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Morgarella [4.7K]
2 years ago
14

2) A man squeezes a pin between his thumb and finger, as shown in Fig. 6.1.

Physics
1 answer:
Salsk061 [2.6K]2 years ago
8 0
<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

<h3>pressure = 1.53 pascals</h3>

hope that helps and please tell me if i am wrong :)

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Certain meteorites have been examined and found to carry samples of which molecules?
ki77a [65]

Answer:

Sugars...

Explanation:

Several meteorites have been found to carry molecules of sugars that are essential for life. These sugars include Ribose, Arabinose and Xylose. These are found in meteorites that are rich in carbon. These significant discoveries can pave way in finding the origin of life on Earth.

6 0
2 years ago
A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro
aleksley [76]

Answer:

Potential difference though which the electron was accelerated is 2.67\times 10^{-6}\ uV\  .

Explanation:

Given :

De Broglie wavelength , \lambda=750\ nm.

Plank's constant , h=6.626\times 10^{-34}\ J.s \ .

Charge of electron , e=-1.6\times 10^{-19}\ C.

Mass of electron , m=9.11\times 10^{-31}\ kg.m=9.11\times 10^{-31}\ kg.

We know , according to de broglie equation :

\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}

Putting all values in above equation we get ,

V=2.67\times 10^{-6}\ uV .

Hence , this is the required solution.

5 0
2 years ago
Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1110N. 
From the geometry, the normal component of Fp can be calculated: 
Fpn = Fp * cos(θp) 
= 1016.31 N * cos(53) 
= 611.63 N 

The total normal force Fn then is: 
Fn = Fg + Fpn 
= 1110 + 611.63
= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
= (Fp * sin(θp)) + 0 
= 1016.31 N * sin(53) 
= 811.66 N</span>

 

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
Two sinusoidal waves are identical except for their phase. When these two waves travel along the same string, for which phase di
Kamila [148]

Answer:

zero or 2π is maximum

Explanation:

Sine waves can be written

      x₁ = A sin (kx -wt + φ₁)

     x₂ = A sin (kx- wt + φ₂)

When the wave travels in the same direction

      Xt = x₁ + x₂

      Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

     a = kx + wt

     Xt = A [sin (a + φ₁) + sin (a + φ₂)

We develop breasts of double angles

     sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a

    sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a

Let's make the sum

     sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)

to have a maximum of the sine function, the cosine of fi must be maximum

     cos φ₁ + cos φ₂ = 1 +1 = 2

the possible values ​​of each phase are

     φ1 = 0, π, 2π  

     φ2 = 0, π, 2π,  

so that the phase difference of being zero or 2π is maximum

6 0
2 years ago
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