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3 years ago

Iron has a density of 7.9 g/cm3. What is the mass of a cube of iron with the length of one side equal to 55.0 mm?

2 answers:

4 0

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. To find the mass of the substance, we need to multiply the volume occupied by the substance.

Mass = Density x Volume
Mass = 7.9 g/cm^3 (5.5 cm)^3
Mass = 43.45 g

mezya [45]3 years ago

3 0

Answer:

Mass, m = 1.31 kg

Explanation:

It is given that,

Density of iron, $d=7.9\ g/cm^3$

Side of cube, a = 55 mm = 5.5 cm

We need to find the mass of the cube of iron. Mass per unit volume is called the density of the substance. It is given by :

$d=\dfrac{m}{V}$

$d=\dfrac{m}{a^3}$

$m=d\times a^3$

$m=7.9\times (5.5)^3$

m = 1314.36 grams

m = 1.31 kg

So, the mass of the iron cube is 1.31 kg. Hence, this is the required solution.

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An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off

QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

$V=\frac{\Delta W}{\Delta q}$

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

$I=\frac{\Delta q}{\Delta t}$

Where:

I is the current (2 A)

Δt is the time (20 s)

$2=\frac{\Delta q}{20}$

$q=40 C$

Then, we can put this value of charge in the voltage equation.

$V=\frac{4600}{40}=115 V$

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0

3 years ago

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is

cricket20 [7]

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

$T = 2\pi \sqrt{\frac{M}{k} }$

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

5 0

3 years ago

If a football player collides with a goal post, what forces are at work?

AlekseyPX

When a footballer collides with the goal post, the forces at work are the action and reaction forces. The player will exert an action force on the goal post, and then a reaction force from the goal post will stop the player. The reaction force call will cause pain and even injury to the player.

7 0

3 years ago

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

$z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}$

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

4 0

3 years ago

Calculate the average drift speed of electrons traveling through a copper wire with a crosssectional area of 80 mm2 when carryin

Vedmedyk [2.9K]

Answer:

The correct answer is $2.8*10^{-5}ms^{-1}$

Explanation:

The formula for the electron drift speed is given as follows,

$u=I/nAq$

where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.

Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is $8.93/63.5 molcm^{-3}$ . Converting this number to m³ using very elementary unit conversion we get $140384molm^{-3}$ . If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,

$n=140384*6.02*10^{23} = 8.45*10^{28}electrons.m^{-3}$

if we convert the area from mm³ to m³ we get $A=80*10^{-6}m^{2}$ .So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,

$u=30/(8.45*10^{28}*80*10^{-6}*1.602*10^{-19})\\u=2.8*10^{-5}m.s^{-1}$

which is our final answer.

6 0

3 years ago