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Firdavs [7]
3 years ago
5

A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to

the plane. If the wheel has a radius of 0.5m and the moment of inertia 0.5kgm?, calculate the translational velocity acquired after travelling 12m up the plane, assuming the wheel is initially at rest (6 marks)​
Physics
1 answer:
Len [333]3 years ago
5 0

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

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Answer:The rate of ejection of photoelectrons will increase

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If the frequency of incident monochromatic light is held constant and its intensity is increased, the rate of ejection of photoelectrons from the metal surface increases with increase in intensity of the monochromatic light. More current flows due to more ejection of photoelectrons.

4 0
3 years ago
PLEASE HELP!
Anna [14]

Answer:

Explanation:

At constant pressure , work done by gas = P x ΔV where P is pressure and ΔV is change in volume

ΔV = 9.2 - 5.6 = 3.6 L

3.6 L = 3.6 x 10⁻³ m³

ΔV = 3.6 x 10⁻³ m³

P = 3.7 x 10³ Pa

So work done

= 3.7 x 10³ x 3.6 x 10⁻³ J

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( c ) is the answer , because work is done by the gas so it will be positive.

5 0
3 years ago
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What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?
Alik [6]
1/2mv^2

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The kinetic energy is 600J
3 0
3 years ago
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URGENT. Please help.
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6 0
3 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
3 years ago
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