Can you please help me with this question

Which of the following is correct? *

Natalija [7]

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

5 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

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

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

4 years ago

A new company is making security cameras. These security cameras will be different than other security cameras because instead o

babunello [35]

Answer:

b

Explanation:

4 0

2 years ago

In 1929 Edwin Hubble used a telescope to discover the movement of galaxies. True or false

larisa86 [58]

Explanation :

Edwin Hubble used a telescope to discover the movement of galaxies in 1929. He was an American Astronomer.

The telescope invented by him is called a Hubble telescope. It has a 2.4-meter mirror.

He gives the relation between the distance to a galaxy and its recessional velocity. This is known as the Hubble law. The relation is as follows :

$v=H_0\ r$

where,

$v$ is recessional velocity.

$H_0$ is Hubble constant.

$r$ is distance.

<em>So, the given statement is True.</em>

4 0

3 years ago

Read 2 more answers

The 1.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine

arlik [135]

Answer:

Explanation:

check attached image for figure, there is supposed to be a figure for this question containing a distance(height of collar at position A) but i will assume 0.2m or 200mm

Consider the energy equilibrium of the system

$U_{A-B}=\bigtriangleup T\\\\F\cos 30°\times h_A - F\sin30°\times h_A + Wh_A=\frac{1}{2}m(v^2_B-v^2_A)\\\\v_B=\sqrt{\frac{2Fh_A(\cos 30° - \sin30°)+mgh_A}{m+v^2_A}}$

Here, F is the force acting on the collar, $h_A$ is the height of the collar at position A, m is the mass of the collar C, g is the acceleration due to gravity, $v_B$ is the velocity of the collar at position B, and $v_A$ is the velocity of the collar at A

Substitute 14.4N for F, 0.2m for $h_A$ , 1.5kg for m, $9.81m/s^2$ for g and 0 for $v_A$

$v_B=\sqrt{\frac{2(14.4\times 0.2(\cos 30° - \sin30°)+1.5\times 9.81\times 0.2}{1.5+0}}\\\\=\sqrt{\frac{6.618}{1.5}}\\\\=4.412m/s$

Therefore, the velocity at which the collar strikes the end B is 4.412m/s

8 0

3 years ago