A circuit is a continuous loop. a. True b. False
2 answers:
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Answer:
34.6 m/s
Explanation:
From conservation of momentum, the sum of initial and final momentum are equal. Momentum is a product of mass and velocity. Initial mass will be 42.8+31.5+25.9=100.2 kg
Final mass will be 31.5+25.9=57.4 kg
From formula of momentum
M1v1=m2v2
Making v2 the subject of the formula then
Substitute 100.2 kg for M1, 19.8 m/s fkr v1 and 57.4 kg for m2 then
The displacement vector (SI units) is
![\vec{r} =At\hat{i}+A[t^{3}-6t^{2}]\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7Br%7D%20%3DAt%5Chat%7Bi%7D%2BA%5Bt%5E%7B3%7D-6t%5E%7B2%7D%5D%5Chat%7Bj%7D)
The speed is a scalar quantity. Its magnitude is
Answer: At√(t⁴ - 12t³ + 36t² + 1)
The speed is a scalar quantity. Its magnitude is
Answer: At√(t⁴ - 12t³ + 36t² + 1)
Answer:
the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94
Explanation:
Given that ;
the top speed of Cheetahs is almost 60 mph
In cornering abilities ; the maximum centripetal acceleration of a cheetah was measured to be = 19 m/s^2
The objective of this question is to determine the what minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?
From the knowledge of Newton's Law;
we knew that ;
Force F = mass m × acceleration a
Also;
The net force = frictional force
so we can say that;
m×a =
where;
the coefficient of static friction is:
= 1.94
Hence; the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94