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2 years ago

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A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.14 m. The disk rotates about an axis through its

center and perpendicular to its face with a constant angular acceleration of 130 rad/s2. Determine the mass of the disk.

1 answer:

sammy [17]2 years ago

4 0

Answer: 3.27kg

Explanation:

Inertia can be said to be the resistance of an object to a change in its motion. This includes a change in its direction. An object will stay still or keep moving at the same speed and in a straight line, except it is acted upon by an unbalanced external force.

Given F = 30N

r = 0.14m

a = 130rad/s²

Then, T = Fr

T = 30*0.14

T = 4.2Nm

Also, T = inertia * acceleration

Inertia = 4.2/a

Inertia = 4.2/130

Inertia = 0.032

Also, inertia = mr²/2

0.032 = m * (0.14²)/2

0.032 = m * 0.0098

m = 0.032/0.0098

m = 3.27kg

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Answer:

Spring constant, k = 5483.11 N/m

Explanation:

It is given that,

Mass of the organ, m = 2 kg

The natural period of oscillation is, T = 0.12 s

Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :

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$k=\dfrac{4\pi^2 m}{T^2}$

$k=\dfrac{4\pi^2 \times 2\ kg}{(0.12\ s)^2}$

k = 5483.11 N/m

So, the spring constant for the spring in the scientist's model is 5483.11 N/m.

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Calculate the kinetic energy of a 4.00kg object traveling at 16.0 m/s

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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

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$F_y = Fsin30$

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

$F_f + F_x = 0$

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To sterilize a 53.0 g glass baby bottle, we must raise its temperature from 25.0°C to 92.0°C. How much heat transfer (in J) is r

Stella [2.4K]

The heat energy required is 2983 J

Explanation:

When an amount of energy Q is supplied to a certain sample of substance of mass m, the temperature of the substance increases by an amount of $\Delta T$ according to the equation :

$Q=mC_s \Delta T$

where

Q is the heat supplied

m is the mass of the object

$C_s$ is the specific heat capacity of the substance

$\Delta T$ is the change in temperature

For the glass bottle in this problem:

$m=53.0 g$

$\Delta T = 92.0 - 25.0 =67.0^{\circ}C$ is the change in temperature

$C_s = 0.84 J/g^{\circ}C$ is the specific heat of glass

Substituting into the equation, we find

$Q=(53.0)(0.84)(67)=2983 J$

Learn more about specific heat capacity:

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