Answer:
Collect more fuel than you think you can use; you may need more than you estimate. Pile fine twigs, grass, or bark shavings loosely as a base.
Answer:
Explanation:
Given that,
The length of the beam L = 3.10m
The mass of the steam beam 
= 430kg
The mass of worker 
= 69.0kg
The distance from the fixed point to centre of gravity of beam = 
and our length of beam is 3.10m
so the distance from the fixed point to centre of gravity of beam is
Then the net torque is
is the weight of steel rod
is the weight of the worker
Torque can now be calculated
≅ 9169Nm
<h3>Therefore,the magnitude of the torque is 9169Nm</h3>
Answer:
The minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
Anywhere else on the loop, the speed needed to keep the car in the loop is obtained from the force that keeps the body in circular motion around the loop which has to just match the force of gravity on the car. (Given that frictional force = 0)
mv²/r = mg
v² = gr = 9.8 × 25 = 245
v = 15.65 m/s
But at the top, the change in kinetic energy of the car must match the potential energy at the very top of the loop-the-loop
Change in kinetic energy = potential energy at the top
Change in kinetic energy = (mv₂² - mv₁²)/2
v₁ = velocity required to stay in the loop anywhere else = 15.65 m/s
v₂ = minimum velocity the car must have at the top of the loop to not fall
And potential energy at the top of the loop = mgh (where h = the diameter of the loop)
(mv₂² - mv₁²)/2 = mgh
(v₂² - v₁²) = 2gh
(v₂² - (15.65)²) = 2×9.8×50
v₂² - 245 = 980
v₂² = 1225
v₂ = 35 m/s
Hence, the minimum speed the car must have at the top of the loop to not fall = 35 m/s
Explanation:
First find the displacement in the x direction:
dₓ = 449 cos 66° + 1112 cos 169° + 1571 cos 26°
dₓ = 182.6 − 1091.6 + 1412
dₓ = 503 km
Next, find the displacement in the y direction:
dᵧ = 449 sin 66° + 1112 sin 169° + 1571 sin 26°
dᵧ = 410.2 + 212.2 + 688.7
dᵧ = 1311 km
The magnitude is:
d² = dₓ² + dᵧ²
d² = (503)² + (1311)²
d = 1404 km
The angle is:
tan θ = dᵧ / dₓ
tan θ = 1311 / 503
tan θ = 2.61
θ = 69°
1404 km and 69° north of east from New Orleans is approximately Toledo.
Here's how I would do it:
How far does he have to go to catch the bug ? <u>23.8 meters</u>
How soon does he want to get there ? <u>1.8 seconds</u>
What speed does he need ? (23.8 m) / (1.8 sec) = <u>13.222 m/sec</u>
What speed is he flying now? <u>3.7m/s</u>
How much does he need to increase it ? (13.222 - 3.7) = <em>9.5 m/s</em> faster
