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ankoles [38]
3 years ago
15

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​

Physics
1 answer:
Scilla [17]3 years ago
7 0

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

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3.3 kg block is on a perfectly smooth ramp that makes an angle of 52° with the horizontal. (a) What is the block's acceleration
Alexus [3.1K]

Answer:

a)  a = 7.72 m / s²,  N = 19.9 N  and b)   F = 25.5 N

Explanation:

To solve this problem we will use Newton's second law, let's set a reference system with an axis parallel to the plane and gold perpendicular axis. Let's break down the weight (W)

    sin52 = Wx / W

    cos52 = Wy / W

    Wx = W sin52

    Wy = w cos 52

Let's write them equations

X axis

    Wx = ma

Y Axis

    N-Wy = 0

    N = Wy

a) Let's calculate the acceleration

    a = W sin52 / m = mg sin 52 / m

    a = g sin 52

    a = 9.8 sin52

    a = 7.72 m / s²

The force of the ramp is normal

    N = Wy = mg cos 52

    N = 3.3 9.8 cos 52

    N = 19.9 N

b) For the block to move at constant speed the sum of force on the axis must be zero,

    F - Wx = 0

    F = Wx

    F = mg sin52

   F = 3.3 9.8 sin 52

   F = 25.5 N

Parallel to the plane and going up

3 0
3 years ago
One kilogram-meter per second squared is also equal to what unit
yan [13]
Metres/second² is acceleration. maybe a bit more clarification?

8 0
3 years ago
Question C) needs to be answered, please help (physics)
Zarrin [17]

(a) Differentiate the position vector to get the velocity vector:

<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>

<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>

<em></em>

(b) The velocity at <em>t</em> = 2.00 s is

<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>

<em></em>

(c) Compute the electron's position at <em>t</em> = 2.00 s:

<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>

The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:

||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m

(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that

tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s)   ==>   <em>θ</em> ≈ -79.4º

or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.

3 0
3 years ago
9. A 2 liter bottle of Coke weighs about 2 kilograms. If that bottle were combined with an equal amount of anti-Coke, how many J
givi [52]

Answer:

The amount of energy that would be released is equal to 4182 Joules.

Explanation:

Total amount of coke = 2 kg = 2000 g

1 calorie per gram is equal to 4.184 Joules of energy

4.184 J/gC*2000g = 8368 J

1 food calorie is roughly equal to 4186 J

8368 - 4186

Therefore, the amount of energy that would be released is equal to 4182 Joules.

3 0
3 years ago
A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra
Tatiana [17]

Answer:

The centripetal acceleration of the car is 8\ m/s^2.

Explanation:

Let the mass of the car, m=10^3\ kg

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(20\ m/s)^2}{50\ m}

a=8\ m/s^2

So, the centripetal acceleration of the car is 8\ m/s^2. Hence, this is the required solution.

4 0
3 years ago
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