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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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3 years ago
A particle with charge -5.60 nC is moving in a uniform magnetic field →B=−(1.25T) k The magnetic force on the particle is measur
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Answer:

Explanation:

Force = q ( v x B)

- 5.6 x 10⁻⁹ (v x - 1.25 k )

- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j

Let v = ai+bj +ck

Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]

= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )

= - 7 a j + 7 b i

( 7bi - 7aj ) x 10⁻⁹

Comparing with given force

7b x 10⁻⁹ b = - 3.4 x 10⁻⁷

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- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷

a = - 105.7

velocity

= -105.7 i - 48.57 j + ck

b ) Component along k can not be obtained .

c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ

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= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷

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4 0
3 years ago
A diver shines a flashlight upward from beneath the water at a 35.2° angle to the vertical. at what angle does the light leave t
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n_r is the refractive index of the second medium (in this case, water, so n_r = 1.33)
\theta _i is the angle of incidence of the light, with respect to the vertical, so \theta_i = 35.2^{\circ}
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Re-arranging the equation and using the data of the problem, we can find the the angle of refraction of the light inside the water:
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3 0
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