Answer:
Explanation:
Given the height reached by a balloon after t sec modeled by the equation
h=1/2t²+1/2t
a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t
If h(t)=1/2t²+1/2t
h(40) = 1/2(40)²+1/2 (40)
h(40) = 1600/2 + 40/2
h(40) = 800 + 20
h(40) = 820 feet
The height of the balloon after 40 secs is 820 feet
b) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
when v = 0sec
v(0) = 0 + 1/2
v(0) = 1/2 ft/sec
at v = 30secs
v(30) = 30 + 1/2
v(30) = 30 1/2 ft/sec
average velocity = v(30) - v(0)
average velocity = 30 1/2 - 1/2
average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec
c) Velocity is the change of displacement of a body with respect to time.
v = dh/dt
v(t) = 2(1/2)t²⁻¹ + 1/2
v(t) = t + 1/2
The velocity of the balloon after 30secs will be;
v(30) = 30+1/2
v(30) = 30.5ft/sec
The velocity of the balloon after 30 secs is 30.5 feet/sec
Because the gravitational force, which points downward, is perfectly balanced by the normal reaction of the floor of the bowling lane, which points upward. The two forces are equal in magnitude, so the net force acting vertically on the bowling ball is zero, therefore there is no acceleration along this direction. Moreover, since the ball is moving in the horizontal direction, the gravitational force has no component along this direction, so it does not change the velocity of the ball.
Answer:
2.068 x 10^6 m / s
Explanation:
radius, r = 5.92 x 10^-11 m
mass of electron, m = 9.1 x 10^-31 kg
charge of electron, q = 1.6 x 10^-19 C
As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.
centripetal force = 
Electrostatic force = 
where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2
So, balancing both the forces we get
v = 2.068 x 10^6 m / s
Thus, the speed of the electron is give by 2.068 x 10^6 m / s.
Answer:
(a) 6.567 * 10^15 rev/s or hertz
(b) 8.21 * 10^14 rev/s or hertz
Explanation:
Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)
Where Fn is frequency at all levels of n.
Z = 1 (nucleus)
e = 1.6 * 10^-19c
m = 9.1 * 10^-31 kg
h = 6.62 * 10-34
K = 9 * 10^9 Nm2/c2
(a) for groundstate n = 1
Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s
(b) first excited state
n = 1
We multiple the groundstate answer by 1/n^3
6.567 * 10^15 rev/s/ 2^3
F2 = 8.2 * 10^ 14 rev/s
<span>1.7 rad/s
The key thing here is conservation of angular momentum. The system as a whole will retain the same angular momentum. The initial velocity is 1.7 rad/s. As the person walks closer to the center of the spinning disk, the speed will increase. But I'm not going to bother calculating by how much. Just remember the speed will increase. And then as the person walks back out to the rim to the same distance that the person originally started, the speed will decrease. But during the entire walk, the total angular momentum remained constant. And since the initial mass distribution matches the final mass distribution, the final angular speed will match the initial angular speed.</span>
