Answer:
T=7.4 N hence T<30 N
Explanation:
The figure is likely to be similar to the one attached. Writing the equation for forces we have
F-T=Fa/g where F is the force, T is tension, a is acceleration and g is acceleration due to gravity. Substituting the figures we have the first equation as
30 N - T = (30/9.81)a
Also, we know that T=F*a/g and substituting 10N for F we obtain the second equation as
T = (10/9.81)a
Adding the first and second equations we obtain
30 = 4.077471967
a Hence
and T=a hence
T is approximately 7.4 N
Answer:
426.84 m
Explanation:
initial velocity u = 0
time t = 3.3 s
distance travelled s = 53.4 m
acceleration due to gravity = g
s = ut + 1/2 g t²
53.4 = 0 + 1/2 g x 3.3²
g = 9.8 m /s²
For the whole length of fall
distance travelled = h
total time = 6.6 + 3.3 = 9.9 s
h = ut + 1/2 g t²
u again = 0
h = .5 x 9.8 x 9.9²
= 480.24 m
distance travelled in last 6.6 s
= 480.24 - 53.4
= 426.84 m
Angle, θ2 at which the light leaves mirror 2 is 56°
<u>Explanation:</u>
Given-
θ1 = 64°
So, α will also be 64°
According to the figure:
α + β = 90°
So,
β = 90° - α
= 90° - 64°
= 26°
β + γ + 120° = 180°
γ = 180° - 120° - β
γ = 180° - 120° - 26°
γ = 34°
γ + δ = 90°
δ = 90° - γ
δ = 90° - 34°
δ = 56°
According to the law of reflection,
angle of incidence = angle of reflection
θ2 = δ = 56°
Therefore, angle θ2 at which the light leaves mirror 2 is 56°
Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
