If one half-life is 60 seconds, how old is the sample after 5 half-lives?

How can tidal force from the moon affect our earth?

VARVARA [1.3K]

Answer:

sorry but I can understand the question

5 0

2 years ago

Read 2 more answers

The element in an incandescent light bulb that releases light energy is

Leviafan [203]

A thin tungsten filament

5 0

2 years ago

A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

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#LearnwithBrainly

5 0

3 years ago

An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below

grin007 [14]

<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let $P$ be the pressure at a point.

Let $p$ be the density fluid at a point.

Let $v$ be the velocity of fluid at a point.

Bernoulli's equation states that $P+\frac{1}{2}pv^{2}+pgh=constant$ for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let $p_{up}$ be the pressure of a point just above the wing.

Let $p_{do}$ be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

$\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}$

So, $p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa$

Force is given by the product of pressure difference and area.

Given that area is $27ms^{2}$ .

So,lifting force is $16398.48\times 27=442758.96N$

6 0

3 years ago

After you enlarge a map, which one of the following scale remains correct?

9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

6 0

3 years ago