Answer:
The gazelles top speed is 27.3 m/s.
Explanation:
Given that,
Acceleration = 4.2 m/s²
Time = 6.5 s
Suppose we need to find the gazelles top speed
The speed is equal to the product of acceleration and time.
We need to calculate the gazelles top speed
Using formula of speed

Where, v = speed
a = acceleration
t = time
Put the value into the formula


Hence, The gazelles top speed is 27.3 m/s.
 
        
             
        
        
        
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:
5.959 m/s
Explanation:
m = Mass of gymnast
u = Initial velocity
v = Final velocity
 = Initial height
 = Initial height
 = Final height
 = Final height
From conservation of Energy



Velocity of gymnast at bottom of swing is 5.959 m/s
 
        
             
        
        
        
784 Newtons or 176.37 lbs
        
             
        
        
        
Answer:
 mph
 mph
Explanation:
 = Speed of bird in still air
 = Speed of bird in still air 
 = Speed of wind = 44 mph
 = Speed of wind = 44 mph
Consider the motion of the bird with the wind
 = distance traveled with the wind = 9292 mi
 = distance traveled with the wind = 9292 mi
 = time taken to travel the distance with wind
 = time taken to travel the distance with wind 
Time taken to travel the distance with wind is given as 

 eq-1
                              eq-1
Consider the motion of the bird with the wind
 = distance traveled against the wind = 6060 mi
 = distance traveled against the wind = 6060 mi
 = time taken to travel the distance against wind
 = time taken to travel the distance against wind 
Time taken to travel the distance against wind is given as 

 eq-2
                              eq-2
As per the question,
Time taken with the wind = Time taken against the wind 





 mph
 mph