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Simora [160]
4 years ago
10

A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?

Physics
1 answer:
Alenkasestr [34]4 years ago
8 0

accn = - 26m/s divided by 13 secs

accn = 2m/s/s

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The width of the central maxima, formed from light of wavelength 575 nm behind a single slit that has a width of 115 μm, is 1.15
Lady bird [3.3K]

Answer:

 L  = 1.15 m

Explanation:

The diffraction phenomenon is described by the equation

        a sin θ = m λ

Where a is the width of the slit, λ  the wavelength and m is an integer, the order of diffraction is left.

The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry

          tan θ = y / L

          tan θ = sint θ / cos θ≈ sin θ

We substitute in the first equation

           a (y / L) = m λ

The first maximum occurs for m = 1

The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is

         y = 1.15 / 2 = 0.575 cm

         y = 0.575 10⁻² m

Let's clear the distance to the screen (L)

       L = a y / λ  

Let's calculate

     L = 115 10⁻⁶  0.575 10⁻² / 575 10⁻⁹

     L  = 1.15 m

3 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
Sometimes students are taught that "air expands as it is heated". And sometimes they are taught that "as air expands it cools".
hram777 [196]

Answer:

Both are true under specific circumstances. And are related to Boyle's law. volume and pressure in a gas are inversely proportional.

Explanation:

There is a tendency to entropy in our reality, that is, in particular true and visible with gases, they tend to occupy the whole space where they are confined, when we heat a volume of gas, then the movement of the particles and in consequence the pressure of the gas increases and to compensate this the volume tends to be increased too, according to Boyle's law. And the opposite happens when the volume is increased, then the pressure is  relieved and since the particles are further one from each other, then the temperature is lower, and therefore it cools down.

7 0
4 years ago
The density of mobile electrons in copper metal is 8.4 × 1028 m-3. Suppose that i= 4.4 × 1018 electrons/s are drifting through a
neonofarm [45]

Answer: 405.3 minutes

Explanation: In order to explain this problem we have to use the following:

Fisrtly we calculate the volume of the wire, this is given by:

Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.

Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3

then the number of the total electrons in tthe wire volume is given by;

n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23

Finally, considering the current in the wire equal to 4.4*10^18 electrons/s

the time consuming to extract all the electrons from the wire is given by:

t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s

equivalent to 405.3 minutes

4 0
3 years ago
A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the
belka [17]
We know, F = 1/4πε * q₁q₂ / r²
Here, q₁ = 5 * 10⁻⁶ C
q₂ = 2 * 10⁻⁶ C
r = 3 * 10⁻² m west

Substitute their values, 
F = (9 * 10⁹) (5 * 10⁻⁶) (2 * 10⁻⁶)  / (3 * 10⁻²)²

F = 100 N  [ East of positive charge ]

Hope this helps!
5 0
4 years ago
Read 2 more answers
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