Answer:
No, neutrons have about the same mass as a proton, but both have more mass than electrons.
Hope this helps a bit,
Flips
<span>you can look at magnesium, it can react with oxygen to form oxides. (chemical) it is malleable and a solid at room temperature. (physical)
</span><span>to measure its density, the mass and volume can be worked out and from this density too. look up the equation, it is quite easy :)
</span><span>physical changes -- it can be melted, and oxidized
</span><span>the chemical changes of oxidation magnesium looses electrons to form oxides, this is a chemical reaction- chemical change..--- use to get the density use (rho) or density D = M/V</span>
<span>The ball clears by 11.79 meters
Let's first determine the horizontal and vertical velocities of the ball.
h = cos(50.0)*23.4 m/s = 0.642788 * 23.4 m/s = 15.04 m/s
v = sin(50.0)*23.4 m/s = 0.766044 * 23.4 m/s = 17.93 m/s
Now determine how many seconds it will take for the ball to get to the goal.
t = 36.0 m / 15.04 m/s = 2.394 s
The height the ball will be at time T is
h = vT - 1/2 A T^2
where
h = height of ball
v = initial vertical velocity
T = time
A = acceleration due to gravity
So plugging into the formula the known values
h = vT - 1/2 A T^2
h = 17.93 m/s * 2.394 s - 1/2 9.8 m/s^2 (2.394 s)^2
h = 42.92 m - 4.9 m/s^2 * 5.731 s^2
h = 42.92 m - 28.0819 m
h = 14.84 m
Since 14.84 m is well above the crossbar's height of 3.05 m, the ball clears. It clears by 14.84 - 3.05 = 11.79 m</span>
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]](https://tex.z-dn.net/?f=r_%7Be%7D%20%3D%20distance%20earth%20to%20the%20astronaut%20%5Bm%5D.%5C%5Cr_%7Bm%7D%20%3D%20distance%20moon%20to%20the%20astronaut%20%5Bm%5D%5C%5Cr_%7Bt%7D%20%3D%20total%20distance%20%3D%203.84%2A10%5E8%5Bm%5D)
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_{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]](https://tex.z-dn.net/?f=F_%7Bm%7D%20%3DG%2A%5Cfrac%7Bm_%7Bm%7D%2Am_%7Ba%7D%20%20%7D%7Br_%7Bm%7D%20%5E%7B2%7D%20%7D%20%5C%5Cwhere%3A%5C%5CG%20%3D%20gravity%20constant%20%3D%206.67%2A10%5E%7B-11%7D%5B%5Cfrac%7BN%2Am%5E%7B2%7D%20%7D%7Bkg%5E%7B2%7D%20%7D%20%5D%20%5C%5Cm_%7Be%7D%3D%20earth%27s%20mass%20%3D%205.98%2A10%5E%7B24%7D%5Bkg%5D%5C%5C%20m_%7Ba%7D%3D%20astronaut%20mass%20%3D%20100%5Bkg%5D%5C%5Cm_%7Bm%7D%3D%20moon%27s%20mass%20%3D%207.36%2A10%5E%7B22%7D%5Bkg%5D)
When we match these equations the masses cancel out as the universal gravitational constant
To solve this equation we have to replace the first equation of related with the distances.
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]](https://tex.z-dn.net/?f=r_%7Bm1%2C2%7D%3D%5Cfrac%7B-b%2B-%20%5Csqrt%7Bb%5E%7B2%7D-4%2Aa%2Ac%20%7D%20%20%7D%7B2%2Aa%7D%5C%5C%20%20where%3A%5C%5Ca%3D80.25%5C%5Cb%3D768%2A10%5E%7B6%7D%20%5C%5Cc%20%3D%20-1.47%2A10%5E%7B17%7D%20%5C%5Creplacing%3A%5C%5Cr_%7Bm1%2C2%7D%3D%5Cfrac%7B-768%2A10%5E%7B6%7D%2B-%20%5Csqrt%7B%28768%2A10%5E%7B6%7D%29%5E%7B2%7D-4%2A80.25%2A%28-1.47%2A10%5E%7B17%7D%29%20%7D%20%20%7D%7B2%2A80.25%7D%5C%5C%5C%5Cr_%7Bm1%7D%3D%2038280860.6%5Bm%5D%20%5C%5Cr_%7Bm2%7D%3D-2.97%2A10%5E%7B17%7D%20%5Bm%5D)
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]](https://tex.z-dn.net/?f=a%20%3D%20G%2A%5Cfrac%7Bm_%7Be%7D%20%7D%7Br_%7Be%7D%20%5E%7B2%7D%20%7D%20%5C%5Ca%20%3D%206.67%2A10%5E%7B11%7D%20%2A%5Cfrac%7B5.98%2A10%5E%7B24%7D%20%7D%7B%28345.72%2A10%5E%7B6%7D%29%5E%7B2%7D%20%20%7D%20%5C%5Ca%3D3.33%2A10%5E%7B19%7D%20%5Bm%2Fs%5E2%5D)
The charge present determines a force to be attractive or repulsive.
The charges acquired by two bodies determines the Force as Attractive Or Repulsive.
Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.
Examples of the experiments and observations:
- On combing hair through a comb and then keeping it close to small pieces of paper shows attraction of paper pieces towards the comb.
This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.
- In both Gravitational Force and Coulomb force, the force remains inversely proportional to the square of the distance following the Inverse Square Law being the Central Force system. This only differs by the fact that in Gravitational Force, masses are used and in Coulomb force, charges are used.
The more the distance between the charges, the less is the Electric Force.
The lesser the distance between the charges, the more is the Electric Force.
If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.
Hence, the charge present determines a force to be attractive or repulsive.
Learn more about Coulomb Force here, brainly.com/question/15451944
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