, v= velocity

mass of electron = 9.11 * 10^-31 Kg

Mass of proton = 1.67 * 10^-27 Kg

since m is a constant,

Δv = h/Δxm4π

For proton;

Δv = 6.6 * 10^-34/4 * 3.14 * 1.67 * 10^-27 * 1 * 10^-10

Δv = 315 ms-1

For electron;

Δv = 6.6 * 10^-34/4 * 3.14 * 9.11 * 10^-31 * 1 * 10^-10

Δv = 577000 ms-1

Ratio of uncertainty of electron to that of proton = 577000 ms-1/315 ms-1= 1832

7 0
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