Answer:

8.57181 s
84.0894561 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Let distance from ground be x
From equation of motion we have

Here, distance covered while the stone is falling will be 

The equation is 
At the ground x = 0

The time taken by the stone to fall to the ground is 8.57181 s

The velocity of the stone when it reaches the ground is 84.0894561 m/s
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
Answer:
The other two small angles are 45° each
Explanation:
Given data in the problem:
The triangle is a right triangle
thus,
one of the angle is 90°
now,
let the other two angles be x and y
thus,
it is given that:
x = 2y - 45°
also in a triangle
sum of all the angles = 180°
thus,
x + y + 90° = 180°
or
x + y = 90°
now, substituting the value of x from the above relation between x and y, we get
2y - 45° + y = 90°
or
3y = 135°
or
y = 45°
also,
x = 2y - 45°
or
x = 2 × (45°) - 45°
or
x = 45°
hence, <u>the other two small angles are 45° each</u>
U = 0, initial vertical velocity
Neglect air resistance, and g = 9.8 m/s².
The time, t, required for the pen to attain a vertical velocity of 19.62 m/s is given by
19.62 m/s = 0 + (9.8 m/s²)*(t s)
t = 19.62/9.8 = 2.00 s
Answer: 2.0 s
Answer:

Explanation:
<u>Given Data:</u>
Momentum = P = 700 kg m/s
Velocity = v = 10 m/s
<u>Required:</u>
Mass = m = ?
<u>Formula:</u>
P = mv
<u>Solution:</u>
m = P / v
m = 700 / 10
m = 70 kg
![\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AnonymousHelper1807</h3>