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Lesechka [4]
3 years ago
14

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. What total distance does the m

ass travel in 16 seconds?
Physics
1 answer:
harina [27]3 years ago
6 0

Explanation:

It is given that,

A mass oscillates up and down on a vertical spring with an amplitude of 3 cm and a period of 2 s. It is a case of simple harmonic motion. If the amplitude of a wave is T seconds, then the distance cover by that object is 4 times the amplitude.

In 2 seconds, distance covered by the mass is 12 cm.

In 1 seconds, distance covered by the mass is 6 cm

So, in 16 seconds, distance covered by the mass is 96 cm

So, the distance covered by the mass in 16 seconds is 96 cm. Hence, this is the required solution.

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2 years ago
Read 2 more answers
If two children, with masses of 16 kg and 24 kg , sit in seats opposite one another, what is the moment of inertia about the rot
Elena-2011 [213]

Answer:

The moment of inertia about the rotation axis is 117.45 kg-m²

Explanation:

Given that,

Mass of one child = 16 kg

Mass of second child = 24 kg

Suppose a playground toy has two seats, each 6.1 kg, attached to very light rods of length r = 1.5 m.

We need to calculate the moment of inertia

Using formula of moment of inertia

I=I_{1}+I_{2}

I=(m+m_{1})\times r^2+(m+m_{2})\times r^2

m = mass of seat

m₁ =mass of one child

m₂ = mass of second child

r = radius of rod

Put the value into the formula

I=(16+6.1)\times(1.5)^2+(24+6.1)\times(1.5)^2

I=117.45\ kg-m^2

Hence, The moment of inertia about the rotation axis is 117.45 kg-m²

8 0
3 years ago
In comparing sunspots to other areas of the sun's surface one can say that sunspots are...?
taurus [48]

Answer:

D. Cooler

Explanation:

The Sun's surface has a temperature ranging from 6 000 Kelvin, thus it is very hot. The radioactive reaction in the core of the Sun is a nuclear fusion reaction. This reaction ensures continuous release of high energy from the surface of the Sun.

But during the reaction, some parts becomes cooler than other parts on its surface. Which is due to the release of high amount of energy into space. The Sun's spot can be found in the cooler part of the Sun.

8 0
2 years ago
Calculate the self-inductance (in mH) of a 45.0 cm long, 10.0 cm diameter solenoid having 1000 loops. mH (b) How much energy (in
Karo-lina-s [1.5K]

Answer:

(a) The self inductance, L = 21.95 mH

(b) The energy stored, E = 4.84 J

(c) the time, t = 0.154 s

Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

where;

N is the number of turns = 1000 loops

μ is the permeability of free space = 4π x 10⁻⁷ H/m

l is the length of the inductor, = 45 cm = 0.45 m

A is the area of the inductor (given diameter = 10 cm = 0.1 m)

A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

3 0
2 years ago
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