Answer:
3) D: 31 m/s
4) D: 84.84 metres
Explanation:
3) Initial velocity along the x-axis is;
v_x = v_o•cos θ
Initial velocity along the y-axis is;
v_y = v_o•sin θ
Plugging in the relevant values, we have;
v_x = 31 cos 60
v_x = 31 × 0.5
v_x = 15.5 m/s
Similarly,
v_y = 31 sin 60
v_y = 31 × 0.8660
v_y = 26.85 m/s
Thus, magnitude of the initial velocity is;
v = √(15.5² + 26.85²)
v ≈ 31 m/s
4) Formula for horizontal range is;
R = (v² sin 2θ)/g
R = (31² × sin (2 × 60))/9.81
R = 84.84 m
Answer: final Velocity v = 10.2m/s
Explanation:
Final speed v(t) is given as
v(t) = u + at .......1
Where; u = the initial speed
a = acceleration
t = time taken
The total distance travelled d is given as
d = ut + 1/2(at^2)
Given
d = 5.0m
u = 2.0m
a = g = 10m/s2 (acceleration due to gravity)
Substituting into the equation above we have
5 = 2t + 5t^2
5t^2 +2t -5 = 0
Applying the quadratic formula. We have;
t = 0.82s & t = -1.22s
t cannot be negative
t = 0.82s
From equation 1 above
v = 2.0m/s + 10(0.82)m/s
v = 10.2m/s
True the elements are ordered in the atomic number
The correct answer should be C
Answer:
The range of powers is 
Explanation:
From the question we are told that
The far point of the left eye is 
The near point of the left eye is 
The near point with the glasses on is 
From these parameter we can see that with the glass on that for near point the
Object distance would be 
Image distance would be 
To obtain the focal length we would apply the lens formula which is mathematically represented as

substituting values


converting to meters


Generally the power of the lens is mathematically represented as

Substituting values


From these parameter we can see that with the glass on that for far point the
Object distance would be 
Image distance would be 
To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

substituting values


converting to meters

Generally the power of the lens is mathematically represented as

Substituting values


This implies that the range of powers of the lens in his glass is
