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Aneli [31]
3 years ago
14

if motorbike can travel 90 kilometre on 2 litre of petrol find the distance it can travel on 5 litre of petrol​

Physics
1 answer:
ella [17]3 years ago
4 0
The motor bike can travel 225 kilometers on 5 liters of petrol
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One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
What is the chemical name for lime
Sergio039 [100]
Calcium Oxide
CaO I think
5 0
3 years ago
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the
Zepler [3.9K]

Answer:

51.94°

Explanation:

I_0 = Unpolarized light

I_2 = Light after passing though second filter = 0.19I_0

Polarized light passing through first filter

I_1=\frac{I_0}{2}

Polarized light passing through second filter

I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}

The angle between the two filters is 51.94°

5 0
3 years ago
What color could be created by mixing cyan and yellow pigment?
Cloud [144]

Answer:

green pigment

Explanation:

green pigment

3 0
2 years ago
Read 2 more answers
8a. What is the equivalent resistance of the following circuit?
ollegr [7]

Answer: Take your pick

Explanation:

if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω

if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω

if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω

if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω

If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω

If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω

If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω

If they are all in series 100 + 300 + 50 = 450 Ω

4 0
2 years ago
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