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julsineya [31]
3 years ago
6

A 140-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

bout the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.800 rev/s in 2.00 s
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

The constant force is 263.55 newtons

Explanation:

There's a rotational version of the Newton's second law that relates the net torque on an object with its angular acceleration by the equation:

\tau = I\alpha (1)

with τ the net torque and α the angular acceleration. It’s interesting to note the similarity of that equation with the well-known equation F=ma. I that is the moment of inertia is like m in the linear case. The magnitude of a torque is defined as

\tau = Fr\sin \theta

with F the force applied in some point, r the distance of the point respect the axis rotation and θ the angle between the force and the radial vector that points toward the point the force is applied, in our case θ=90 and sinθ=1, then (1):

Fr = I\alpha (2)

Because the applied force is constant the angular acceleration is constant too, and for constant angular acceleration we have that it's equal to the change of angular velocity over a period of time:

\alpha=\frac{0.800}{2.00}=0.40 \frac{rev}{s^{2}}

It's important to work in radian units so knowing that 1rev=2\pi rad

\alpha=2.51 \frac{rad}{s^{2}} (3)

The moment of inertia of a disk is:

I=\frac{MR^{2}}{2} (4)

with M the mass of the disk and R its radius, then

I=\frac{(140)(1.50)^{2}}{2}=157.5 kg*m^2

using the values (3) and (4) on (2)

Fr = (157.5)(2.51) (2)

Because the force is applied about the rim of the disk r=R=1.50:

F= \frac{(157.5)(2.51)}{1.50}=263.55 N

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Answer:

646.8 N

Explanation:

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3 years ago
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P
MArishka [77]

Answer:

F \approx 19.5 N

Explanation:

From the question we are told that:

Pressure of  P_{CO_2}=1.50 * 105 Pa.

Bottle cap area A_b= 4.40 * 10-4 m^2

 

Generally the equation for Resultant pressure P_r is give as is mathematically given by

P_r=P_{CO_2}-P_a

Where

P_a=atmospheric\ pressure = 1.013*10^5 pa

P_r=1.50 * 105 Pa-1.013*10^5 pa

P_r=0.487*10^5 pa

Generally the equation for Force exerted by screw F is give as is mathematically given by

F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N

F \approx 19.5 N

4 0
2 years ago
A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is
malfutka [58]

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

4 0
3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

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