All categories

3 years ago

Can laws in science change

2 answers:

6 0

Laws reflect scientific knowledge that experiments have repeatedly verified (and never falsified). Their accuracy does not change when new theories are worked out, but rather the scope of application, since the equation (if any) representing the law does not change.

Verizon [17]3 years ago

4 0

No, Scientific Laws cannot change. Laws have been proven, while theories on the other hand, cannot. Theories can be changed over time, but Laws are “set in stone”.

You might be interested in

Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

<u />

The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

3 years ago

Point charge A is located at point A and point charge B is at point B. Points A and B are separated by a distance r. To determin

fredd [130]

Answer:

B. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges.

Explanation:

Total electric potential is the sum of all the electric potential. And because electric potential is a scalar quantity you have to account for the signs.

3 0

3 years ago

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

Tju [1.3M]

Answer:

$_T}=24.57Nm$

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

$_T } = Fr$

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

$_T}=(0.795)*(30.9)$

$_T}=24.57Nm$

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

$_T}=I \omega$

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

$I = mr^2$

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

$_T}=I \omega$

Making angular acceleration the subject of the formula; we have;

$\omega = \frac{_T}{I}$

ω = $\frac{24.57}{707.51}$

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a = ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0

3 years ago

What happens to particle spacing when temperature increase, What we call this process?

Slav-nsk [51]

Answer:

The atoms start vibrating really fast, making the space between them larger. It's an increase in Kinetic Energy.

6 0

4 years ago

A 1400 kg car moving +13.7 m/s makes an elastic collision with a 3200 kg truck, initially at rest. What is the velocity of the c

marishachu [46]

Answer:

-13.7m/s

Explanation:

If the car has a mass of 1400kg with a speed of +13.7m/s -> hits [email protected] rest and bounces with an elastic collision:

fm = -fm

Therefore, the speed the car will possess will be at a negative state.

= -13.7m/s

6 0

3 years ago