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Eduardwww [97]
2 years ago
13

Please help!

Physics
1 answer:
Dovator [93]2 years ago
6 0
  • Work=W=5.0×10^-7J
  • Charge=8.0nC=Q
  • Vf=24V

We need potential difference V first

\\ \sf\longmapsto V=\dfrac{W}{Q}

\\ \sf\longmapsto V=\dfrac{5\times 10^{-7}J}{8\times 10^{-9}C}

\\ \sf\longmapsto V=0.625\times 10^2V

\\ \sf\longmapsto V=62.5V

Now

\\ \sf\longmapsto V=V_f-V_i

\\ \sf\longmapsto V_i=V-V_f

\\ \sf\longmapsto V_i=62.5-24=38.5V

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A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15 minutes, w
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Answer:

The average speed of the bus, v = 1.55 m/s

Explanation:

Given that,

Number of blocks traveled by bus towards east = 6

Number of blocks traveled by bus towards north = 8

Length of each block = 100 m

Distance traveled by bus towards east 6 x 100 = 600 m

Distance traveled by bus towards north 8 x 100 = 800 m

The total distance traveled, d = 600 + 800 = 1400

Time taken by the bus to travel is, t = 15 minutes

The velocity is given by the formula

                                  v = d/t  m

Substituting the values in the above equation

                                   v = 1400 m /(15 x 60) s

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Hence, the average speed of the bus, v = 1.55 m/s

6 0
3 years ago
A Cat is on a balcony floor (90cm below the railing), keenly eyeing a butterfly hovering 60 cm above the railing. With what spee
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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:

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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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All describe the relationship of the parts of an electrical current except: _________.
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A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component o
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Explanation:

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