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Westkost [7]
3 years ago
5

I need to speak with zuka

Physics
1 answer:
Alex787 [66]3 years ago
3 0

Answer:

DMs are not accessible anymore. I assume Zuka is a staff member? the only way to talk to a staff member anymore is to report something, but even then, the probably won't even look at what they're deleting :/

May I have brainliest please? :)

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A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves the bat at 30.0 m/s in a direction
lara31 [8.8K]
<h3>Answer</h3><h3>7 Ns</h3><h3>Explanation</h3>

Given in the question,

mass of foul ball = 0.140 kg

initial speed with which ball was hit with the bat = 30 m/s

final speed  = 40 m/s

According to the scenario the whole scene is making a right angle triangle

So, to the solve the question we will use pythagorus theorem

<h3>Hypotenuse² = base² + height²</h3>

Here,

Hypotenuse= Magnitude of impulse

Base = 1st change of momentum

height = 2nd change of momentum

 

1st impulse (1st change of momentum)

p = m(1)v(1) = (0.14 kg)(40.0 m/s) = 5.6 kg m / s = 5.6 N s

2nd impulse (2nd change of momentum)

p = m(2)v(2) = (0.14 kg)(30.0 m/s) = 4.2 kg m / s = 4.2 N s

Magnitude of impulse (hypotenuse of triangle)

impulse² = (5.6)² + (4.2)²

impulse² = 31.36 + 17.64

impulse² = 49

impulse² = √49

impulse = 7.0 N s

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Conditional probability involves events that are independent. true or false?
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Its false I got It right .


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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
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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>

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