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4 years ago

Please help step by step also

Physics

1 answer:

NARA [144]4 years ago

8 0

There's really only one step to this problem . . . REMEMBERING Ohm's law.

Current = Voltage / Resistance

The rest is just arithmetic, either on paper or on your calculator.

Current = (1.5 V) / (3 ohms)

<em>Current = 0.5 Ampere</em>

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Another unfortunate bug splatters on the windshield of a moving car. Which has the greater change in momentum---the bug or the c

bija089 [108]

Answer:

The change in momentum for the bug and the car will be equal, impulses will be equal in opposite directions and the bug will have a greater acceleration compared to the car, because it has a smaller mass.

Explanation:

Hope this helps..

4 0

3 years ago

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the

Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.

3 0

3 years ago

What is the critical angle of light traveling from benzene (n=1.501) into air?

Taya2010 [7]

The critical angle formula should be: sin^-1(1/n)

where "n" is 1.501 into the air

<span>The critical angle of light travelling from benzene, happens because the larger angles of incidence from the inside of the benzene has experienced the total internal reflection. </span>

4 0

4 years ago

Elle is playing with a ball in a bus that moves in a straight line with constant velocity. What can you say about the motion of

kramer

Given that they are all on the same bus that is travelling in a straight line at the same velocity, when Elle throws the ball directly upwards, the ball will simply fall back to her. This is because the bus, Elle, and the ball are all travelling in the same direction and at the same speed. Among the choices, the correct answer is A.

8 0

3 years ago

I need to solve this problem

castortr0y [4]

Answer:

λ = V / f the wavelength versus the frequency

V = f λ and V (speed) proportional λ for a fixed frequency

F = f^2 * (M / L) * λ^2 = (f * λ)^2 * (M / L)^2 force (tension) on string at a given frequency

F2 / F1 = (λ2 / λ1)^2 other items are constant

Let λ1 = 6 then λ2 must be 3/2 λ1 for a constant length

F2 / F1 = (6 / 4)^2 = 9/4

The tension must be increased to 9 / 4 of the original tension

Check: if the frequency is fixed then V will be larger for a larger wavelength (situation 2)

One can also write V = (F / (M / L))^1/2

Then for fixed M L

F2 / F1 = (V2 / V1)^2

Since V = f λ Velocity is proportional to λ for a fixed frequency

Then if V2 / V1 = 3 / 2 F2 = 9/4 F1

4 0

2 years ago

Please help step by step also​

Please help step by step also