Answer:
67.17 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 32 m/s
Angle of projection (θ) = 20°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =?
The range can be obtained as follow:
R = u²Sine 2θ / g
R = 32² × Sine (2×20) / 9.8
R = 1024 × Sine 40 / 9.8
R = (1024 × 0.6428) / 9.8
R = 67.17 m
Thus, the range of the ball is 67.17 m
Answer:0.318 revolutions
Explanation:
Given
Initially Propeller is at rest i.e. ![\omega _0=0 rad/s](https://tex.z-dn.net/?f=%5Comega%20_0%3D0%20rad%2Fs)
after ![t=10 s](https://tex.z-dn.net/?f=t%3D10%20s)
![\omega =10 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D10%20rad%2Fs)
using ![\omega =\omega _0+\alpha t](https://tex.z-dn.net/?f=%5Comega%20%3D%5Comega%20_0%2B%5Calpha%20t)
![10=0+\alpha \cdot 10](https://tex.z-dn.net/?f=10%3D0%2B%5Calpha%20%5Ccdot%2010)
![\alpha =1 rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D1%20rad%2Fs%5E2)
Revolutions turned in 2 s
![\theta =\omega _0t+\frac{\alpha t^2}{2}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%5Comega%20_0t%2B%5Cfrac%7B%5Calpha%20t%5E2%7D%7B2%7D)
![\theta =0+\frac{1\times 2^2}{2}](https://tex.z-dn.net/?f=%5Ctheta%20%3D0%2B%5Cfrac%7B1%5Ctimes%202%5E2%7D%7B2%7D)
![\theta =2 rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D2%20rad)
To get revolution ![\frac{\theta }{2\pi }](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctheta%20%7D%7B2%5Cpi%20%7D)
=![\frac{2}{2\pi}=0.318\ revolutions](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B2%5Cpi%7D%3D0.318%5C%20revolutions)
<u>Answer:</u>
<em>The distance between Pluto and Charon is
</em>
<u>Explanation:</u>
Force of gravitation between two objects
Where
are the masses of the objects,r is the distance between the objects
G is the universal gravitational constant=
Here mass of pluto = ![1.3 \times 10^2^2 kg](https://tex.z-dn.net/?f=1.3%20%5Ctimes%2010%5E2%5E2%20%20kg)
mass of charon =
Force of gravitation
![F_g= \frac {G(M_1 M_2 )}{r^2}](https://tex.z-dn.net/?f=F_g%3D%20%5Cfrac%20%7BG%28M_1%20M_2%20%29%7D%7Br%5E2%7D)
![r^2= \frac {G(M_1 M_2 )}{F_g }](https://tex.z-dn.net/?f=r%5E2%3D%20%5Cfrac%20%7BG%28M_1%20M_2%20%29%7D%7BF_g%20%7D)
![r= \sqrt \frac{(G(M_1 M_2 )}{F_g}](https://tex.z-dn.net/?f=r%3D%20%5Csqrt%20%5Cfrac%7B%28G%28M_1%20M_2%20%29%7D%7BF_g%7D)
![=\sqrt \frac {((6.674\times 10^-^1^1) \times 1.3 \times 10^2^2 \times 1.6\times 10^2^1)}{(3.61 \times 10^1^8 )}](https://tex.z-dn.net/?f=%3D%5Csqrt%20%5Cfrac%20%7B%28%286.674%5Ctimes%2010%5E-%5E1%5E1%29%20%5Ctimes%201.3%20%5Ctimes%2010%5E2%5E2%20%5Ctimes%201.6%5Ctimes%2010%5E2%5E1%29%7D%7B%283.61%20%5Ctimes%2010%5E1%5E8%20%29%7D)
=![\sqrt \frac {(13.88*10^32)}{(3.61*10^18)}](https://tex.z-dn.net/?f=%5Csqrt%20%5Cfrac%20%7B%2813.88%2A10%5E32%29%7D%7B%283.61%2A10%5E18%29%7D)
=
Answer:
(a) The magnitude of the force electric force F = k ×(+Q) × (-2·Q)/r₀² = -2·Q²/(r₀²) which can be written as follows;
![F = \dfrac{k \times (+ Q) \times (-2\cdot Q)}{r_0^2} = -\dfrac{k \times 2 \times Q^2}{r_0^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7Bk%20%5Ctimes%20%28%2B%20Q%29%20%5Ctimes%20%28-2%5Ccdot%20Q%29%7D%7Br_0%5E2%7D%20%3D%20%20-%5Cdfrac%7Bk%20%5Ctimes%202%20%5Ctimes%20Q%5E2%7D%7Br_0%5E2%7D)
Given that the charge of Y is twice the charge on X, we have the charge X will move towards the charge Y which is the +x direction
(b) The force, F, acting between the charges is given as follows;
![F = \dfrac{k \times Q_1 \times Q_2}{r_0^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7Bk%20%5Ctimes%20Q_1%20%5Ctimes%20Q_2%7D%7Br_0%5E2%7D)
When, Q₁ = X = -4Q, Q₂ = Y = -2Q, and r₀ = 2·r₀ we have;
![F = \dfrac{k \times (-4\cdot Q) \times (-2\cdot Q)}{(2 \cdot r_0)^2} = \dfrac{k \times 8 \times Q^2}{4 \cdot r_0^2} = \dfrac{k \times 2 \times Q^2}{r_0^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7Bk%20%5Ctimes%20%28-4%5Ccdot%20Q%29%20%5Ctimes%20%28-2%5Ccdot%20Q%29%7D%7B%282%20%5Ccdot%20r_0%29%5E2%7D%20%3D%20%20%5Cdfrac%7Bk%20%5Ctimes%208%20%5Ctimes%20Q%5E2%7D%7B4%20%5Ccdot%20r_0%5E2%7D%20%3D%20%5Cdfrac%7Bk%20%5Ctimes%202%20%5Ctimes%20Q%5E2%7D%7Br_0%5E2%7D)
Therefore, the electric force exerted on object Y by the object X is the same and acts in an opposite direction to the force of X on Y in (a)
The net charge of the two objects in part (a) is Q - 2Q = -Q
The net charge of the two objects in part (b) is -4Q - 2Q = -6Q
Therefore, the net charge increases by a factor of 6 in part (b)
Explanation: