A science student observes three lines in an emission spectrum, one red, one yellow, and one blue. Which one was caused by an el
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Answer:
X-Positions: Y-Positions
x(0) = 0 y(0) = 0
x(2) = 120 m y(2) = 19.6 m
x(4) = 240 m y(4) = 78.4 m
x(6) = 360 m y(6) = 176.4 m
x(8) = 480 m y(8) = 313 m
x(10) = 600m y (10) = 490 m
Explanation:
X-Positions
- First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
- After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
- So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:
- It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.
Y-Positions
- We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
- As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
- At any moment, it is subject to the acceleration of gravity, g.
- As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:
- Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
- y(2) = 2* 9.8 m/s2 = 19.6 m
- y(4) = 8* 9.8 m/s2 = 78.4 m
- y(6) = 18*9.8 m/s2 = 176.4 m
- y(8) = 32*9.8 m/s2 = 313.6 m
- y(10)= 50 * 9.8 m/s2 = 490.0 m
Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r =
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = = 0.0328 m, or 3.28 cm.
Refer to the diagram shown below.
Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s
The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.
Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)
Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)
Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s
The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.
Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)
Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)