Answer:
option (b) 4900 N
Explanation:
m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R
F = G Me x m / (R + h)^2 
F = G Me x m / 2R^2
F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2
F = 4900 N
 
        
             
        
        
        
Answer:
y = 0.0233 m
Explanation:
In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:
Δx = λL/d
where,
Δx = distance between fringes
λ = wavelength of light
L = Distance between screen and slits
d = Slit Separation
Now, for initial case:
λ = 425 nm = 4.25  x 10⁻⁷ m
y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m
Therefore,
8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d
L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)
L/d = 2.08 x 10⁴ 
using this for λ = 560 nm = 5.6 x 10⁻⁷ m:
Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)
Δx = 0.0116 m
and,
y = 2Δx
y = (2)(0.0116 m)
<u>y = 0.0233 m</u>
 
        
             
        
        
        
Answer:
Explanation:
Given
Distance between two loud speakers 
Distance of person from one speaker 
Distance of person from second speaker 
Path difference between the waves is given by

for destructive interference m=0 I.e.




frequency is given by 

where 

For next frequency which will cause destructive interference is
i.e.  and
 and 


frequency corresponding to this is 

for 


Frequency corresponding to this wavelength

 
                         
 
        
             
        
        
        
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

<h3>Further explanation</h3>
Let's recall Elastic Potential Energy formula as follows:

where:
<em>Ep = elastic potential energy ( J )</em>
<em>k = spring constant ( N/m )</em>
<em>x = spring extension ( compression ) ( m )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
<u>Asked:</u>
kinetic energy = Ek = ?
<u>Solution:</u>
<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>






<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>







<h3>Learn more</h3>

<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity
 
        
                    
             
        
        
        
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres 
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C. 
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region? 
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula 
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately 
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2 
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres