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bazaltina [42]
2 years ago
8

In a game of egg-toss, you and a partner are throwing an egg back and forth trying not to break it. Given your knowledge of mome

ntum, what hint could you give to your partner to keep the force of impact on the egg as low as possible? Clearly explain your answer.
Physics
1 answer:
lutik1710 [3]2 years ago
3 0
F=dP/dt.  So you want the momentum to change as slowly as possible in time to minimize the force.  So as you catch the egg, let your hand move backward with it for awhile, slowly bringing it to a stop.  If you hold your hand steady when you catch it the force due to the impact could break it.
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Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2
GarryVolchara [31]

Answer:

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

Explanation:

Given vibrating system is

u''+\frac{1}{4}u'+2u= 2cos \omega t

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) =  - A ω² cosωt -B ω² sin ωt

Putting this in given equation

-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t

Equating the coefficient of sinωt and cos ωt

\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2

\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0.........(1)

and

\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0

\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0........(2)

Solving equation (1) and (2) by cross multiplication method

\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}

\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}

\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}   and        B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}

Therefore the required solution is

U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t

5 0
2 years ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

3 0
2 years ago
Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp
MatroZZZ [7]

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

6 0
1 year ago
Identify whether the features and functions belong to the Hubble Space Telescope or the ALMA radio telescope. 1.installed on lan
tester [92]
Hublle Space Telescope
- Orbits around  Earth.
- Receive light before it enters Earth's Atmosphere. 
- Get information about stars before they are born.

Alma Radio Telescope
- Installed on land.
- Receive light after it passes through Earth's Atmosphere.
- Get information about stars after they are born. 

5 0
2 years ago
Read 2 more answers
Write difference between upstroke and downstroke of water pump
lakkis [162]

Explanation:

Upstroke is a mechanism which helps to raise the plunger and downstroke helps to help lower the plunger. On the up-stroke of the plunger, the lower valve opens and the upper valve is closed. ... Whereas, on the downstroke, the lower valve closes and the upper one opens.

7 0
2 years ago
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