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Westkost [7]
3 years ago
11

Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given poin

t in the field is 80 V, what is the voltage at a point 1.0 m directly east of the point
Physics
1 answer:
qaws [65]3 years ago
5 0

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V

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Temperature Dependence of the pH of pure Water
Mamont248 [21]

Answer:

At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

Explanation:

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8 0
3 years ago
You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee
Archy [21]

Answer:

V = 4.81 V

Explanation:

  • As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
  • We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

        V_{rint} = I* r_{int}

  • The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

       V = V_{b} - V_{rint}  = 5.03 V = 6.0 V - 0.383 A* r_{int}

  • We can solve for rint, as follows:

         r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega

  • When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

       V = V_{b} - V_{rint}  = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V

  • As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.
5 0
3 years ago
As an object rolls downhill, some of the energy is
Charra [1.4K]
When the object is at the top of the hill it has the most potential energy. If it is sitting still, it has no kinetic energy. As the object begins to roll down the hill, it loses potential energy, but gains kinetic energy. The potential energy of the position of the object at the top of the hill is getting converted into kinetic energy. Hope this helped. :)


7 0
3 years ago
Read 2 more answers
PLEASE HELP!! DUE SOON!!!
max2010maxim [7]

Answer:

See attached file :)

Hope this helps!

All the love, Ya boi Fraser :)

4 0
2 years ago
Read 2 more answers
A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground. How fast is th
zysi [14]

Answer:

0.83 m/s

Explanation:

FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:

h=ut+\frac{1}{2}gt^2

where

h is the initial height

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting h = 1.8 m and solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.8)}{9.8}}=0.61 s

We know that the horizontal distance travelled by the ball is

d = 0.5 m

Therefore, we can find the horizontal velocity (which is constant during the whole motion):

v= \frac{d}{t}=\frac{0.5}{0.60}=0.83 m/s

4 0
2 years ago
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