Answer:
A. would be the correct answer to this problem
Step-by-step explanation:
Usually, Δ stands for a difference: if you have two quantities a₂ and a₁, their difference a₂-a₁ can be shortened as Δa.
This said, your formula is not set up correctly: the linear speed can be found with the formula:
v = ω·r
where r is the radius and <span>ω is the angular frequency, which is given by:
</span>ω = Δα / Δt
Substituting this into the one above, you find the correct formula:
<span>v = (Δα / Δt) · r
The problem gives you directly </span>Δα, which is 1/3 rad, because does not say at what angle the point started moving and at what angle it stopped.
Similarly, for the time you have Δt, which is 20 s.
Therefore, plugging in the numbers you get:
v = (Δα / Δt) <span>· r = (1/3 </span>÷ 20) × 10 = 1/6 = 0.167 cm/s
Would it be 43? 62-19=43? Sorry if this isn’t helpful I’m just trying to catch up on homewokr so I gotta answer some questions
Answer:
a) weight of the car = 2816,1 lbs
b) 2773 lbs
Step-by-step explanation:
The equilibrium force is 490 lbs. That force keep the car at rest, then
∑ Fy = 0 and ∑Fx = 0
Forces acting on the car:
The external force 490 lbs
weight of the car uknown
Normal force
sin∠10° = 0,174
cos∠10° = 0.985
∑Fx = 0 mg*sin10°- 490 = 0 ∑Fy = 0 mg*cos10° - N = 0
mg*0,174= 490
mg = 490 / 0,174
mg = 2816,1 lbs
weight of the car = 2816,1 lbs
The Normal force
mg*cos10° - N = 0 2816,1 * 0,985 = N
N = 2773 lbs
Then equal force in magnitude and in opposite direction will car exets on the driveway