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marin [14]
3 years ago
5

How many examples are there for oscillatory motion (i) Motion of football players (ii) String of a guitar (iii) Motion of a chil

d on a swing (iv) Rotating fan (f) Surface of drums (vi) Moving of bicycles on road (vii) Rolling of a ball on ground
is Apurva account ban ??​
Physics
1 answer:
andrew11 [14]3 years ago
7 0

Answer:

yesssssss mere followers mein dekho Mrs Queen I'd hai uska follow her ☺

I'm a baby so I don't know the answer

You might be interested in
A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as V_1, the relation with the final volume as

V_2 = V_1 +0.163\% V_1

V_2 = V_1 +0.00163V_1

V_2 = 1.00163V_1

Initial temperature = 21.1\°C

Let T be the temperature after expanding by the formula of volume expansion

we have,

V_2 = V_1 (1+\gamma \Delta t)

Where \gamma is the volume coefficient of copper 5.1*10^{-5}/C

1.00163V_1 = V_1(1+\gamma(T-21.1\°))

1.00163 = 1+5.1*10^{-5}(T-21.1\°)

0.00163 = 0.000051T-0.0010761

T = 53.0608\°C

Therefore the temperature is 53.06°C

7 0
4 years ago
About how much of Earth's water is available for humans to use?
lesya692 [45]

Answer:

0.3 %

Explanation:

Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.

5 0
3 years ago
Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
3 years ago
The force of attraction between any two objects is known as
Leona [35]

Answer:

Gravity :)

Explanation:

7 0
4 years ago
Read 2 more answers
A tank of volume 0.25 ft is designed to contain 50 standard cubic feet of air. The temperature is 80° F.Calculate the pressure i
slavikrds [6]

Answer:

The inside Pressure of the tank is 4499.12 lb/ft^{2}

Solution:

As per the question:

Volume of tank, V = 0.25 ft^{3}

The capacity of tank, V' = 50ft^{3}

Temperature, T' = 80^{\circ}F = 299.8 K

Temperature, T = 59^{\circ}F = 288.2 K

Now, from the eqn:

PV = nRT                      (1)

Volume of the gas in the container is constant.

V = V'

Similarly,

P'V' = n'RT'                       (2)

Also,

The amount of gas is double of the first case in the cylinder then:

n' = 2n

\]frac{n'}{n} = 2

where

n and n' are the no. of moles

Now, from eqn (1) and (2):

\frac{PV}{P'V'} = \frac{nRT}{n'RT'}

P' = 2P\frac{T'}{T} = 2\times 2116\times \frac{299.8}{288.2} = 4499.12 lb/ft^{2}

7 0
4 years ago
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