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sergij07 [2.7K]
2 years ago
15

Discuss how the hardness or softness of the landing surface is related to the time required to stop the egg

Physics
1 answer:
monitta2 years ago
3 0

Answer

To understand this concept it is necessary to understand Newton's Second Law

According to Newtons Second law applied force is equal to rate of change of momentum of a body.

Mathematically,

F=\frac{dp}{dt}

here, \frac{dp}{dt} is rate of change of momentum with respect to time

It means If two eggs fall from same height,one on softer surface and other on hard surface that time the momentum of both eggs will remain equal at both the surfaces. But, impact time will be different. On hard surface egg will stop almost instantly so impact time will be small and hence the force on egg will be large therefore the egg will breakup.

On the other hand on the soft surface like a cotton, egg will not stop instantly but it will slow down for a few seconds and then stop due to which the time of impact will increase. Therefore the force on egg will be less and it won't break up.


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A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f
Setler79 [48]

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒  Magnetic force = Copper wire's weight

So,

⇒   B\times I\times L=M\times g

On putting the estimated values, we get

⇒  B\times 50\times 1=7.037\times 10^{-3}\times 9.81

⇒  50B=69.03297\times 10^{-3}

⇒  B=1.38\times 10^{-3} \ T

(b)...

As we know,

⇒  m=\delta\times L\times \frac{\pi \ d^2}{4}

⇒      =8960\times 1\times \frac{\pi \ (0.001)^2}{4}

⇒      =2240\times \pi \ 0.000001

⇒      =7.037\times 10^{-3} \ kg

7 0
2 years ago
Ugonna stands at the top of an incline and pushes a 100−kg crate to get it started sliding down the incline. The crate slows to
Anna [14]

Answer:(a)891.64 N

(b)0.7

Explanation:

Mass of crate m=100 kg

Crate slows down in s=1.5 m

initial speed u=1.77 m/s

inclination \theta =30^{\circ}

From Work-Energy Principle

Work done by all the Forces is equal to change in Kinetic Energy

W_{friction}+W_{gravity}=\frac{1}{2}mv_i^2-\frac{1}{2}mv_f^2

W_{gravity}=mg(0-h)=mgs\sin \theta

W_{gravity}=-mgs\sin \theta

W_{gravity}=-100\times 9.8\times 1.5\sin 30=-735 N

change in kinetic energy=\frac{1}{2}\times 100\times 1.77^2=156.64 J

W_{friction}=156.64+735=891.645

(b)Coefficient of sliding friction

f_r\cdot s=W_{friciton}

891.645=f_r\times 1.5

f_r=594.43 N

and f_r=\mu mg\cos \theta

\mu 100\times 9.8\times \cos 30=594.43

\mu =0.7

5 0
2 years ago
You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
zaharov [31]

Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
2 years ago
Un cuerpo se mueve en línea recta segun la ecuación x=10+20t-4.9t2 (x está expresado en metros y t en segundos). ¿Cuál es la lon
lora16 [44]

Answer:

La longitud del camino recorrido es de 25.9 [m]

Explanation:

Se reemplaza el valor de tiempo en segundos en la ecuación dada de desplazamiento

x=10+20*(3) - 4.9*(3)^2

x= 25.9 [metros]

4 0
2 years ago
Read 2 more answers
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
</span><span> Solve for a 
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
</span><span> m = 1.67*10^-27 kg 
</span><span> E = -150 N/C (- because it is down and up is positive) 
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward) 
</span> answer
 a = -6,4*10^5 m/s^2 (downward) 
3 0
2 years ago
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