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Dimas [21]
3 years ago
14

A rod of mass M = 4 kg, length L = 1.8 meters, and moment of inertia ML2/12 is free to move on a frictionless surface. The rod i

s at rest when a puck of mass m = 0.4 kg approaches with a speed vi = 20 m/s perpendicular to the rod's length and strikes the rod at a point d = 0.3 meters from its lower end. After the collision, the puck moves backward with a speed vf = 10 m/s. The rod's center of mass moves forward with speed v and rotates with angular speed ω. (a) What is the speed v of the rod's center of mass after the collision?
Physics
1 answer:
Pachacha [2.7K]3 years ago
8 0

Answer:

The speed of the rod's center of mass after the collision is 6 m/s.

Explanation:

Given that,

Mass of rod = 4 kg

Length l = 1.8 m

Moment of inertia I=\dfrac{ML^2}{12}

Mass of puck = 0.4 kg

Initial speed= 20 m/s

Distance = 0.3 m

Final speed = 10 m/s

(a). We need to calculate the speed v of the rod's center of mass after the collision

As there is no external force acting on the system so, linear and angular momentum of the system will be conserved.

Using conservation of momentum

m_{i}v_{i}=m_{f}v_{f}+Mv

Put the value into the formula

0.4\times20=-0.4\times10+2v

v=\dfrac{8+4}{2}

v=6\ m/s

Hence, The speed of the rod's center of mass after the collision is 6 m/s.

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Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

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To practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.7 kg and length 3.0 m
Zanzabum

Answer:

T_{1} = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about T_{1} we have:

(M + m) g * 0.5L - T_{2}(L - d) = 0

⇒ T_{2} = [(M + m) g * 0.5L] ÷ (L - d)

T_{2} = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

T_{2}= 59.535 ÷ 2.4

T_{2} = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

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Using sum of equilibrium in the vertical direction, we have:

T_{1} + T_{2} = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

T_{1} + 24.81 = 26.46 + 13.23

T_{1} = <u>14.88</u> N

8 0
3 years ago
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