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3 years ago

How do newton's laws apply to an egg drop

1 answer:

5 0

It applies to Newton's 3rd law because there will be a simultaneous reaction but in the opposite direction so when the egg lands, an upward force will act on it.

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A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0

3 years ago

The escape velocity at the surface of Earth is approximately 11 km/s. What is the mass, in units of ME (the mass of the Earth),

SVETLANKA909090 [29]

Answer:11 km/s

Explanation:

Given

Escape velocity at the surface of earth is 11 km/s

Escape velocity is given by

$V_e=\sqrt{\frac{2GM}{R}}$

Escape velocity at the surface of earth

$11=\sqrt{\frac{2GM}{R}}$ --------------------1

If Escape velocity is three times and the radius is also the three times

$V_e_2=\sqrt{\frac{2G(3M)}{3R}}$

$V_e_2=\sqrt{\frac{2GM}{R}}=V_e$

i.e. $V_e_2=11 km/s$

5 0

3 years ago

An antigen is a protein made by your body to respond to a specific foreign molecule.

Tom [10]

hope this helps it's F

8 0

4 years ago

Alexandra is attempting to drag her 32.6 kg golden retriever across the wooden floor by applying a horizontal force. What force

vichka [17]

Answer:

230.26 N

Explanation:

Since the speed is constant, acceleration is zero hence the net force will be given by the product of mass, coefficient of friction and acceleration due to gravity

F=0.72*32.6*9.81=230.26 N

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3 years ago

Electromagnetic radiation consists of particles called:

Westkost [7]

Photons are particles of electromagnetic radiation.

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3 years ago