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dedylja [7]
3 years ago
9

A bob of mass of 0.18 kilograms is released from a height of 45 meters above the ground level. What is the value of the kinetic

energy gained by the bob at the ground level
Physics
1 answer:
goblinko [34]3 years ago
3 0
Plug into the equation K=1/2mv^2 sooo.. (1/2)(.18)v(you don't have velocity so use the equation for potential energy because you haveheight)(U=mgh so (.18kg)(9.8m/s)(45)=U (set U equal to K)  
(.18)(9.8)(45m)=1/2(.18)v^2
solve for v
don't forget to take the square root of v
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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part
marta [7]

Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

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let compare x axis component

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−3.30×10−7N = -1.2qy

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y= -3.3×10^-7/-1.2×-5.10×10^-9)

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Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

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V•F is

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V•F=0

d. Angle between V and F

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x= arccos(0)

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Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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Can anybody help me solve this problem? Thank you so much!
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at bottom box has only kinetic energy
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h = height at top
a = slope angle
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work = Fs = umgh x cot(a)
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