1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sveta_85 [38]
2 years ago
13

Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric e

ld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
Physics
1 answer:
kvv77 [185]2 years ago
5 0

This question is incomplete, the complete question is;

A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.

Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.

<em>Hint </em>: ΔV = Ed <em> </em>

Answer:

the required potential difference, between the capacitor plates is 600 V

Explanation:

Given the data in the question;

B = 0.60 T

d = 2.0 mm = 0.002 m

v = 5.0 × 10⁵ m/s.

since particle pass straight through without deflection.

F_{net = 0

so, F_E = F_B

qE = qvB

divide both sides by q

E = vB

we substitute

E = (5.0 × 10⁵) × 0.6

E = 300000 N/C

given that; potential difference ΔV = Ed

we substitute

ΔV = 300000 × 0.002

ΔV = 600 V

Therefore, the required potential difference, between the capacitor plates is 600 V

You might be interested in
In pairs figure-skating competition, a 65-kg man and his 45-kg female partner stand facing each other both at rest on the ice. I
Lina20 [59]

Answer:

The final velocity of her partner is approximately -1.04 m/s or 1.04 m/s in the opposite direction to her direction of motion

Explanation:

The given parameters are;

The mass of the man, m₁ = 65 kg

The mass of the woman, m₂ = 45 kg

Taking the relative initial velocity of the man and the woman as 0 m/s, we have;

The initial velocity of the man, v₁₁ = 0 m/s

The initial velocity of the man, v₁₂ = 0 m/s

The final velocity of the woman, v₂₂ = 1.5 m/s

The final velocity of the man = v₂₁

Therefore, we have, by the conservation of momentum principle;

The total initial momentum = The total final momentum

Which gives;

m₁ × v₁₁ + m₂ × v₁₂ = m₁ × v₂₁ + m₂ × v₂₂

Substituting the known values;

65 × 0 + 45 × 0 = 65 × v₂₁ + 45 × 1.5

∴ 65 × v₂₁ + 45 × 1.5 = 0

45 × 1.5 = - 65 × v₂₁

v₂₁ = 45 × 1.5/(-65) ≈ -1.04 m/s

The final velocity of the man, her partner = v₂₁ ≈ -1.04 m/s.

6 0
3 years ago
Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th
brilliants [131]

From the case we know that:

  1. The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
  2. The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
  3. The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0
1 year ago
What connects the upper motor neurons to lower motor neurons?
skad [1K]

Answer:

The upper motor neurons synapse in the spinal cord connect with anterior horn cells of lower motor neurons, usually via interneurons. The anterior horn cells are the cell bodies of the lower motor neurons and are located in the grey matter of the spinal cord.

Explanation:

Interneurons are the central nodes of neural circuits, enabling communication between the upper motor neurons, sensory or motor neurons located in the brain and spinal cord and they send signals to lower motor neurons or central nervous system (CNS) in the brain stem and spinal cord . When they get a signal from the upper motor neurons, they send another signal to your muscles to make them contract. They play vital roles in reflexes, neuronal oscillations, and neurogenesis in the adult mammalian brain.

Renshaw cells are among the very first identified interneurons. They are excited by the axon collaterals of the motor neurons. In addition, Renshaw cells make inhibitory connections to several groups of motor neurons.

7 0
3 years ago
You want to move a 4- kg bookcase to a different place in the living room. If u push with a force of 65 n and the bookcase accel
IrinaK [193]

Answer:

1.65

Explanation:

The equation of the forces along the horizontal direction is:

F-F_f = ma (1)

where

F = 65 N is the force applied with the push

F_f is the frictional force

m = 4 kg is the mass

a=0.12 m/s^2 is the acceleration

The force of friction can be written as F_f = \mu R (2), where

\mu is the coefficient of kinetic friction

R is the normal force exerted by the floor

The equation of forces along the vertical direction is

R-mg=0 (3)

since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find

F-\mu mg = ma

And solving for \mu,

\mu = \frac{F-ma}{mg}=\frac{65-(4)(0.12)}{4(9.8)}=1.65

7 0
2 years ago
If a man weight 155 lb on earth, specify
Natali [406]
For the answer to the question above, on Earth, a one-pound object has a mass of about 0.453592 kilograms. 

<span>Therefore the man's mass is 155 * 0.453592 = 70.30676 kilograms. </span>

<span>The part about the Moon's gravity is irrelevant. While the weight of a person or object would be different on the Moon, the mass would be the same.</span>
3 0
2 years ago
Other questions:
  • Help please me please
    14·1 answer
  • you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the win
    8·1 answer
  • Why do the sun and moon seem to move across the sky each day? the Earth’s rotation the sun’s rotation the moon’s revolution the
    15·1 answer
  • A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h
    8·1 answer
  • A beam of unpolarized light in air strikes a flat piece of glass at an angle of incidence of 54.2 degrees. If the reflected beam
    14·2 answers
  • In the visible spectra of stars, absorption lines of hydrogen are produced when atoms are excited from n = 2 to higher levels (t
    5·1 answer
  • The diagram shows the process used in gene therapy.
    12·2 answers
  • A mnemonic device in which phrases or poems use the first letter of each word to help a person remember the information is an ex
    11·2 answers
  • What the energy transformation of an solar panel?
    13·1 answer
  • A car travels eastwards at 60km/h for 2h, then travels northwards at 20km/h for 8h. Find,
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!