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2 years ago

5

When a piece of iron or mass 78 gm is put in a graduated cylinder containing 100 cm cube of water the reading of the cylinder be

come 110cm cube
Calculate the density of iron.

1 answer:

Wittaler [7]2 years ago

7 0

Answer:

Ro = 7.8 [g/cm³]

Explanation:

According to the principle of Archimedes, the volume of a body immersed in a liquid is equal to the volume displaced by water. That is, in this problem The displacement volume is equal to the new volume minus the original volume.

$V_{n}=110[cm^{3} ]\\V_{o}=100[cm^{3} ]\\V_{d}=110-100 = 10 [cm^{3} ]$

We now know that density is defined as the relationship between mass and volume.

$Ro = m/V_{d}$

where:

Ro = density [g/cm³]

m = mass = 78 [g]

Vd = displacement volume [cm³]

$Ro = 78/10\\Ro = 7.8 [g/cm^{3} ]$

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Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

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3 years ago

Convert 1nanosecond in to its SI init

SOVA2 [1]

<em>Convert 1nanosecond in to its SI init</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.</em>

<em>In SI units, nano is 1000th part of micro which in turn is 1000th part of mini which in turn is 1000th part of main unit. Now, when you affix nano to any unit, here in case, second, it means that you are referring to 1000th part of 1000th part of 1000th part of second or in short, 1000000000th(10^9) part of a second.So to convert nanosecond into second, just multiply the nanosecond with 0.000000001 or (10^-9)</em>

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2 years ago

Red: Distance = 3 x Time

sukhopar [10]

2 because that’s correct I I I I I I I I I I I I I I I I I I I

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3 years ago

Plain electromagnetic wave (in air) has a frequency of 1 MHz and its B-field amplitude is 9 nT a. What is the wavelength in air?

Norma-Jean [14]

Answer:

Part a)

$\lambda = 300 m$

Part b)

$E = 2.7 N/C$

Part c)

$I = 9.68 \times 10^{-3} W/m^2$

$P = 3.22 \times 10^{-11} N/m^2$

Explanation:

Part a)

As we know that frequency = 1 MHz

speed of electromagnetic wave is same as speed of light

So the wavelength is given as

$\lambda = \frac{c}{f}$

$\lambda = \frac{3\times 10^8}{1\times 10^6}$

$\lambda = 300 m$

Part b)

As we know the relation between electric field and magnetic field

$E = Bc$

$E = (9 \times 10^{-9})(3\times 10^8)$

$E = 2.7 N/C$

Part c)

Intensity of wave is given as

$I = \frac{1}{2}\epsilon_0E^2c$

$I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)$

$I = 9.68 \times 10^{-3} W/m^2$

Pressure is defined as ratio of intensity and speed

$P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}$

$P = 3.22 \times 10^{-11} N/m^2$

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3 years ago

What do diffraction and refraction have in common? pick only one

evablogger [386]

B. they both involve wave interaction.

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2 years ago

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