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3 years ago

Moira is drawing the electric field lines around a pair of charges. One charge is positive, and the other charge is negative.

1 answer:

8 0

The best answer is B. If you curve them away, they have a less tendency to be attracted to the pair of charges and would stay around the pair instead of interacting with them.

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A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)

Kobotan [32]

Answer: A)

Explanation: when an electron is placed in a magnetic field, it experiences a force.

This force is given below as

F=qvB*sinθ

F = force experienced by charge.

q = magnitude of electronic charge

v = speed of electron

B= strength of magnetic field

θ = angle between magnetic field and velocity.

What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

F = qvB * 0 = 0.

The charge being at rest has nothing to do with the angle between magnetic field strength and velocity.

3 0

3 years ago

An endothermic reaction is defined as a reaction that _____.

Dmitry_Shevchenko [17]

Answer:

B.) Releases heat or light

Explanation:

5 0

3 years ago

Read 2 more answers

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

6 0

3 years ago

Read 2 more answers

Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a wa

creativ13 [48]

Answer:

in oil film λ = 303.57 10⁻⁹ m

in the water film λ = 319.55 10⁻⁹ m

Explanation:

When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,

when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship

v = λ f

in the void we have

c = λ₀ f

we divide the two expression

c / v = λ₀ / λ

the refractive index is

n = c / v

n = λ₀ /λ

λ = λ₀ / n

let's calculate

in oil film

λ = 425 10⁻⁹ / 1.40

λ = 303.57 10⁻⁹ m

in the water film

λ = 425 10⁻⁹ / 1.33

λ = 319.55 10⁻⁹

those wavelengths are in the ultraviolet

3 0

2 years ago