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dsp73
3 years ago
6

It was a children versus grown-ups competition at school. One event required the adult to throw a basketball as far as he could.

When it was the child's turn, he had to throw a baseball instead. The event coordinator said, "It's only fair!" Based on Newton's laws, was this event fair?
Physics
2 answers:
Anika [276]3 years ago
7 0

Answer:

Based on Newton's Law, the event is not fair because the ball which have large mass and have less acceleration will have less displacement if both the balls are thrown with same force.

Explanation:

Size of basketball is bigger than the size of baseball and the mass of basketball is greater than the mass of baseball.

We know from Newton's second law

F= m\times a

or

a=\frac{F}{m}

from above equation it is clear that ball with higher mass will have less acceleration if force is constant.

and if acceleration is less than according to Newtons second equation of motion, displacement of ball can be given as

s=ut+\frac{at^2}{2}

From above equation it is clear that the ball which have large mass and have less acceleration will have less displacement if both the balls are thrown with same force.

aleksklad [387]3 years ago
4 0
Based on Newton's Law, the event is not fair because:

<span>No, because a basketball is bigger than a baseball, and objects that are bigger accelerate slower.


I hope my answer has come to your help. God bless and have a nice day ahead!</span>

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lord [1]

(a) 3675 N

Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:

F_x = m a_x

where

F_x is the horizontal component of the force

m is the mass of the passenger

a_x is the horizontal component of the acceleration

Here we have

m = 75.0 kg

a_x = 49.0 m/s^2

Substituting,

F_x=(75.0)(49.0)=3675 N

(b) 3748 N, 11.3 degrees above horizontal

In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:

R=mg=(75.0)(9.8)=735 N

where we used

g=9.8 m/s^2 as acceleration of gravity.

So, this is the vertical component of the force exerted by the seat on the passenger:

F_y = 735 N

and therefore the magnitude of the net force is

F=\sqrt{F_x^2+F_y^2}=\sqrt{3675^2+735^2}=3748 N

And the direction is given by

\theta = tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{735}{3675})=11.3^{\circ}

4 0
3 years ago
A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
mihalych1998 [28]
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
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3 years ago
A doodler effect occurs when a source of sounds move
Alla [95]

Answer:

yes its true

Explanation:

8 0
3 years ago
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You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am
emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz

v₀ = velocity at which the sound source is moving towards the reference point = ?

v = velocity of sound waves = 343 m/s

8.61 = 7.87/(1 - (v₀/v))

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7 0
3 years ago
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
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