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2 years ago

A speed versus time graph is shown:

Physics

1 answer:

erma4kov [3.2K]2 years ago

6 0

In the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

In the given graph we can deduce the following;

at the time interval, 0 s to 3.5 s, the speed of the object = 1 cm/s

when the time, t= 4 s, the speed of the object = 0 cm/s

at the time interval, 4.0 s to 8.0 s, the speed of the object = 0 cm/s

When the speed of an object is zero (0), the object is simply at rest.

Thus, we can conclude that in the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

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Answer:

Check body of the explanation

Explanation:

Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about $10 ms^-^2$ , the density of whatever you're sinking in, and the depth at which you are. In formula, $p(h) = p_0 + \rho g h$ , and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - $1000 * 10 * 1 = 10^4 Pa$ , while in B is $1000*10*2 = 2*10^4 Pa$ , so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is $800*10*2 = 1,6*10^4 Pa$ : Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so $4m*3m*2m = 24m^3$

2b. Density is defined as mass over volume, so you simply multiply the volume you found earlier by the density of paraffine: $800* 24 = 1,92 *10^4kg$

2c. Weight is defined as the mass of something times the acceleration due to gravity, in our case it's $1.92 *10^4 kg * 10 ms^{-2} = 1.92 * 10^5 N$

2d. $\rho gh$ again, what a surprise! $800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa$

3. Yet again, $\rho gh$ . $1000 {kg \over m^3} * 10 {N \over kg}} * 2 m = 2* 10^4 {N\over m^2} =2*10^4 Pa$

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Scorpion4ik [409]

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3 years ago

A spring is compressed 0.035 m inside a dart gun. (K=500 N/m). The spring has elastic energy. Calculate it.

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The elastic potential energy of the spring is 0.31 J

Explanation:

The elastic potential energy of a spring is given by

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For the spring in this problem, we have:

k = 500 N/m (spring constant)

x = 0.035 m (compression)

Substituting, we find the elastic potential energy:

$E=\frac{1}{2}(500)(0.035)^2=0.31 J$

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