Answer:
64 m
Explanation:
Using the following symbols
x: distance
v: velocity
a: constant acceleration
t: time
v₀: initial velocity
x₀: initial position
The equations of motion for a constant acceleration are given by:
(1) x = 0.5at²+v₀t+x₀
(2) v = at+v₀
From equation (2) you can calculate the time t it takes the car to come to a complete stop.
(3) t = (v-v₀)/a
Now you plug equation (3) in equation(1):
(4) x = 0.5a((v-v₀)/a)²+v₀((v-v₀)/a)+x₀
In equation (4) the position x is the only unknown.
Answer:
The value is 
Explanation:
From the question we are told that
The width of the slit is 
The distance of the screen from the slit is D = 1.25 m
The width of the central maximum is 
Generally the width of the central maximum is mathematically represented as
Here m is the order of the fringe and given that we are considering the central maximum, the order will be m = 1 because the with of the central maximum separate's the and first maxima
So
=> 
=> 
=>
Answer:
Kinetic Energy = 0
Explanation:
No matter if energy is totally conserved during the fall or when it hits the ground, the kinetic energy of the silver sphere will be zero because the velocity component in 
becomes zero, the over all energy becomes zero.
Answer:
C has 5 electrons
Explanation:
Given:
The data acquired from the experiment performed by Millikan:
Q_a = 3.20 x10^{-19} C
Q_b = 4.80 x10^{-19} C
Q_c = 8.00 x 10^{-19} C
Q_d = 9.60 x 10^{-19} C
Find:
How many Electrons were present in drop C
Solution:
It is known that the charge of an electron e = 1.602 *10^-19 C / electron.
Hence the number of electrons n in drop C will be:
n = Q_c / e
n = 8.00 x 10^{-19} / 1.602*10^-19
n = 4.99 = 5 electrons
Answer: The drop C contains 5 electrons.
<span>Answer:
Chlorine has 17 electrons therefore for 1+ and 2+ we need the atoms that have 18 and 19 electrons these are argon and potassium: Ar+ and K 2+
For 1- and 2- we need the atoms that have 16 and 15 electrons and these are sulfur and phosphorus, S- and P 2-.
Note that + ions imply the loss of electrons and - ions equate to the gain of electrons.</span>
