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deff fn [24]
3 years ago
10

Sir Dudley the Dextrous is loading cannons on top of the castle's tallest towers in preparation for an expected attack. He slips

, and a cannonball of mass 14.0 kg 14.0 kg rolls to the edge of the tower, which has a perfectly horizontal surface, and falls to the ground below. It takes the cannonball 3.15 s 3.15 s to fall to the ground, and air resistance is negligible. How much work is done by gravity on the falling cannonball?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

6667.9 J

Explanation:

We are given that

Mass of cannon.m=14 kg

Time,t=3.15 s

Initial velocity,u=0

We have to find the work done by gravity on the falling cannon ball.

g=9.8 m/s^2

h=ut+\frac{1}{2}gt^2

Substitute the values

h=0+\frac{1}{2}\times 9.8\times (3.15)^2=48.6 m

Work done,W=mgh

Using the formula

W=14\times 9.8\times 48.6=6667.9 J

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8 0
3 years ago
A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the
Vikki [24]

We can use the formula of the moment of inertia given by:

r\cdot F=I\alpha

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}

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8 0
1 year ago
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be
Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂²  → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT  , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² %  = 0,036 %

4 0
3 years ago
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