Answer:
No. of laps of Hannah are 7 (approx).
Solution:
According to the question:
The total distance to be covered, D = 5000 m
The distance for each lap, x = 400 m
Time taken by Kara, 
Time taken by Hannah, 
Now, the speed of Kara and Hannah can be calculated respectively as:


Time taken in each lap is given by:



t = 500 s
So, Distance covered by Hannah in 't' sec is given by:


No. of laps taken by Hannah when she passes Kara:

≈ 7 laps
The speed is 0.956 m / s.
<u>Explanation</u>:
The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.
K.E = 1/2
m

where K.E represents the kinetic energy,
m represents the mass,
v represents the velocity.
K.E = 1/2
m

1.10
10^42 = 1/2
3.26
10^31

= (1.10
10^42
2) / (3.26
10^31)
v = 0.956 m / s.
Answer:
The ball was in air for 3.896 s
Explanation:
given,
g = 9.8 m/s², acceleration due to gravity,
If the launch angle is 45°, the horizontal range will be maximum.
The horizontal and vertical launch velocities are equal, and each is equal to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The time to attain maximum height is one half of the time of flight.
v = u + at ∵ v = 0 (max. height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The time of flight is twice of the maximum height time
2 t₁ = 3.896 s
The horizontal distance traveled is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in air for 3.896 s
Answer:
A. 261.6 hz.
B. 0.656 m.
Explanation:
A.
When yhe tube is open at one end and closed at the other,
F1 = V/4*L
Where,
F1 = fundamental frequency
V = velocity
L = length of the tube
When the tube is open at both ends,
F'1 = V/2*L
Where
F'1 = the new fundamental frequency
Therefore,
V/2*L x V/4*L
F'1 = 2 * F1
= 2 * 130.8
= 261.6 hz.
B.
F1 = V/4*L
Or
F'1 = V/2*L
Given:
V = 343 m/s
F1 = 130.8
L = 343/(4 * 130.8)
= 0.656 m.