Answer:
a)r' = 4 r
b)
Explanation:
Given that car is moving with uniform speed.
At the verge of sliding
Friction force = Force due to circular motion
So we can say that
-----1
Where is the coefficient of friction
r is the radius of circular path
V is the velocity of car
a)
if the car speed become double
It means that new speed of car =2V
lets tale new radius of circular path is r'
---------2
From equation 12 and 2 we cay that
r' = 4 r
It means that we have to increase radius 4 times to avoid sliding
b)
In the above expression there is no any terms of mass ,it means that sliding speed does not depends on the weight of car.
So the sliding speed will remain same.
Answer:
a) 20.81 J
b) 8.29 J
Explanation:
V = iR + L di/dt
where
i = a(1-e^-kt)
for large t
i = V/R
i = 24 / 9.4
i = 2.55 A
so
i = 2.55(1-e^-kt)
di/dt = 2.55 k e^-kt
24 = 24-24e^-kt + 6.4(2.55)k e^-kt
24 = 6.4(2.55) k
k = 24 / (6.4 * 2.55)
k = 24 / 16.32
k = 1.47 = R/L
so
i = 2.55(1-e^-(Rt/L))
current is maximum at great t
i max = 2.55 - 0
energy = (1/2) L i^2
E = (1/2)(6.4)2.55^2
E = 20.81 Joules
one time constant T = L/R and e^-(Rt/L) = 1/e = .368
i = 2.55 (1 - 0.368)
i = 2.55 * 0.632
i = 1.61 amps
energy = (1/2)(6.4)1.61^2
E = 8.29 Joules
Answer:
x=45
Explanation:
by taking 42 and dividing it by 14 you get 1/3, because 14 is 1/3 of 42 you can then see that the ratio is multiplied by 3. si then you can just multiply 15 by 3 to get 45
Answer:
False
Explanation:
It is a vector because of the acceleration being a vector. F=ma.
Answer:
7.3 newtons to the west
Explanation:
3.7kg × 11a - 3.7kg × ? = 3.7n