Answer:
t=1.4hours
Explanation:
The half life is 1hour
At t=0 he has a mass of 4kg
So he want it to be 1kg, so that his weapon can work.
Applying the exponential function of decay
M=Cexp(-kt)
Where,
M is the mass at any time
C is a constant of integration
k is the rate of decay
Given that it has an half life of 1 hours.
Then k is 1
At t =0 the mass is 4kg
Therefore
4=Cexp(0)
C=4
M=4exp(-kt)
Since rate of decay is 1, then k=1
M=4exp(-t)
We need to find t at M=1kg
1=4exp(-t)
1=4exp(-t)
1/4=exp(-t)
0.25=exp(-t)
Take In of both sides
In(0.25)=-t
-1.3863=-t
Then, t=1.386hour
Then it will take about 1.4 hours to get to 1kg.
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
<u>K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J</u>
Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Answer:
Both A and C
Explanation:
I just got it correct on Edg
