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3 years ago

7

An airtight box has a removable lid of area 1.10 10-2 m2 and negligible weight. The box is taken up a mountain where the air pre

ssure outside the box is 7.00 104 Pa. The inside of the box is completely evacuated. What is the magnitude of the force required to pull the lid off the box

1 answer:

andrezito [222]3 years ago

4 0

Answer:

F= $7.7\times10^2 N$

Explanation:

The magnitude of force required to pull the lid off the box by air pressure.

We know that Pressure, P= Force(F)/Area(A)

Force, F= P×A

Given: A= $1.10\times10^{-2} m^2$

P= $7\times10^{4} Pa$ .

Therefore, F= $7\times10^{4}\times1.10\times10^{-2}$ .

F= $7.7\times10^2 N$

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If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.

grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution. By definition is defined as:

$\theta = 1.22\frac{\lambda}{d}$

Where,

$\lambda$ = Wavelength

d = Width of the slit

$\theta$ = Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

$r = l\theta$

Relacing $\theta$

$r = l(\frac{1.22\lambda}{d})$

$r = 1.22\frac{l\lambda}{d}$

Replacing with our values we have that,

$r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})$

$r = 3680.91m$

Therefore the diameter of the beam on the moon is

$d = 2r$

$d = 2 * (3690.91)$

$d = 7361.8285m$

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0

3 years ago

A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside

vitfil [10]

Answer:

The value is $B = 0.0452 \ T$

Explanation:

From the question we are told that

The number of turns is N = 1000

The length is L = 50 cm = 0.50 m

The radius is r = 2.0 cm = 0.02 m

The current is I = 18.0 A

Generally the magnetic field is mathematically represented as

$B = \mu_o * \frac{N }{L} * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0$

=> $B = 0.0452 \ T$

3 0

3 years ago

Water can form large dewdrops in nature how would droplets made of vegetable oil instead of water be different

Natasha2012 [34]

Answer:d

Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water

3 0

2 years ago

Read 2 more answers

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east.

Anna71 [15]

Answer:

Time, t = 12 minutes

Explanation:

It is given that,

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

$d=16-8+8-32+11.2=-4.8\ km$

d = 4800 m

Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s

Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :

$v=\dfrac{d}{t}$

$t=\dfrac{d}{v}$

$t=\dfrac{4800\ m}{6.66667\ m/s}$

t = 719.99 seconds

t = 720 seconds

or

t = 12 minutes

So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.

7 0

3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

3 years ago