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3 years ago

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Jim stands at the front of a space ship at 90% of the speed of light and fires a laser gun forward. Jim’s ship is moving past St

eve who sees the ship moving at 90% of the speed of light away from him. From Steve’s point of view the laser light leaving the gun moves at? A. 190% speed of light B. 10% C. 90% D. 100%

1 answer:

klasskru [66]3 years ago

3 0

It's 100% (D). The speed of light is the same, in all frames of reference,
and no matter whether the source or the observer is moving, or how fast,
or in what direction, or for how long, or what day it is.

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2. An electron and a proton are separated by 5 cm:

Diano4ka-milaya [45]

Answer:

a) the charge of an electron is equivalent to the magnitude of the elementary charge but barring a negative sign since the side of the elementary charge is roughly 1.602 * 10 - 19 Columbus then the charge of the electronic is-1.602 * 10 - 19

b) b=2T on the electron moving in the magnetic field

7 0

3 years ago

Two light bulbs are 2.0 m apart. From what distance can these light bulbs be marginally resolved by a small telescope with a 4.5

andrezito [222]

Answer:

R = 1.2295 10⁵ m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

how angles are measured in radians

θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

$\frac{y}{R} = 1.22 \frac{ \lambda}{D}$

R = $\frac{ y \ D}{1.22 \lambda}$

let's calculate

R = $\frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}$

R = 1.2295 10⁵ m

3 0

3 years ago

The carbon isotope 14C is used for carbon dating of archeological artifacts. 14C(mass 2.34×10−26kg) decays by the process known

Nookie1986 [14]

Answer:

2240.92365 m/s

Explanation:

$m_1$ = Mass of electron = $9.11\times 10^{−31}\ kg$

$v_1$ = Speed of electron = $5.7\times 10^7\ m/s$

$p_2$ = Neutrino has a momentum = $7.3\times 10^{-24}\ kg m/s$

M = total mass = $2.34\times 10^{-26}\ kg$

In the x axis as the momentum is conserved

$Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s$

In the y axis

$Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s$

The resultant velocity is

$R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s$

The recoil speed of the nucleus is 2240.92365 m/s

3 0

3 years ago

An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal

alexandr1967 [171]

Answer

given,

resistance = 0.05 Ω

internal resistance of battery = 0.01 Ω

electromotive force = 12 V

a) ohm's law

V = IR

and volage

$V = \epsilon - Ir$

now,

$IR = \epsilon - Ir$

$I(R+r) = \epsilon$

$I= \dfrac{\epsilon}{R+r}$

inserting the values

$I= \dfrac{12}{0.05+0.01}$

I = 200 A

b) Voltage

V = I R

V = 200 x 0.05

V = 10 V

c) Power

P = I V

P = 200 x 10 = 2000 W

d) total resistance = 0.05 + 0.09 = 0.14 Ω

$I= \dfrac{\epsilon}{R+r}$

$I= \dfrac{12}{0.14+0.01}$

I = 80 A

V = 80 x 0.05 = 4 V

P = 4 x 80 = 320 W

6 0

3 years ago

Read 2 more answers

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

7 0

3 years ago