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liraira [26]
3 years ago
7

A proton in a uniform electric field moves along a straight line with constant acceleration. Starting from rest it attains a vel

ocity of 1,000,000 m/s in a distance of 0.01 m. a.) What is the acceleration? b.) What time is required to reach the given velocity?
Physics
1 answer:
Sav [38]3 years ago
5 0

a) The acceleration of the proton is 5.0\cdot 10^{13} m/s^2

b) The time required to reach the given velocity is 2\cdot 10^{-8}s

Explanation:

a)

This is a motion at constant acceleration, so we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the proton in this problem, we have:

v=1,000,000 m/s is the final velocity

u=0 is the initial velocity (it starts from rest)

s = 0.01 m is the distance covered

Solving for a, we find the acceleration:

a=\frac{v^2-u^2}{2s}=\frac{(1,000,000)^2-0}{2(0.01)}=5.0\cdot 10^{13} m/s^2

b)

For this part, we can use the following suvat equation instead:

v=u+at

where:

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken for the velocity to change from u to v

We have here the following data:

v=1,000,000 m/s is the final velocity

u=0 is the initial velocity (it starts from rest)

a=5.0\cdot 10^{13} m/s^2

Solving for t, we find

t=\frac{v-u}{a}=\frac{1,000,000}{5.0\cdot 10^{13}}=2\cdot 10^{-8}s

Learn more about accelerated motion here:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid
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Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

Explanation:

For xenon fluoride

105/2 = 52.5

For F-F

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Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole

For tetraflouride

284/4 = 71

For F-F

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For hexaflouride

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A Ferris wheel turns at a constant 150.0 revolutions per hour. (a) Express this rate of rotation in units of radians per second.
RUDIKE [14]

Answer:

(a) 0.261 rad/s

(b) 1007.72 m

Explanation:

The angular velocity of the Ferris wheel is 150.0 revolutions per hour.

(a) To calculate the angular velocity of the wheel in units of radians per second, you take into account the following equivalence:

1 hour = 3600 seconds

1 revolution = 2π radians

You use the previous conversion factors:

150.0\ \frac{rev}{h}*\frac{2\pi \ rad}{1\ rev}*\frac{1\ h}{3600\ s}=0.261\frac{rad}{s}

In units of radians per seconds the wheel turns at 0.261 rad/s

(b) To find the arc length described by the wheel, you first calculate the angle described by the  wheel in the time t, by using the following formula:

\theta=\omega t     (1)

ω: angular velocity = 0.261 rad/s

t: time = 4.95 min

You first convert the time to units of seconds

4.95min*\frac{60s}{1min}=297s

Next, you replace the values of the parameters in the equation (1):

\theta=(0.261\frac{rad}{s})(297s)=77.51rad

Next, you use the following formula for the arc length:

s=r\theta     (2)

r: radius of the wheel = 13.0 m

You replace the values of the parameters in the equation (2):

s=(13.0m)(77.51rad)=1007.72m

The arc length described by the wheel is 1007.72m

4 0
3 years ago
A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its moti
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Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

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