The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
<h3>What is Columb's law?</h3>
The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Similar charges repel each other, whereas charges that are opposed attract each other.
Given data;
Electric force,F = 9 × 10 ⁻⁹ N
Distance between charges,d = 7 × 10⁻⁴ m
Chrge,q₁ = q₂ =q C
From Columb's law;
![\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9} = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\ q = 7 \times 10^{-13} \ C](https://tex.z-dn.net/?f=%5Crm%20F%20%3D%20K%20%5Cfrac%7Bq_1q_2%7D%7Bd%5E2%7D%20%5C%5C%5C%5C%209%20%5Ctimes%2010%5E%7B-9%7D%20%20%3D%209%20%5Ctimes%2010%5E9%20%5Cfrac%7Bq%5E2%7D%7B%287%20%5Ctimes%2010%5E%7B-4%7D%29%5E2%7D%20%5C%5C%5C%5C%20q%5E2%20%3D%204.9%20%5Ctimes%2010%5E%7B-25%7D%20%5C%5C%5C%5C%20%20q%20%3D%207%20%5Ctimes%2010%5E%7B-13%7D%20%5C%20C)
Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C
To learn more about Columb's law refer to the link;
brainly.com/question/1616890
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Impulse = Change in momentum.
The ball was moving with a momentum of 0.45 * 22 = 9.9
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9
Answer:
![2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
29010.53917 m
Explanation:
= Density of asteroid = 2 g/cm³
V = Volume
d = Diameter = 10 km
r = Radius = ![\dfrac{d}{2}=\dfrac{10}{2}=5\ km](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%7D%7B2%7D%3D%5Cdfrac%7B10%7D%7B2%7D%3D5%5C%20km)
v = Velocity = 11 km/s
= Heat vaporization of water = ![2.26\times 10^6\ J/kg](https://tex.z-dn.net/?f=2.26%5Ctimes%2010%5E6%5C%20J%2Fkg)
= Change in temperature = 100-20
Mass is given by
![m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg](https://tex.z-dn.net/?f=m%3D%5Crho%20V%5C%5C%5CRightarrow%20m%3D%5Crho%5Cdfrac%7B4%7D%7B3%7D%5Cpi%20r%5E3%5C%5C%5CRightarrow%20m%3D2000%5Cdfrac%7B4%7D%7B3%7D%5Ctimes%20%5Cpi%5Ctimes%205000%5E3%5C%5C%5CRightarrow%20m%3D1.0472%5Ctimes%2010%5E%7B15%7D%5C%20kg)
The kinetic energy is
![K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J](https://tex.z-dn.net/?f=K%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%5C%5C%5CRightarrow%20K%3D%5Cdfrac%7B1%7D%7B2%7D1.0472%5Ctimes%2010%5E%7B15%7D%5Ctimes%2011000%5E2%5C%5C%5CRightarrow%20K%3D6.33556%5Ctimes%2010%5E%7B22%7D%5C%20J)
Heat is given by
![Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%2BmH_v%5C%5C%5CRightarrow%206.33556%5Ctimes%2010%5E%7B22%7D%3Dm%5Ctimes%20%284186%5Ctimes%20%28100-20%29%2B2.26%5Ctimes%2010%5E6%29%5C%5C%5CRightarrow%20m%3D%5Cdfrac%7B%206.33556%5Ctimes%2010%5E%7B22%7D%7D%7B4186%5Ctimes%20%28100-20%29%2B2.26%5Ctimes%2010%5E6%7D%5C%5C%5CRightarrow%20m%3D2.44156%5Ctimes%2010%5E%7B16%7D%5C%20kg)
Mass of water is ![2.44156\times 10^{16}\ kg](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B16%7D%5C%20kg)
Volume is ![\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=%5Cdfrac%7B2.44156%5Ctimes%2010%5E%7B16%7D%7D%7B10%5E3%7D%3D2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
Amount of water is ![2.44156\times 10^{13}\ m^3](https://tex.z-dn.net/?f=2.44156%5Ctimes%2010%5E%7B13%7D%5C%20m%5E3)
If it were a cube
![h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m](https://tex.z-dn.net/?f=h%3DV%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20h%3D%282.44156%5Ctimes%2010%5E%7B13%7D%29%5E%7B%5Cdfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20h%3D29010.53917%5C%20m)
The height of the water would be 29010.53917 m
Answer:
(a) Magnitude of Vector = 207.73 m
(b) Direction = 65.48°
Explanation:
(a)
The formula to find out the magnitude of a resultant vector with the help of its x and y components is given as follows:
![Magnitude\ of\ Vector = \sqrt{d_{x}^{2} + d_{y}^{2}} = \sqrt{(86.2\ m)^{2} + (189\ m)^{2}}\\\\](https://tex.z-dn.net/?f=Magnitude%5C%20of%5C%20Vector%20%3D%20%5Csqrt%7Bd_%7Bx%7D%5E%7B2%7D%20%2B%20d_%7By%7D%5E%7B2%7D%7D%20%3D%20%5Csqrt%7B%2886.2%5C%20m%29%5E%7B2%7D%20%2B%20%28189%5C%20m%29%5E%7B2%7D%7D%5C%5C%5C%5C)
<u>Magnitude of Vector = 207.73 m</u>
(b)
For the direction of the vector we have the formula:
![Direction = tan^{-1}(\frac{y}{x})\\\\Direction = tan^{-1}(\frac{189\ m}{86.2\ m})\\\\](https://tex.z-dn.net/?f=Direction%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7By%7D%7Bx%7D%29%5C%5C%5C%5CDirection%20%3D%20tan%5E%7B-1%7D%28%5Cfrac%7B189%5C%20m%7D%7B86.2%5C%20m%7D%29%5C%5C%5C%5C)
<u>Direction = 65.48°</u>