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3 years ago

When a 25000-kg fighter airplane lands on the deck of the aircraft carrier, the carrier sinks 0.23cm deeper into the water.

Physics

1 answer:

Genrish500 [490]3 years ago

4 0

Answer:

10604 square meters

Explanation:

0.23 cm = 0.0023m

Assume the carrier has a shape of a rectangular box. When the carrier sinks 0.0023m deeper into water, the extra volume submerged into water is the same as the extra water volume being replaced. This extra volume would add an additional buoyant force to counter balance the extra weight created by the 25000 kg fighter.

With g being constant, the mass of the extra water displaced is the same as the mass of the fighter airplane $m_f$ . And mass of the water displaced is its volume V times its density $\rho$

$V\rho = m_f$

$1025V = 25000$

$V = 25000/1025 = 24.39 m^3$

We assume the carrier has a shape of a rectangular box, so this extra displaced volume is the extra depth d = 0.0023 m times cross-section area A

$V = dA$

$24.39 = 0.0023A$

$A = 24.39 / 0.0023 = 10604 m^2$

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Dafna11 [192]

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3 years ago

Space-shuttle astronauts experience accelerations of about 35 m/s2 during takeoff. What force does a 75 kg astronaut experience

amm1812

Answer:

<h3>The answer is 2625 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 75 × 35

We have the final answer as

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2 years ago

How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to

Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.

3 0

2 years ago

Read 2 more answers

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

2 years ago

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on

alekssr [168]

Answer:

The change in temperature is $\Delta T = 1795 K$

Explanation:

From the question we are told that

The temperature coefficient is $\alpha = 4 * 10^{-3 }\ k^{-1 }$

The resistance of the filament is mathematically represented as

$R = R_o [1 + \alpha \Delta T]$

Where $R_o$ is the initial resistance

Making the change in temperature the subject of the formula

$\Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]$

Now from ohm law

$I = \frac{V}{R}$

This implies that current varies inversely with current so

$\frac{R}{R_o} = \frac{I_o}{I}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]$

From the question we are told that

$I = \frac{I_o}{8}$

Substituting this we have

$\Delta T = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]$

=> $\Delta T = \frac{1}{3.9 * 10^{-3}} (8 -1 )$

$\Delta T = 1795 K$

6 0

3 years ago