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Agata [3.3K]
3 years ago
7

The specific gravity of a substance is given by G = DS/DW, where DS is the density of the substance in kg/m3 and DW is the densit

y of water, which is known to be 1000 kg/m3. The density of a particular substance is measured to be DS = 500±5 kg/m3. Estimate the specific gravity, and find the uncertainty in the estimate
Physics
1 answer:
katen-ka-za [31]3 years ago
3 0

Answer:

G=0.5\pm0.005

Explanation:

The specific gravity is given by,

G=\frac{D_S}{D_W}=\frac{500}{1000} =0.5

Now, in order to calculate the uncertainty (relative error) in G, we must first take log (base e) on both sides of the equation,

lnG=ln(\frac{D_S}{D_W} )=lnD_S-lnD_W

Differentiating the above equation,

\frac{dG}{G}=\frac{dD_S}{D_S}

The second term is zero because it is known that D_W=1000kg/m^3 and hence a constant.

Putting the appropriate values, we get,

\frac{dG}{G}=\frac{dD_S}{D_S}=\frac{5}{500} =0.01

Therefore, uncertainty in G = 0.01\times0.5=0.005

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does increasing the frequency of a wave also increase its wavelength if not how are these quantities related​
Aleks [24]

Answer: Increasing the frequency does not increase the wavelength. They are inversely related.

Explanation:

As wavelength increases, frequency decreases. If you look at a transverse wave and it has a long wavelength, there only a few waves produce. Which means there is less frequency produced. So as wavelength increases, frequency decreases. The other way around can work to. As frequency increases, wavelength decreases. They are inversely related.

8 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
Please help! I'm not sure what equation or the process to do this question.
lions [1.4K]

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

5 0
2 years ago
HELP PLEASEEEEEEEEEEEEEEEE
prisoha [69]
Solid has vibrating molecules that barely move to keep it's shape

liquid moves at an average speed and keeps it's volume but not it's shape

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2 years ago
1pt If the room is the frame of reference for the study of an object moving in three dimensions, from where should distances be
oksian1 [2.3K]

Answer:

Option B, two walls and the floor

Explanation:

The distance should be measured from the point where at least the three axed meet.

Two walls and the floor are equivalent to three axes.

Vertical wall 1 = Y Axis

Horizontal wall2 = X Axis

Floor = Z Axis

Thus, the distance should be measured from the point where two walls and one floor meet.

Option , B is correct

5 0
3 years ago
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