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3 years ago

Which of the following statements about the force on a charged particle due to a magnetic field are not valid?

1 answer:

5 0

The correct answer is "None of the above; all of these statements are valid." All the statements namely, it depends on the particle's charge, it depends on the strength of the external magnetic field, it depends on the particle's velocity, and it acts at right angles to the direction of the particle's motion are all valid. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

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A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

$a=\frac{qvB}{m}$

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

8 0

3 years ago

When traveling on narrow mountain roads _______________. A. honk your horn if you cannot see at least 200 ft ahead B. expect oth

yarga [219]

Answer:

The correct option is;

A. honk your horn if you cannot see at least 200 ft ahead

Explanation:

According the California Driver Handbook on Safe Driving Practices, it is required of the driver driving on a narrow mountain road without clear visualization of what is 200 ft ahead of her or him to honk the horn of the vehicle.

The sounding of the horn will alert those ahead of the driver of the possible danger due to her or his oncoming vehicle so that they (those ahead of the driver's oncoming vehicle) can react appropriately.

6 0

4 years ago

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axi

Hitman42 [59]

Answer:

the balls reached a height of 4.9985 m

Explanation:

Given the data in the question;

mass one m = 3.8 kg

mass two M = 2.1 kg

Initial velocities

u = 22 m/s

U = { moving downward} = 12 m/s

Now, using the law conservation of linear moment;

mu + MU = v( m + M )

we solve for "v" which is the velocity of the ball s after collision;

v = (mu + MU) / ( m + M )

so we substitute our given values into the equation

v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )

v = ( 83.6 - 25.2 ) / 5.9

v = 58.4 / 5.9

v = 9.898 m/s

Now, we determine required height using the following relation;

v"² - v² = 2gh

where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²

0 - v² = 2gh

v² = -2gh

so we substitute

( 9.898 )² = -2 × -9.8 × h

97.97 = 19.6 × h

h = 97.97 / 19.6

h = 4.9985 m

Therefore, the balls reached a height of 4.9985 m

8 0

3 years ago

Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of

vesna_86 [32]

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

8 0

3 years ago

A hot rod can accelerate from 0 to 60 km/h in 5.4 s.

mylen [45]

Answer:

<h2>a) Acceleration is 3.09 m/s²</h2><h2>b) Distance traveled is 45.05 m</h2><h2>c) Time taken to travel 250 m is 12.72 s</h2>

Explanation:

a) We have equation of motion v = u + at

Initial velocity, u = 0 km/hr = 0 m/s

Final velocity, v = 60 km/hr = 16.67 m/s

Time, t = 5.4 s

Substituting

v = u + at

16.67 = 0 + a x 5.4

a = 3.09 m/s²

Acceleration is 3.09 m/s²

b) We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 3.09 m/s²

Time, t = 5.4 s

Substituting

s = ut + 0.5 at²

s = 0 x 5.4 + 0.5 x 3.09 x 5.4²

s = 45.05 m

Distance traveled is 45.05 m

c) We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 3.09 m/s²

Displacement, s = 0.25 km = 250 m

Substituting

s = ut + 0.5 at²

250 = 0 x t + 0.5 x 3.09 xt²

t = 12.72 s

Time taken to travel 250 m is 12.72 s

6 0

3 years ago