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2 years ago

Help me

Physics

1 answer:

Anni [7]2 years ago

7 0

Answer:

Explanation:

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A 0.5 m diameter wagon wheel consists of a thin rim having a mass of 7 kg and six spokes, each with a mass of 1.2 kg. 1.2 kg 7 k

Arte-miy333 [17]

Explanation:

It is given that,

Mass of the rim of wheel, m₁ = 7 kg

Mass of one spoke, m₂ = 1.2 kg

Diameter of the wagon, d = 0.5 m

Radius of the wagon, r = 0.25 m

Let I is the the moment of inertia of the wagon wheel for rotation about its axis.

We know that the moment of inertia of the ring is given by :

$I_1=m_1r^2$

$I_1=7\times (0.25)^2=0.437\ kgm^2$

The moment of inertia of the rod about one end is given by :

$I_2=\dfrac{m_2l^2}{3}$

l = r

$I_2=\dfrac{m_2r^2}{3}$

$I_2=\dfrac{1.2\times (0.25)^2}{3}=0.025\ kgm^2$

For 6 spokes, $I_2=0.025\times 6=0.15\ kgm^2$

So, the net moment of inertia of the wagon is :

$I=I_1+I_2$

$I=0.437+0.15=0.587\ kgm^2$

So, the moment of inertia of the wagon wheel for rotation about its axis is $0.587\ kgm^2$ . Hence, this is the required solution.

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3 years ago

HELP ASAP PLSSSS WILL BE MARKED BRAINLIEST

vladimir1956 [14]

Answer:

Nails rusting over time as they are exposed to oxygen

Explanation:

New substance with new property is formed

It have also changed the chemical property of the substance

It is difficult to reverse the change etc...

4 0

2 years ago

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alexdok [17]

Answer:

Honey how can i draw you cant draw here

Explanation:

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2 years ago

Help please it's urgent

insens350 [35]

I cant read it, i could most likely help if i could read it.

7 0

3 years ago

You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the

podryga [215]

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

$F_x = Fcos30$

$F_y = Fsin30$

Now the normal force on the block is given as

$N = Fsin30 + mg$

$N = 0.5F + (18\times 9.8)$

$N = 0.5F + 176.4$

now the friction force on the cart is given as

$F_f = \mu N$

$F_f = 0.625(0.5F + 176.4)$

$F_f = 110.25 + 0.3125F$

now if cart moves with constant speed then net force on cart must be zero

so now we have

$F_f + F_x = 0$

$Fcos30 - (110.25 + 0.3125F) = 0$

$0.866F - 0.3125F = 110.25$

$F = 199.2 N$

so the force must be 199.2 N

8 0

3 years ago