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zhannawk [14.2K]
3 years ago
10

A 2 kilogram object falls a distance of 5 meters. How much potential energy does this object have before it falls? How much work

is done on it by gravity as it falls
Physics
1 answer:
Y_Kistochka [10]3 years ago
5 0

1) Potential energy before the fall: 98 J

The gravitational potential energy of an object is given by:

U=mgh

where m is the mass of the object, g is the gravitational acceleration and h the heigth of the object above the ground.

In this problem, m = 2 kg, g = 9.8 m/s^2 and h = 5 m: substituting into the equation, we find

U=(2 kg)(9.8 m/s^2)(5 m)=98 J


2) Work done by gravity: 98 J

The work done by a force is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

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

In this problem:

F=mg=(2 kg)(9.8 m/s^2)=19.6 N

d = 5 m

\theta=0^{\circ}

Therefore,

W=(19.6 N)(5 m)(cos 0^{\circ})=98 J

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A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th
Ber [7]

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

7 0
3 years ago
The foot pedal on hydraulic brake system exerts a force of 45 lb on a piston with a diameter of 1.00 cm. The brake fluid in the
ElenaW [278]

Answer:

the correct answer is C,   281 lb

Explanation:

This is an exercise in fluid mechanics specifically on Pascal's principle, which states that the pressure in a fluid is equal to all points that are at the same depth

             P = \frac{F_{1} }{A_{1} } = \frac{F_{2} }{A_{2} }

If we use the subscript 1 for the piston diameter mentor (d = 1.00vm) which is r = 0.05 cm, the force is F1 = 45 lb

          F₂ = \frac{F_{1} }{A_{1} }  F₁

in area of ​​a circules

         A = π r²

we substitute

          F₂ = \frac{r_{2}^2 }{r_{1}^2}  F_{1}

let's calculate

           F2 = 2.50² / 1.00²  45

           F2 =281  lb

therefore the correct answer is C

3 0
3 years ago
Explain what voltage,current, resistance are and how they relate to Ohms law. Then give examples
SOVA2 [1]
Voltage is the difference in charge between two points.
Current is the rate the charge flows
Resistance is the tendency a material has to resist the flow of charge (current)
Combining voltage resistance and current Ohm developed the formula
V (Voltage)= I (Current) x R (Resistance)
3 0
4 years ago
Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc
oee [108]

Answer:

The value is \Delta  \phi   =   4.12 \ rad

Explanation:

From the question we are told that

    The frequency of each sound is  f_1 = f_2 = f =  540 \  Hz

      The speed of the sounds is  v = 330 \  m/s

       The  distance of the first source from the point considered is  a = 4.40 \  m

        The distance of the second source from the point considered is  b  = 4.00  \  m

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_a =  2 \pi [\frac{a}{\lambda}  + ft]

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

           \phi_b =  2 \pi [\frac{b}{\lambda}  + ft]          

Here b is the distance o f the first wave from the considered point  

Gnerally the phase diffencence is mathematically represented as  

           \Delta \phi= \phi_a - \phi_b  =  2 \pi [\frac{ a}{\lambda}  + ft ] - 2 \pi [\frac{b}{\lambda}  + ft ]      

=>      \Delta  \phi   =   \frac{2\pi [ a - b]}{ \lambda }

Gnerally the wavelength is mathematically represented as

        \lambda  =  \frac{v}{f}

=>     \lambda  =  \frac{330}{540}

=>     \lambda  =  0.611 \ m

=>    \Delta  \phi   =   \frac{2* 3.142 [ 4.40 - 4.0 ]}{  0.611  }

=>    \Delta  \phi   =   4.12 \ rad

     

5 0
3 years ago
The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un
Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

5 0
4 years ago
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