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luda_lava [24]
3 years ago
8

A merry-go-round starts from rest and reaches the angular speed of 4 rpm in the first two minutes. what would be the angular acc

eleration of the merry-go-round in rpm/min?
Physics
1 answer:
Anarel [89]3 years ago
7 0

initially marry go round is at rest

after t= 2 minutes its final angular speed is 4 rpm

w_f = 2 \pi *\frac{4}{60}

w_f = 0.42 rad/s

now by using kinematics we will have

w_f = w_0 + \alpha * t

0.42 = 0 + \alpha * (2*60)

\alpha = 3.5 * 10^{-3} rad/s^2

\alpha = 3.5 * 10^{-3} * \frac{60*60}{2\pi}

\alpha = 2 rev/min^2

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Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet
Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times(5\times10^{-2})^2

A=7.85\times10^{-3}\ m^2

We need to calculate the capacitor

Using formula of capacitor

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}

C=6.94\times10^{-12}\ F

We need to calculate the resistance of the wire

Using formula of resistivity

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}

R=4.27\times10^{-3}\ \Omega

We need to calculate the voltage

Using formula of charge

q=CV

V=\dfrac{q}{C}

Put the value into the formula

V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}

V=1.801\times10^{3}\ V

(a). We need to calculate the current

Using formula of current

I=\dfrac{V}{R}

I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}

I=421779.85\ A

I=4.217\times10^{5}\ A

(b). We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}

E=11.2\times10^{5}\ N/C

The electric field in the wire is 11.2\times10^{5}\ N/C.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

E=\dfrac{1}{2}CV^2

Put the value into the formula

E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2

E=1.126\times10^{-5}\ J

The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Hence, (a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

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3 years ago
May I get some help please?
ziro4ka [17]
Arrows point away from both north and south.
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3 years ago
A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
amid [387]
A= f/m

a= 19/2

a= 9.5m/s^2
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What is the magnitude of the Box's Acceleration?​
mojhsa [17]

The Box's Acceleration : g sin θ

<h3>Further explanation  </h3>

Newton's 2nd law explains that the acceleration produced by the resultant force on an object is proportional and in line with the resultant force and inversely proportional to the mass of the object  

∑F = m. a  

F = force, N  

m = mass = kg  

a = acceleration due to gravity, m / s²  

We plot the forces acting on the block (picture attached) according to the y-axis and the x-axis.

Because the motion of the block is in the same direction as the x-axis, ignoring the friction force with the inclined plane, then

\tt \sum F_x=m.a\\\\W.sin\theta=m.a\\\\mgsin\theta=m.a\\\\a=gsin\thet\theta

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3 years ago
The friction force between a bag and the floor is 4 N. what is the net force acting on the bag? What is the acceleration of the
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Well, if the bag is moving across the floor at a constant velocity. Then I would assume that the Fnet force is equal to 0, and thus the value of acceleration would also be equal to 0.
6 0
2 years ago
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