T is in seconds (s)
<span>2pi is dimensionless </span>
<span>L is in meters (m) </span>
<span>g is in meters per second squared (m/s^2) </span>
<span>so you can write the equation for the period of the simple pendulum in its units... </span>
<span>s=sqrt(m/(m/s^2)) </span>
<span>simplify</span>
<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>
<span>s=sqrt(s^2) </span>
<span>s=s </span>
<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
Answer:
The potential energy of the rock = 10.5 kN
Explanation:
Mass of rock = 25 kg
Acceleration due to gravity = 10 m/s²
Height = 42 m
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity and h is the height.
PE = 25 x 10 x 42 = 10500 N = 10.5 kN
The potential energy of the rock = 10.5 kN
Answer:
35 km/hr
Explanation:
Average speed = (total of the speed)/(the sets of speeds given)
Direction does not matter in this instance since speed is only magnitude,
Average speed = (30 + 40)/2
Average speed = 70 ÷ 2
= 35 km/hr
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
