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2 years ago

How does Newton's three laws of motion apply to the egg drop project

2 answers:

6 0

Answer:
Newton's third law, which states that every action has an equal and opposite reaction, can be related to an egg drop because the forces pushing the egg up and down must follow the law.

Explanation:
Newton's third law states that for each action, there is an equal and opposite reaction.
An egg by itself will splat against the concrete if dropped from a height. The point of an egg drop is to keep this from happening. In other words, you must make sure that the downward force of gravity has an opposite force, great enough to slow the egg down enough to make sure it doesn't break.
If you tie a parachute to save the egg, air drag pushes the parachute up, but the force of gravity on the egg pushes it down still. However, the air drag works opposite gravity and slows the egg's descent enough to ensure it doesn't break on impact.

BabaBlast [244]2 years ago

4 0

<span> Newtons First Law is applied on my egg experiment because it will not move or change it's acceleration until a force acts upon it. Newton's Second Law is applied because of the acceleration caused by natural forces as the egg is plummeting to the earth. And the amount of acceleration the egg has will be largely affected by the amount of force used to hurl the egg to the ground. The Third Law of Motion affects the egg because the amount of force the egg hits the ground with, the ground pushes back with equal and opposite force.</span>

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A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

A 0.15 kg project hits the ground with a speed of 11 m/s. The project comes to rest in 0.015 seconds. What is the net force?

Blababa [14]

Answer:

Explanation:

ACCORDING TO NEWTONS SECOND LAW;

F = mass * acceleration

F = m(v-u/t)

m is the mass = 0.15kg

v is the final velocity = 11m/s

u is the initial velocity = 0m/s

t is the time = 0.015

Substitute;

F = 0.15(11-0)/0.015

F = 0.15(11)/0.015

F = 1.65/0.015

F = 110N

Hence the net force is 110N

4 0

2 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago

Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure

Lynna [10]

Answer:

yes it was a constant speed and the car traveled 10 meters in 20 seconds.

Explanation:

8 0

3 years ago