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Butoxors [25]
3 years ago
12

When momentum is conserved it is called _____. (multiple choice)

Physics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

Based off the word "conserved" I would say

A. Conservation of Momentum.

Explanation:

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A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten
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Alguien me ayuda con calculos estequiometricos?
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A baseball pitcher throws the ball in a motion where there is rotation of the forearm about the elbow joint as well as other mov
Brums [2.3K]

Answer: 330.88 J

Explanation:

Given

Linear velocity of the ball, v = 17.1 m/s

Distance from the joint, d = 0.47 m

Moment of inertia, I = 0.5 kgm²

The rotational kinetic energy, KE(rot) of an object is given by

KE(rot) = 1/2Iw²

Also, the angular velocity is given

w = v/r

Firstly, we calculate the angular velocity. Since it's needed in calculating the Kinetic Energy

w = v/r

w = 17.1 / 0.47

w = 36.38 rad/s

Now, substituting the value of w, with the already given value of I in the equation, we have

KE(rot) = 1/2Iw²

KE(rot) = 1/2 * 0.5 * 36.38²

KE(rot) = 0.25 * 1323.5

KE(rot) = 330.88 J

6 0
3 years ago
A jogger accelerates at a constant rate as she travels 5.0
VladimirAG [237]

Answer:

2.0s

Explanation:

...

3 0
3 years ago
A constant force is exerted on a cart that is initially at rest on a frictionless air track. The force acts for a short time int
Inessa05 [86]

Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of

<em>F</em> = <em>m a</em>   →   <em>a</em> = <em>F </em>/ <em>m</em>

so that the cart's final speed is

<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>

<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>

<em />

If we force is halved, so is the accleration:

<em>a</em> = <em>F</em> / <em>m</em>   →   <em>a</em>/2 = <em>F</em> / (2<em>m</em>)

So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give

(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>

3 0
3 years ago
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