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photoshop1234 [79]
2 years ago
8

What is the speed of a car traveling 31,450 meters in 68450 seconds?

Physics
1 answer:
atroni [7]2 years ago
3 0

Answer:

2,152,752,500 meters per second

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A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.
elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

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3 years ago
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A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d
oksano4ka [1.4K]

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

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2 years ago
Diagram the cross section of a graduated cylinder, illustrating how to read the meniscus.
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<span>When taking the reading you must see where the lower meniscus lies . The value at which the lower meniscus lies is your reading .</span>
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2 years ago
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A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)
egoroff_w [7]

Explanation:

The given data is as follows.

             q = 6.0 nC = 6 \times 10^{-9} C

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            \sigma = \frac{q_{in}}{A} .......... (1)

Since, area of the sphere is as follows.

               A = 4 \pi r^{2} ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      \sigma = \frac{q_{in}}{4 \pi r^{2}}

                   = \frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}            

                   = 0.477 \times 10^{-5}

or,               = 4.77 \mu C/m^{2}

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 \mu C/m^{2}.

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2 years ago
How can we avoid water pollution​
Alexandra [31]

Explanation:

1.Pick up litter and throw it away in a garbage can.

2.Blow or sweep fertilizer back onto the grass if it gets onto paved areas. ...

3.Mulch or compost grass or yard waste. ...

4.Wash your car or outdoor equipment where it can flow to a gravel or grassed area instead of a street.

5.Don't pour your motor oil down the storm drain.

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3 years ago
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