Complete Question
Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?
Answer:
Go-cart A is faster
Explanation:
From the question we are told that
The length of the track is
The speed of A is
The uniform acceleration of B is
Generally the time taken by go-cart A is mathematically represented as
=>
=>
Generally from kinematic equation we can evaluate the time taken by go-cart B as
given that go-cart B starts from rest u = 0 m/s
So
=>
=>
Comparing we see that is smaller so go-cart A is faster
Answer:
Fc=5253
N
Explanation:
Answer:
Fc=5253
N
Explanation:
sequel to the question given, this question would have taken precedence:
"The 86.0 kg pilot does not want the centripetal acceleration to exceed 6.23 times free-fall acceleration. a) Find the minimum radius of the plane’s path. Answer in units of m."
so we derive centripetal acceleration first
ac (centripetal acceleration) = v^2/r
make r the subject of the equation
r= v^2/ac
ac is 6.23*g which is 9.81
v is 101m/s
substituing the parameters into the equation, to get the radius
(101^2)/(6.23*9.81) = 167m
Now for part
( b) there are two forces namely, the centripetal and the weight of the pilot, but the seat is exerting the same force back due to newtons third law.
he net force that maintains circular motion exerted on the pilot by the seat belts, the friction against the seat, and so forth is the centripetal force.
Fc (Centripetal Force) = m*v^2/r
So (86kg* 101^2)/(167) =
Fc=5253
N
Friction occurs when the surfaces and heat of two surfaces grind against each other.
I think these are the answers
<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:
Ff = UsN
where Us is the coefficient of static friction and N is the normal force.
In order to get the crate moving you must first apply enough force to overcome the static friction:
Fapplied = Ff
Since Fapplied = 43 Newtons:
Fapplied = Ff = 43 = UsN
and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11
43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
Answer:
Specific heat capacity is an intensive property and does not depend on sample size.
Explanation: