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vichka [17]
3 years ago
7

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

turns with negligible friction.
(a) A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?
Physics
1 answer:
daser333 [38]3 years ago
7 0

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

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sergejj [24]

Answer:

y = 2.45 x

Explanation:

given,

Let 'x' be the number of cans.

and 'y' be the total area of the room

now,

From the given statement in the question

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y = \dfrac{15}{7}\times \dfrac{8}{7} \times x

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From the Above equation we can state that

To pain the entire room 2.45 cans of paint is required.

8 0
3 years ago
1. In Bohr's model of the atom, where are the electrons and protons located? (1 point)
r-ruslan [8.4K]
Well, it's really dangerous to try and visualize physical models for things
of this size.  But if you must, then it's something like this:

-- The nucleus is a tight-packed bunch of protons and neutrons, located
at the center of each atom.

-- The electrons live all around the nucleus, in a space far from it. 
A description of the relative sizes that I read more than 60 years ago
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and the electrons whizzing around it have a size-relationship that's
about the same as a bunch of grapes in the middle of the state of Texas.

This also tells us that matter is mostly empty space ! 

-- In Bohr's model of the atom, he described the whole thing very much
like a miniature solar system ... the electrons are tiny, solid little balls,
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We learned later that it's impossible to talk about things like "how big is
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the electron have".  The best we can do is talk about a 'cloud' around the
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This is NOT because we don't have good enough technology yet to
zoom in on the electrons, and at some time in the future we'll be able
to sharply see where they are and how fast they're moving.  It's because
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"Location" is described in terms of probability, objects behave like solid
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Weird ?     Hard to understand ?     You said it !

BTW ... the answer to the question is ' A ' .
5 0
3 years ago
Which of the following is Ohm's Law?<br> OR=xR<br> O1-V/R<br> OVR/<br> OV-/R
kodGreya [7K]

Answer:

I=V/R

Explanation:

Ohm's law gives the mathematical relationship between, current, voltage and the resistance of the circuit. Its mathematical form is given by :

V = IR ....(1)

V = voltage (in volts)

I = current (in Ampere)

R = resistance (in ohms)

Equation (1) can be rewritten as :

I=\dfrac{V}{R}

Hence, the correct relation is I=V/R.

8 0
3 years ago
A 763 kg car moving at 26 m/s brakes to a stop. The brakes contain about 15 kg of iron that absorb the energy. What is increase
Neko [114]

Answer:

\Delta T=38.20^{\circ}

Explanation:

It is given that,

Mass of the car, m = 763 kg

Speed of the car, v = 26 m/s

Mass of the iron, m' = 15 kg

Specific heat of iron, c = 450 J/kg

When the car is in motion, it will possess kinetic energy. It is given by :

K=\dfrac{1}{2}mv^2

K=\dfrac{1}{2}\times 763\times (26)^2

K = 257894 J

Since, energy is absorbed by the brakes. The kinetic energy of the car is absorbed by the brakes. So,

K=mc\Delta T

\Delta T is the increase in temperature of the brakes

\Delta T=\dfrac{K}{m'c}

\Delta T=\dfrac{257894}{15\times 450}

\Delta T=38.20^{\circ}

So, the increase in temperature of the brakes is 38.20 degrees Celsius. Hence, this is the required solution.

3 0
3 years ago
A 15 g ball at the end of 2.4 m string is swung in the horizontal circle. It revolves once every 2.09 s. What is the magnitude o
Mrrafil [7]

Answer:

0.325 N

Explanation:

From the question,

T = 4π²rm/t²............................ Equation 1

Where T = Tension, r = radius or length of the string, m = mass of the string, t = time.

Given: r = 2.4 m, m = 15 g = 0.015 kg, t = 2.09 s.

Constant: π = 3.14

Substitute these values into equation 1

T = 4(3.14²)(2.4)(0.015)/2.09²

T = 1.4198/4.3681

T = 0.325 N

3 0
3 years ago
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