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3 years ago

What is an example of strong nuclear force

Physics

2 answers:

LuckyWell [14K]3 years ago

6 0

Answer: Example of strong nuclear force is the force acting between nucleons inside the nucleus .

Explanation: This force holds the nucleus together . The range of strong nuclear force is very small and is effective only in the range of 1 fermi meter . This force can exist between protons and protons, neutrons and protons or neutrons and neutrons . These forces are 10 million times stronger than chemical binding force .

Gnom [1K]3 years ago

3 0

Dropping a nuke on another country.

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What's the gravity?

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Answer:

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Explanation:

what is gravity made of?

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2 years ago

003 (part 1 of 2) 10.0 points

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2 years ago

Planck’s constant is a ratio between which two quantities?

Readme [11.4K]

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Explanation:

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2 years ago

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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

Jill can use a force of 12.0 N to lift a single box up 5.00 m. How many of these boxes must she lift it in a minute to use 60.0

Mama L [17]

Answer:

60 boxes

Explanation:

The work done by lifting a single box is equal to the force applied (the weight of the box) times the displacement of the box:

$W_1 = Fd=(12.0 N)(5.00 m)=60 J$

Power is related to the work done by the equation:

$P=\frac{W}{t}$

where W is the work done and t is the time. In this problem, we are told that the power used is P=60.0 W, while the time taken is t = 1 min = 60 s, so the total work done must be

$W=Pt=(60.0 W)(60 s)=3600 J$

Therefore, the number of boxes that she has to lift in order to use this power is the total work divided by the work done in lifting each box:

$N=\frac{W}{W_1}=\frac{3,600 J}{60 J}=60$

5 0

3 years ago