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Sergeeva-Olga [200]
3 years ago
11

A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x6y − 5x. Find the force that

acts at the point (x, y). (Use the following as necessary: x and y.)
Physics
1 answer:
liq [111]3 years ago
7 0

Answer:

Force, F=((-18x^5y+5)i+(-3x^6)j)\ N

Explanation:

Given that,

A potential energy function for a system in which a two-dimensional force acts is of the form of :

U=3x^6y-5x

We need to find the force that acts at the point (x, y). The force in 2 dimensional with components is given by :

F=(\dfrac{-dU}{dx},\dfrac{-dU}{dy})\\\\F=(\dfrac{-d(3x^6y-5x)}{dx},\dfrac{d(3x^6y-5x)}{dy})\\\\F=(-(18x^5y-5)),(-(3x^6))\\\\F=((-18x^5y+5)i+(-3x^6)j)\ N

So, the force acting at the point (x,y) is ((-18x^5y+5)i+(-3x^6)j)\ N. Hence, this is the required solution.

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Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they h
omeli [17]

Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

= 3 x 10⁶ m /s

Kinetic energy of electron = 1/2 m v²

= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²

= 40.95 x 10⁻¹⁹ J

Kinetic energy in electron comes from lose of electrical energy equal to

Ve where V is potential difference under which electron is accelerated and e is electronic charge .

V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

V = 25.6 Volt .

6 0
3 years ago
What are the answers please help
aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.)  10.31 cm

e.)  10.24 cm  

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.)  If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm  

3 0
3 years ago
2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate
adoni [48]

Answer:

<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

  • v is image distance
  • u is object distance, u is 10 cm
  • f is focal length, f is 5 cm

{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

{ \tt{f + fM = v}} \\  { \tt{f(1 +M) = v }} \\ { \tt{1 +M =  \frac{v}{f}  }} \\  \\ { \boxed{ \mathfrak{formular :  } \: { \tt{ M =  \frac{v}{f}  - 1 }}}}

  • v is image distance, v is 10 cm
  • f is focal length, f is 5 cm
  • M is magnification.

{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

<u> </u><u>»</u><u> </u><u>Nature</u><u> </u><u>of</u><u> </u><u>Image</u><u> </u><u>:</u>

  • Image is magnified
  • Image is erect or upright
  • Image is inverted
  • Image distance is identical to object distance.
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If a moving car speeds up until it is going twice as fast, how much kinetic energy doe s it have compared with its initial kinet
dezoksy [38]
The kinetic energy of an object of mass m and velocity v is given by
K= \frac{1}{2} mv^2

Let's call v_i the initial speed of the car, so that its initial kinetic energy is
K_i =  \frac{1}{2} mv_i^2
where m is the mass of the car. 

The problem says that the car speeds up until its velocity is twice the original one, so 
v_f = 2 v_i
and by using the new velocity we can calculate the final kinetic energy of the car
K_f =  \frac{1}{2} mv_f^2 =  \frac{1}{2}m (2 v_i)^2 = 4 ( \frac{1}{2} mv_i^2)=4 K_i
so, if the velocity of the car is doubled, the new kinetic energy is 4 times the initial kinetic energy.
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Compared to water, most metals have low specific heats.<br> true<br> or<br> false
Alchen [17]
Yes that's a true statement. That's why grandmother put a hot water bottle to warm up her bed, and not a hot bar of steel or lead.
8 0
4 years ago
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