Answer:
A) 3.13 m/s
B) 5.34 N
C) W = 26.9 J
Explanation:
We are told that the position as a function of time is given by;
x(t) = αt² + βt³
Where;
α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³
Thus;
x(t) = 0.21t² + 0.0204t³
A) Velocity is gotten from the derivative of the displacement.
Thus;
v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)
v(t) = 0.42t + 0.0612t²
v(4.5) = 0.42(4.5) + 0.0612(4.5)²
v(4.5) = 3.1293 m/s ≈ 3.13 m/s
B) acceleration is gotten from the derivative of the velocity
a(t) = v'(t) = 0.42 + 2(0.0612t)
a(4.5) = 0.42 + 2(0.0612 × 4.5)
a(4.5) = 0.9708 m/s²
Force = ma = 5.5 × 0.9708
F = 5.3394 N ≈ 5.34 N
C) Since no friction, work done is kinetic energy.
Thus;
W = ½mv²
W = ½ × 5.5 × 3.1293²
W = 26.9 J
The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
Speed of the proton = 5.02 × 10 ⁶ m /a
Angel of between the velocity and the magnetic force = 60 °
The magnitude of magnetic field B = 0.180 T
The magnitude of the magnetic force on the proton is,
![F = q(v \times B)](https://tex.z-dn.net/?f=F%20%3D%20q%28v%20%5Ctimes%20B%29)
![F = qvB \: sin \: θ](https://tex.z-dn.net/?f=F%20%3D%20qvB%20%5C%3A%20sin%20%5C%3A%20%20%CE%B8)
![F = 1.6 \times 10 ^{ - 19} \times 5.02 \times 10 ^{6} \times 0.180 \times \: sin \: 60°](https://tex.z-dn.net/?f=F%20%3D%201.6%20%5Ctimes%2010%20%5E%7B%20-%2019%7D%20%20%5Ctimes%205.02%20%5Ctimes%2010%20%5E%7B6%7D%20%20%5Ctimes%200.180%20%5Ctimes%20%20%5C%3A%20sin%20%5C%3A%2060%C2%B0)
![= 1.25 \times 10 ^{ - 13} \: N](https://tex.z-dn.net/?f=%20%3D%201.25%20%5Ctimes%2010%20%5E%7B%20-%2013%7D%20%20%5C%3A%20N)
Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.
To know more about magnetic force, refer to the below link:
brainly.com/question/23096032
#SPJ4
D. A solution because it dissolves when mixed with water
Explanation:
Given that,
Wavelength of the light, ![\lambda=691\ nm=691\times 10^{-9}\ m](https://tex.z-dn.net/?f=%5Clambda%3D691%5C%20nm%3D691%5Ctimes%2010%5E%7B-9%7D%5C%20m)
(a) Slit width, ![a=3.8\times 10^{-4}\ m](https://tex.z-dn.net/?f=a%3D3.8%5Ctimes%2010%5E%7B-4%7D%5C%20m)
The angle that locates the first dark fringe is given by :
![sin\theta=\dfrac{\lambda}{a}](https://tex.z-dn.net/?f=sin%5Ctheta%3D%5Cdfrac%7B%5Clambda%7D%7Ba%7D)
![sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}](https://tex.z-dn.net/?f=sin%5Ctheta%3D%5Cdfrac%7B691%5Ctimes%2010%5E%7B-9%7D%7D%7B3.8%5Ctimes%2010%5E%7B-4%7D%7D)
![\theta=0.104^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D0.104%5E%7B%5Ccirc%7D)
(b) Slit width, ![a=3.8\times 10^{-6}\ m](https://tex.z-dn.net/?f=a%3D3.8%5Ctimes%2010%5E%7B-6%7D%5C%20m)
The angle that locates the first dark fringe is given by :
![sin\theta=\dfrac{\lambda}{a}](https://tex.z-dn.net/?f=sin%5Ctheta%3D%5Cdfrac%7B%5Clambda%7D%7Ba%7D)
![sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}](https://tex.z-dn.net/?f=sin%5Ctheta%3D%5Cdfrac%7B691%5Ctimes%2010%5E%7B-9%7D%7D%7B3.8%5Ctimes%2010%5E%7B-6%7D%7D)
![\theta=10.47^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D10.47%5E%7B%5Ccirc%7D)
Hence, this is the required solution.
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880
Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000
Lost Energy = 5880 - 3000 = 2880 J
In short, Your Answer would be 2880 Joules
Hope this helps!