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2 years ago

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Megan rode the bus to school, which is located 8 kilometers from her home. If Megan's frame of reference is her house, and it to

ok her 15 minutes to get home, what was the bus' average velocity during the return trip?

1 answer:

Dmitriy789 [7]2 years ago

7 0

Answer:

Explanation: idk sry

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Henri draws a wave that has a 4 cm distance between the midpoint and the trough. Geri draws a wave that has an 8 cm vertical dis

andre [41]

Answer:

Henri’s wave and Geri’s wave have the same amplitude and the same energy

Explanation:

The amplitude of a wave is the distance between the midpoint and the trough (or the crest). This is equivalent to half the distance between the trough and the crest. Therefore:

amplitude of Henri's wave: 4 cm

amplitude of Geri's wave: 8/2 = 4 cm

The energy of a wave is directly proportional to its amplitude.

6 0

2 years ago

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An unspecified force causes a 0.20-kg object to accelerate at 0.40 m/s2. If 0.30 kg is added to the 0.20-kg object and the force

Naddik [55]

Answer:

a = 0.16

Explanation:

given,

mass of the object 1 = 0.2 kg

mass of the object 2 = 0.3 kg

acceleration when force is on 0.2 kg = 0.4 m/s²

acceleration when both mass are combine = ?

F = m a

F = 0.2 × 0.4

F = 0.08 N

force acting is same and total mass = 0.2 + 0.3 = 0.5 Kg

F = m a

$a = \dfrac{F}{m}$

$a = \dfrac{0.08}{0.5}$

a = 0.16 m/s²

the acceleration acting when both the body is attached is a = 0.16

4 0

2 years ago

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

WINSTONCH [101]

Answer:

$=2,012,319.36 \ m/s$

Explanation:

-The only relevant force is the electrostatic force

-The formula for the electrostatic force is:

$F = Eq$

E is the electric field and q is the magnitude of the charge.

#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.

$F_e = F_p\\\\F_e= Force \ on \ electron\\F_p = Force \ on \ proton$

-Applying Newton's 2nd Law:

$F=ma$

$F_e=M_ea_e$

$F_p=M_pa_p$

#equate the two forces:

$F_e = F_p\\\\M_ea_e=M_pa_p\\\\a_e=\frac{M_pa_p}{M_e}$

#The equations for velocity in uniform acceleration:

$V_f^2=V_o^2+2ad\\\\V_o^2=0\\\\\therefore V_f^2=2ad$

#For the proton:

$V_f^2=2a_pd\\\\a_p=\frac{V_f^2}{2d}\\\\a_p=\frac{47000m/s)^2}{2d}$

#For the electron:

$V_f^2=2{a_e}^2\times 2d\\\\A_e=M_p\times A_p/M_e\\\\V_f^2=M_p\times (47000m/s)^2/2d\times2d/M_e\\\\V_f^2=M_p\times (47000m/s)^2/M_e\\\\V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}$

The mass values of the proton and electron are:

$M_p=1.67\times 10^{-27} kg\\\\M_e=9.11\times10^{-31}kg$

The speed of the ion is therefore calculated as:

$V_f=47000m/s\times\sqrt{\frac{M_p}{M_e}}\\\\=47000m/s\times\sqrt{\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\\\\=2,012,319.36 \ m/s$

Hence, the ion's speed at the negative plate is $=2,012,319.36 \ m/s$

7 0

2 years ago

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

Sonbull [250]

The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.

A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?

The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.

I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.

B <<<<====== answer.

5 0

2 years ago

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The speed of light in benzene is 2.00×108 m/s. what is the index of refraction of benzene?

Klio2033 [76]

The index of refraction of a material is the ratio between the speed of light in vacuum, c, and the speed of light in that material, v:
$n= \frac{c}{v}$
where the speed of light in vacuum is $c=3 \cdot 10^8 m/s$ . The speed of light in benzene is $v=2.00 \cdot 10^8 m/s$ , so we can use the previous relationship to find the refractive index of benzene:
$n= \frac{3 \cdot 10^8 m/s}{2.00 \cdot 10^8 m/s}=1.5$

7 0

2 years ago