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nexus9112 [7]
3 years ago
9

Helpp pls

Physics
1 answer:
Kipish [7]3 years ago
5 0

Answer:

The intensity of the electric field is

|E|=10654.37 \:N/C

Explanation:

The electric field equation is given by:

|E|=k\frac{q}{d^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge at 0.4100 m from the balloon
  • d is the distance from the charge to the balloon

As we need to find the electric field at the location of the balloon, we just need the charge equal to 1.99*10⁻⁷ C.

Then, let's use the equation written above.

|E|=(9*10^{9})\frac{1.99*10^{-7}}{0.41^{2}}

|E|=10654.37 \:N/C

I hope it helps you!

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We will have the following:

First, the equation to use is the following:

d=v_ot+\frac{1}{2}at^2

Now, we transform the total distance the cat would need to travel:

90\operatorname{cm}+60\operatorname{cm}=150\operatorname{cm}\cdot\frac{1m}{100\operatorname{cm}}=1.5m

So, the cat would need to travel 1.5 meters. ("d" in the equation).

Now, using the speed given we determine the time it would take the cat to traverse the 1.5 meters:

t=\frac{1.5m\cdot1s}{0.45m}\Rightarrow t=\frac{10}{3}\Rightarrow t=3.333\ldots

So, the time it would take the cat to traverse the distance will be approximately 3.33 seconds.

Now, we know that the acceleration will be given by Earth's gravity, so:

1.5m=v_0(\frac{10}{3}s)+\frac{1}{2}(-\frac{9.8m}{s^2})(\frac{10}{3}s)^2\Rightarrow1.5m=v_0(\frac{10}{3}s)+(-\frac{490}{9}m)\Rightarrow\frac{1007}{18}m=v_0(\frac{10}{3}s)\Rightarrow v_0=\frac{1007}{60}\frac{m}{s}\Rightarrow v_0=16.78333\ldots

So, the initial vvelocity the cat must leave the floor in order to arrive at the butterfly with the optimum pouncing speed of 0.45 m/s is approximately 16.78 m/s or exactly 1007/60 m/s.

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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la
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Answer:

h=53.09m         (2)

v_{min}>5.05m/s

v_{max}

Explanation:

<u>a)Kinematics equation for the first ball:</u>

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=8.9m/s      

The ball reaches the ground, y=0, at t=t1:

0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}

h=1/2*g*t_{1}^{2}-v_{o}t_{1}           (1)

Kinematics equation for the second ball:

v(t)=v_{o}-g*t

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

y_{o}=h       initial position is the building height

v_{o}=0       the ball is dropped

The ball reaches the ground, y=0, at t=t2:

0=h-1/2*g*t_{2}^{2}

h=1/2*g*t_{2}^{2}         (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03              (3)

We solve the equations (1) (2) (3):

1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)

-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)

2.06*gt_{1}-2v_{o}t_{1}=g*1.06

t_{1}=g*1.06/(2.06*g-2v_{o})

vo=8.9m/s

t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s

t2=t1-1.03              (3)

t2=3.29sg

h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m         (2)

b)t_{1}=g*1.06/(2.06*g-2v_{o})

t1 must :   t1>1.03  and t1>0

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1.03>9.81*1.06/(2.06*g-2v_{o})

1.03*(2.06*9.81-2v_{o})

20.8-2.06v_{o}

(20.8-10.4)/2.06

v_{min}>5.05m/s

limit case: t1>0:

g*1.06/(2.06*g-2v_{o})>0

2.06*g-2v_{o}>0

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v_{max}

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