Answer:
The value of F= - 830 N
Since the force is negative, it implies direction of the force applied was due south.
Explanation:
Given data:
Mass = 1000-kg
Distance, d = 240 m
Initial velocity, v1 = 20.0 m/s
Final velocity, v2 = 0 (since the car came to rest after brake was applied)
v2²= v1² + 2ad (using one of the equation of motion)
0= 20² + (2 x a x 240)
0= 400 + 480 a
a = - 400/480
a = - 0.83 m/s²
Then, imputing the value of a into
F = ma
F = 1000 kg x ( - 0.83 m/s²)
F= - 830 N
The car was driving toward the north, and since the force is negative, it implies direction of the force applied was due south.
Answer:
q₁ = -6.54 10⁻⁵ C
Explanation:
Force is a vector quantity, but since all charges are on the x-axis, we can work in one dimension, let's apply Newton's second law
F = F₁₂ + F₂₃
the electric force is given by Coulomb's law
F = k q₁q₂ / r₁₂²
let's write the expression for each force
F₂₃ = k q₂ q₃ / r₂₃²
F₂₃ = 9 10⁹ 34.4 10⁻⁶ 72.8 10⁻⁶ / 0.1²
F₂₃ = 2.25 10³ N
F₁₂ = k q₁q₂ / r₁₂²
F₁₂ = 9 10⁹ q₁ 34.4 10⁻⁶ / 0.1²
F₁₂ = q₁ 3,096 10⁷ N
we substitute in the first equation
225 = q₁ 3,096 10⁷ +2.25 10³
q₁ = (225 - 2.25 10³) / 3,096 10⁷
q₁ = -6.54 10⁻⁵ C
Answer:
(a) Power= 207.97 kW
(b) Range= 5768.6 meter
Explanation:
Given,
Mass of bullet, ![m=0.02 kg](https://tex.z-dn.net/?f=m%3D0.02%20kg)
Kinetic energy imparted, ![K=1200 J](https://tex.z-dn.net/?f=K%3D1200%20J)
Length of rifle barrel, ![d=1 m](https://tex.z-dn.net/?f=d%3D1%20m)
(a)
Let the speed of bullet when it leaves the barrel is
.
Kinetic energy, ![K=\frac{1}{2} mv^{2}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E%7B2%7D)
![v=\sqrt{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
![=\sqrt{\frac{2\times1200}{0.02} }](https://tex.z-dn.net/?f=%3D%5Csqrt%7B%5Cfrac%7B2%5Ctimes1200%7D%7B0.02%7D%20%7D)
![=346.4m/s](https://tex.z-dn.net/?f=%3D346.4m%2Fs)
Initial speed of bullet, ![u=0](https://tex.z-dn.net/?f=u%3D0)
The average speed in the barrel,
![=\frac{0+346.4}{2} \\=173.2 m/s](https://tex.z-dn.net/?f=%3D%5Cfrac%7B0%2B346.4%7D%7B2%7D%20%5C%5C%3D173.2%20m%2Fs)
Time taken by bullet to cross the barrel, ![t=\frac{d}{v}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv%7D%20)
![=\frac{1}{173.2}\\ =0.00577 second](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B173.2%7D%5C%5C%20%3D0.00577%20second)
Power,
![=\frac{1200}{0.00577} \\=207.97kW](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1200%7D%7B0.00577%7D%20%5C%5C%3D207.97kW)
(b)
In projectile motion,
Maximum height, ![H_m=\frac{v^2\sin^2\theta}{2g} \\](https://tex.z-dn.net/?f=H_m%3D%5Cfrac%7Bv%5E2%5Csin%5E2%5Ctheta%7D%7B2g%7D%20%5C%5C)
Range, ![R=\frac{v^2\sin2\theta}{g}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7Bv%5E2%5Csin2%5Ctheta%7D%7Bg%7D)
given that, ![H_m=R](https://tex.z-dn.net/?f=H_m%3DR)
then, ![\frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2%5Csin%5E2%5Ctheta%7D%7B2g%7D%3D%5Cfrac%7Bv%5E2%5Csin2%5Ctheta%7D%7Bg%7D%5C%5C%5Csin%5E2%5Ctheta%3D2%5Csin%5Ctheta%5Ccos%5Ctheta%5C%5C%5C%5C%5Ctan%5Ctheta%3D4%5C%5C%5Ctheta%3D%5Ctan%5E-%5E14%5C%5C%5Ctheta%3D75.96%5E0%5C%5CR%3D%5Cfrac%7Bv%5E2%5Csin2%5Ctheta%7D%7Bg%7D%5C%5C%3D%5Cfrac%7B346.4%5E2%5Ctimes%5Csin%282%5Ctimes75.96%29%7D%7B9.8%7D%5C%5C5768.6%20meter)
Answer:
What's the fórmula For power
Explanation:
Power=Work
Time