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An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e

zimovet [89]

Answer:

a) $v = 6.25\times 10^{5} m/s$

b) $v = 1.73\times 10^{4} m/s$

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is $4.55\times 10^5 m/s$

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

$a =\frac{q E}{m}$

$a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}$

$a = 2.58\times 10^{11} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355$

$v = 6.25\times 10^{5} m/s$

b) for proton

$a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}$

$a = -1.41\times 10^{8} m/s^2$

from equation of motion we have

$v^2 = u^2 + 2aS$

$= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355$

$v = 1.73\times 10^{4} m/s$

3 0

2 years ago

When summer changes to fall, what seasonal changes do plants experience? (IGNORE HIGHLIGHTED ANSWER)

lorasvet [3.4K]

Answer:

C

Explanation:

There is a decrease in temperature and daylight and plants produce less food.

Please mark brainliest!!

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7 0

1 year ago

The acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. Th

Lorico [155]

1/3 the weight than it is on earth, duh

3 0

2 years ago

Read 2 more answers

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

2 years ago

A small airplane with a wingspan of 18.0 m is flying due north at a speed of 63.6 m/s over a region where the vertical component

choli [55]

Answer:

(a) ε = 1373.8.

(b) The wingtip which is at higher potential.

Explanation:

(a) Finding the potential difference between the airplane wingtips.

Given the parameters

wingspan of the plane is = 18.0m

speed of the plane in north direction is = 70.0m/s

magnetic field of the earth is = 1.20μT

The potential difference is given as:

ε = Blv

where ε = potential difference of wingtips

B = magnetic field of earth

l = wingspan of airplane

v = speed of airplane

ε = 1.2 x 18.0 x 63.6

ε = 1373.8

(b) Which wingtip is at higher potential?

The wingtip which is at higher potential.

5 0

2 years ago