Ask question

Ask question

All categories

19 days ago

5

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance

of 1.9 cm. You may want to review (Pages 646 - 648) . Part A What is the electric field strength?

1 answer:

inna [2.7K]19 days ago

7 0

Response:

$1.8\times 105 N/C$

Clarification:

We have the given value

$u=2\times 10^7 m/s$

$v=4\times 10^7 m/s$

d=1.9 cm= $\frac{1.9}{100}=0.019 m$

Knowing that 1m=100 cm

Our goal is to determine the electric field intensity.

$v^2-u^2=2as$

Applying the formula

$(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)$

$16\times 10^{14}-4\times 10^{14}=0.038a$

$0.038a=12\times 10^{14}$

$a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2$

$q=1.6\times 10^{-19} C$

Electron mass, m $=9.1\times 10^{-31} kg$

$E=\frac{ma}{q}$

Plugging in the values

$E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}$

$E=1.8\times 105 N/C$

You might be interested in

Six pendulums of mass m and length L as shown are released from rest at the same angle (theta) from vertical. Rank the pendulums

Sav [2839]

Answer: 1m, 1m, 2m, 2m, 4m, 4m.

It’s important to remember that the masses attached do not influence the number of oscillations.

Explanation:

To determine the number of oscillations (complete cycles), we can apply the formula n = t / T ……equation 1

The variables that impact the period of a simple pendulum are solely its length and gravitational acceleration. The period remains unaffected by factors such as mass.

period (T)= 2 x π x √(L/g) ….equation 2

where π = 3.142, L= rope length, and g = 9.8 m/s (gravitational acceleration)

According to the question, the time (t) is 60 seconds.

By merging equations 1 and 2, we obtain

number of oscillations = time / (2 x π x √(L/g))

Case 1: for L = 4m

number of oscillations = 60 / ( 2 x 3.142 x √(4/9.8))

= 14.9 = 14 complete cycles (the problem specifies complete cycles)

Case 2: for L = 2m

number of oscillations = 60 / ( 2 x 3.142 x √(2/9.8))

= 21.4 = 21 complete cycles

Case 3: where L = 4m, results in the same as case 1, yielding 14 complete cycles

Case 4: where L = 2m, mirrors the outcome in case 2, producing 21 complete cycles

Case 5: in the instance of L = 1m

number of oscillations = 60 / ( 2 x 3.142 x √(1/9.8))

= 30.1 = 30 complete cycles

Case 6: when L = 1m, which repeats case 5, also gives 30 complete cycles

From these findings, the order of the pendulums from the highest to lowest number of complete cycles is as follows: 1m, 2m, 2m, 4m, 4m.

Remember, the number of oscillations is independent of their respective masses.

3 0

1 month ago

Read 2 more answers

Mike recently purchased an optical telescope. Identify the part of the electromagnetic spectrum that is closest to the frequency

kicyunya [2911]

The electromagnetic spectrum spans from radio waves to gamma rays. The picture provided illustrates this entire spectrum. However, the optical telescope is limited to observing only the visible spectrum, which ranges from 400 nm to 700 nm. This segment reflects the colors of ROYGBIV, with red exhibiting the highest frequency and violet the lowest frequency.

8 0

1 month ago

Read 2 more answers

Water is made of two hydrogen atoms and one oxygen atom bonded together. Julia is describing how water undergoes a physical chan

kicyunya [2911]

Steam transforms into gas as it escapes into the atmosphere. Even if you manage to capture the steam as it ascends, it can revert to H2O when it cools down. Due to the evaporation process, the final volume of water will differ from the original amount.

8 0

1 month ago

Read 2 more answers

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Maru [2984]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The worker is positioned at 2.5 m

7 0

28 days ago

In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric

ValentinkaMS [3097]

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass $M = 1.0*10^{13} kg$

Radius $r = 1.6km = 1600 m$

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

$g = \frac{GM}{R^2}$

Where G represents the gravitational constant and M denotes the mass of the planet.

$g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}$

$g = 2.607*10^{-4} m/s^2$

Now, let’s compute the time value.

$h = \frac{1}{2} gt^2$

$t = \sqrt{\frac{2h}{g}}$

$t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}$

$t = 87.58s$

Ultimately, the time for the rock to hit the surface is t = 87.58s.

8 0

1 month ago