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3 years ago

What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?

Physics

1 answer:

Fudgin [204]3 years ago

3 0

Answer:

potential energy at origin is $2.57*10^{6} volt$

Explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

$\Delta r = \sqrt{43^2 +28^2}$

$\Delta r = 51.313 cm$

potential energy per unit charge $\Delta V = - Edr$

$\Delta V = 5*10^6*51.313*10^{-2} J/C$

$\Delta V = 2.57*10^{6} volt$

potential energy at origin is 2.57*10^{6} volt

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.89m/s A vehicle that starts to move from the rest gets an acceleration of 5m/s² within 2 seconds. Calculate the velocity and di

GuDViN [60]

<u>Explanation:</u>

s = ? u = 0m/s v = ? a = 5m/s² t = 2s

v = u + at

= 0 + (5 x 2)

= 10 m/s

s = ut + 1/2 at²

= (0)(2) + x $\frac{1}{2}$ x 5 x 2²

= 10 m

Hope this helps!

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1 year ago

A sample of Bismuth-212 has a mass of 2.64 grams (g) after 121 seconds (s). What was the initial mass of the sample if Bismuth-2

uysha [10]

The mass of a radioactive element at time t is given by
$m(t) = m_0 ( \frac{1}{2} )^{ \frac{t}{t_{1/2}} }$
where $m_0$ is the mass at time zero, while $t_{1/2}$ is the half-life of the element.

In our problem, $m(t)=2.64 g$ , t=121.0 s and $t_{1/2}=60.5 s$ , so we can find the initial mass $m_0$ :
$m_0= \frac{m(t)}{ (\frac{1}{2})^{t/t_{1/2}} } = \frac{2.64 g}{( \frac{1}{2} )^{121/60.5}} =4 \cdot 2.64 g=10.56 g$

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2 years ago

Why we don’t feel atmospheric pressure?

Veseljchak [2.6K]

Answer:

The reason we can't feel it is that the air within our bodies (in our lungs and stomachs, for example) is exerting the same pressure outwards, so there's no pressure difference and no need for us to exert any effort.

8 0

2 years ago

Read 2 more answers

If the thrower takes 0.90 s to complete one revolution, starting from rest, what will be the speed of the discus at release?

Marysya12 [62]

Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled

Let
α = the angular acceleration
ω = the final angular velocity

The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²

The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s

If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s

Answer: 13.963 rad/s

8 0

2 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

2 years ago