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V125BC [204]
3 years ago
12

Two hoops, starting from rest, roll down identical inclined planes. The work done by nonconservative forces, such as air resiste

nce, is zero (Wnc = 0J). Both have the same mass M, but, one hoop has the twice the radius of the other. The moment of inertia for each hoop is I = Mr sequared, where r is the radius. Which hoop, if either has the greater total kinetic energy (transational and rotational) at the bottom of the incline.
Hoop 1 Radius = R and Mass = M
Hoop 2 Radius = 1/2 R and Mass = M
Physics
1 answer:
Maurinko [17]3 years ago
6 0

Answer: both hoops have the same kinetic energy at the bottom of the incline.

Explanation:

If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.

K1 + U1 = K2 + U2

If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then

K1 = 0 and U2 = 0

⇒ ΔK = ΔU = m g. h

If both inclines have the same height, and both hoops have the same mass m, the  change in kinetic energy, must be the same for both hoops.

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Answer:

True.

Explanation:

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The two metallic strips that constitute some thermostats must differ in:_______
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Answer:

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3 years ago
Consider the following system, where F = 80 N, m = 1 kg, and M = 11 kg M m F What is the magnitude of the force with which one b
Lorico [155]

Answer:

Force exerted by the lighter block on the heavier block is 6.63 N

Explanation:

Given Data

F = 80N

m = 1kg

M = 11kg

Solution:

*We assume that there is no friction

Calculating the acceleration of the system

a = \frac{F}{m+M}

a = \frac{80}{1+11}

a = \frac{80}{12}

a = 6.67ms^{-2}

Let's write the Equation of Motion of the heavier block  

F_{1} = F - F_{2}

Ma = F - F_{2}

force exerted by the lighter block on the heavier block is calculated as

 F_{2} = F - Ma

 F_{2} = 80 - (11 x 6.67)  

 F_{2} = 6.63 N

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