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2 years ago

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Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o

f air is 1.29 kg/m3kg/m3, and the density of aluminum is 2.7×103kg/m32.7×103kg/m3.

1 answer:

spin [16.1K]2 years ago

6 0

Answer:

The true weight of the aluminium is $m_{alu} =$ 4.5021 kg

Explanation:

Given data

$m_{app}$ = 4.5 kg

$\rho_{air}$ = 1.29 $\frac{kg}{m^{3} }$

$\rho_{al}$ = 2.7× $10^{3} \frac{kg}{m^{3} }$

The true mass of the aluminium is given by

$m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }$

Put all the values in above equation we get

$m_{alu} = \frac{(2700)(4.5)}{2700-1.29}$

$m_{alu} =$ 4.5021 kg

Therefore the true weight of the aluminium is $m_{alu} =$ 4.5021 kg

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Suppose the circumference of a bicycle wheel is 2 meters. If it rotates at 1 revolution per second when you are riding the bicyc

Debora [2.8K]

Answer:

(c) speed will be 2 m/sec

Explanation:

We have given circumference C = 2 meters

Angular velocity = 1 rev/sec =1×2π rad/sec

We know that circumference $C=2\pi r$ , where r is the radius of circumference

So $2=2\pi \times r$

r = 0.318 meter

Linear velocity $v=r\omega =0.318\times 2\times \pi =2m/sec$

So option C will be the correct option

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2 years ago

The velocity of a 3.00 kg parti- cle is given by :v = (8.00tiˆ + 3.00t2jˆ) m/s, with time t in seconds. At the instant the net f

FromTheMoon [43]

Answer:

Part a)

Direction of net force is

$\theta = 46.7 degree$

Part b)

Direction of the velocity is given as

$\phi = 27.9 degree$

Explanation:

As we know that the velocity of the particle is given as

$v = 8.00 t \hat i + 3.00 t^2\hat j$

now the acceleration is given as

$a = \frac{dv}{dt}$

$a = 8.00 \hat i + 6.00 t\hat j$

now magnitude of net acceleration is given as

$a = \sqrt{64 + 36t^2}$

$F = 3a$

$35 = 3\sqrt{64 + 36t^2}$

$t = 1.41 s$

Part a)

Now direction of net force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{6t}{8}$

$tan\theta = \frac{6(1.41)}{8}$

$\theta = 46.7 degree$

Part b)

Direction of the velocity is given as

$tan\phi = \frac{v_y}{v_x}$

$tan\phi = \frac{3.00 t^2}{8.00 t}$

$tan\phi = \frac{3.00(1.41)}{8.00}$

$\phi = 27.9 degree$

8 0

2 years ago

A 12 kg box sliding on a horizontal floor has an initial speed of 4.0 m/s. The coefficient of friction bctwecn thc box and the f

Hitman42 [59]

Answer:

(D) 96 kg-m/s

Explanation:

Let's start off by first calculating the normal force between the box and the floor.

This will be:

Normal Force = 12 * 9.81 = 117.72 N

We can now use the friction equation to find the frictional force on the box when it is moving:

Frictional force = Coefficient of friction * Normal Force

Frictional force = 0.4 * 117.72 = 47.09 N

Finally, since we have the force on the box, we can find the acceleration:

F = Mass * Acceleration

47.09 = 12 * Acceleration

Acceleration = 3.92 m/s^2

Final speed after 2 seconds:

$V=U+a*t$

$V = 4 +(-3.92)*(2)$

V= -3.84 m/s

Since we know the initial and final speeds, we can calculate the change in momentum:

Change in momentum = Final Momentum - Initial Momentum

Change in momentum = $3.84*12-(-4)*12$

Change in momentum = 94.08 kg*m/s

Thus we can see that option (D) is the closest answer.

6 0

2 years ago

A quarterback throws a football with an initial velocity v at an angle θ above horizontal. Assume the ball leaves the quarterbac

Maru [420]

(a) The y-component or vertical velocity is calculated using:
Vy = Vsin(∅)

(b) The x-component or horizontal velocity is calculated using:
Vx = Vcos(∅)

6 0

2 years ago

A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by

Airida [17]

Answer:

Mobility of the minority carriers, $\mu_{n} =1184.21 cm^{2} /V-sec$

Diffusion coefficient for minority carriers, $D_{n} = 29.20 cm^2 /s$

Verified from Einstein relation as $\frac{D_{n} }{\mu_{n} } = 25 mV$

Explanation:

Length of sample, $l_{s} = 2 cm$

Separation between the two probes, L = 1.8 cm

Drift time, $t_{d} = 0.608 ms$

Applied voltage, V = 5 V

Mobility of the minority carriers ( electrons), $\mu_{n} = \frac{V_{d} }{E}$

Where the drift velocity, $V_{d} = \frac{L}{t_{d} }$

$V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s$

and the Electric field strength, $E = \frac{V}{l_{s} }$

E = 5/2

E = 2.5 V/cm

Mobility of the minority carriers:

$\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec$

The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

$\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV$

Verified.

4 0

2 years ago