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lakkis [162]
3 years ago
8

Help with this physics task pls

Physics
1 answer:
cupoosta [38]3 years ago
7 0

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )  

t=\frac{v}{a} ; d=s*(t-t_{0} )

2)

k=\frac{2*U}{x^{2} }; T_{2}=\frac{P_{2}*V_{2}*T_{1}  }{P_{1}*V_{1}  }  \\

3)

L=\frac{F}{\pi*r*P}; d=\frac{w}{F*cos(o)}

4)

t^{2}=\frac{2*x}{g}  ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} }  \\

5)

h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}

6)

h=\frac{m}{(1/2)*\pi *r^{2} }  ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2}   }

7)

b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}

8)

a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2}  }{M}

9)

v_{o}=\frac{x-\frac{1}{2}*a*t^{2}  }{t}  ; F=\frac{W+uNd}{d*cos(o)}

10)

h=\frac{E-\frac{1}{2}*m*v^{2}  }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2}  }{\frac{1}{2}m }

11)

N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2}   }{\frac{1}{2}*k }

12)

x=x_{o} +\frac{v^{2-v_{o}^{2}  } }{2a}  ;  m=\frac{P*A-F_{1}-F_{2} }{g}

13)

x_{o} = x-\frac{F}{k} ;  u=\frac{cos(o)-\frac{a}{g} }{sin(o)}

14)

t=\frac{d}{v} +t_{o} ; t_{o} = t-(\frac{v-v_{o} }{a} )

15)

F_{2}=\frac{W-F_{1} *d}{d}+F_{3}   ;  v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}

16)

y_{1}=y-\frac{u}{mg}  ; x^{2} = \frac{2W}{k}+x_{o} ^{2}

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Answer:

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Explanation:

The potential energy is transformed to kinetic energy less the frictional energy. Potential energy= mgh where m represent mass, g is acceleration due to gravity and h is the height of cliff

Since we have force of air resistance, work done due to air resistance will be product of force and distance

mgh-Fh= 0.5mv^{2}= KE

Substituting 10 Kg for m, 9.81 for g and 60 m for F then the kinetic energy at the bottom will be

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A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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