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frozen [14]
2 years ago
11

A particle with charge q, mass m, and initial speed v0 in the x direction enters a region where the electric field is uniform in

the x direction. How much time does it take for the particle to cross the region of length d
Physics
1 answer:
Nezavi [6.7K]2 years ago
8 0

Answer:

S = 1/2 Vo t + 1/2 a t^2 = d      time for particle to travel distance d

F = E q         force acting on particle

a = F / m = E q / m

d = Vo t + E q / (2 m) t^2

One would need to solve the quadratic equation shown to find the time t

t^2 + (2 m) / E q * V0 t - (2 m) / E q * d = 0

or t^2 + A V0 t - A d = 0 where A =  (2 m) / E q

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A car traveling 91 km/h is 280 m behind a truck traveling 76 km/h.
Olegator [25]

Answer:

speed of car = 95 km/h

speed of truck = 75 km/h

relative speed of car with respect to truck = 95 - 75 = 20 km/h

now we will convert it into m/s

now time to cross the truck will be given as

time = 19.8 s

so it will take 19.8 s to cross the truck

Explanation:

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3 years ago
The total energy in a substance as a result of the motion and position of all the particles is _________. a. heat b. temperature
bonufazy [111]
The answer would be Thermal Energy.
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How do you calculate a bearing angle and its equivalent angle?
denis-greek [22]

Explanation:

A bearing if an angle is measured clockwise from north direction.

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3 years ago
Hannah has information about an object in circular orbit around Earth.
Elena-2011 [213]

The correct answer is:

the distance of the orbiting object to Earth.

In fact, we know that the gravitational force that keeps the object in circular motion around the Earth is equal to the centripetal force, so we can write:

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If we re-arrange the equation, we find an expression for the tangential speed of the object:

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3 0
3 years ago
Read 2 more answers
At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?
Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$m / s^2

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$m/s^2 is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0
2 years ago
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