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BARSIC [14]
2 years ago
10

Most people can throw a baseball farther than a bowling ball, and most people would find it less painful to catch a flying baseb

all than a bowling ball flying at the same speed as the baseball. Explain these two situations in terms of
Newton’s First Law of Motion
Newton’s Second Law of Motion
Physics
1 answer:
IRISSAK [1]2 years ago
6 0

Based on Newton's first  and second law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.

<h3>What is Newton's first law of motion?</h3>

Newton's first law of motion first law of motion states that a body at rest or uniform motion in a straight line will continue in that path unless acted upon by an external force.

Newton's first law of motion is also called law of inertia because it depends on mass of the object.

An object with a greater mass will require greater force to be stopped or get moving.

Based on Newton's first law of motion most people would find it less painful to catch a flying baseball than a bowling ball flying at the same speed as the baseball because the mass of the baseball is smaller and will require smaller force to be stopped.

Also according to Newton's second law of motion, the force applied to an object is proportional to the product of mass and acceleration of the object. Thus, a baseball with smaller mass will require smaller force to be stopped.

Learn more about Newton's first law of motion here: brainly.com/question/10454047

#SPJ1

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For what value of the ratio r/a of plate radius to separation between the plates does the electric field at the point x=a/2 on t
SIZIF [17.4K]

solution:

E\delta =\frac{R}{\epsilon0}(1-\frac{A}{\sqrt{4R^{2}}-ac}

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4r^{2}/^{_a{2}}+1}})

=\frac{R}{\epsilon0}(1-\frac{1}{\sqrt{4x^2+1}})

x=\frac{r}{a}

infinite case,

Ei=\frac{r}{\epsilon0}

\therefore e\delta =ei(1-\frac{1}{\sqrt{4x^{2}+1}})

we have to find x when,

ei-e\delta =1% ,y=ei=1/100 ei

or,ei-ei+\frac{ei}{\sqrt{4x^2+1}} = 1/100ei

\frac{1}{\sqrt{4x^2+1}}=\frac{1}{100}

4x^2+1 =10^4

x=\frac{\sqrt{\frac{10^4-1}{4}}}=49.99\approx 50

\therefore \frac{r}{a}\approx 50

6 0
3 years ago
Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in
Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

F_1=\frac{Kq_1q_2}{r^2}

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

F_2 = \frac{K2q_12q_2}{r^2} =  \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1

Therefore, the force between them quadruples

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3 years ago
A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: <span>

<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is: 

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

<span>Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)</span>

<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
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<span>x2 = 14 ft  + 15 ft = 29 ft</span>

x1 = 14 ft + 1 ft = 15 ft

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W = 625π * (29^2 - 15^2) 
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3 0
4 years ago
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E_m=E_p
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It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
E_p=mgh=E_m=\frac{1}{2}mv^2
Solving for h gives us:
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