The period of a simple pendulum is given by:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
Answer:
Explanation:
Given that,
The initial speed of a car, u = 0
Time, t = 18 s
Distance, d = 390 m
We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.
or
So, the acceleration of the car is 
.
Answer:
Latent heatnof fusion = 417.5 J
Explanation:
Specific latent heat of fusion of water is 334kJ.kg-1.
The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.
The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.
Latent heat = ML
Latent heat= 1.25 kg * 334kJ.kg-1
Latent heat = 1.25*334 *(J/kg)*kg
Latent heat = 417.5 J
Answer:
The distance covered by the balloon is 47.52 meters.
Explanation:
Given that,
Initial speed of the balloon, u = 1.14 m/
Let us assumed we need to find the distance covered by the balloon after t = 3 second. Let d is the distance covered by the balloon. It can be given by :
Here, a = g
d = 47.52 meters
So, the distance covered by the balloon is 47.52 meters. Hence, this is the required solution.
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ + 
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis
