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2 years ago

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

1 answer:

6 0

Thank you for posting your question here at brainly. But your question seems incomplete. I will assume you based the situation below:

<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T.

The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>

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4. What type of assessments are based on repeatable, measurable data?

Harman [31]

Answer:

OBJECTIVE ASSESSMENTS
SUBJECTIVE ASSESSMENTS
BODY COMPOSITION ASSESSMENTS
COGNITIVE ASSESSMENTS

Explanation:

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2 years ago

When two adjacent lights blink on and off in quick succession, we perceive a single light moving back and forth between them. th

makkiz [27]

This is called the Phi Phenomenon.
This is an illusion of movement created when two or more adjacent lights blink on and off in quick succession; when two adjacent stationary lights blink on and off in quick succession; we perceive a single light moving back and forth between them. It is an optical illusion of perceiving a series of still images, when viewed in rapid succession, as continuous motion.

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3 years ago

A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s a

anastassius [24]

a) y(max) = 337.76 m

b) t₁ = 5.30 s the time for y maximum

c)t₂ = 13.60 s time for y = 0 time when the fly finish

d) vₓ = 30 m/s vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα v₀ₓ = 60*(1/2) v₀ₓ = 30 m/s ( constant )

v₀y = v₀ * sinα v₀y = 60*(√3/2) v₀y = 30*√3 m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y = y₀ + v₀y*t - (1/2)*g*t² (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt = v₀y - g*t

dy/dt = 0 v₀y - g*t₁ = 0 t₁ = v₀y/g

v₀y = 60*sin60° = 60*√3/2 = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s the time for y maximum

And y maximum is obtained from the substitution of t₁ in equation (1)

y (max) = 200 + 30*√3 * (5.30) - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max) = 337.76 m

Total time of flying (t₂) is when coordinate y = 0

y = 0 = y₀ + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂ - 4.9*t₂² 4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for t₂

t = [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t = [51.96 ±√2700 + 3920]/9.8

t = [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8 we dismiss this solution ( negative time)

t₂ = 13.60 s time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60° vₓ = 30 m/s as we said before v₀ₓ is constant

vy = v₀y - g *t vy = 30*√3 - 9.8 * (13.60)

vy = 51.96 - 133.28 vy = - 81.32 m/s

The sign minus means that vy change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60 m

x = 408 m

5 0

2 years ago

sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

$F_{net} = F_1 - F_2$