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2 years ago

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The H line in Calcium is normally at 396.9 nm. However, in a star's spectrum, it is measured at 398.1nm. How fast is the star mo

ving and is it moving towards the Earth or away from the Earth?
1- 904.3 km/s away from the Earth
2- 904.3 km/s towards the Earth
3- 907.0 km/s towards the Earth
4- 907.0 km/s away from the Earth

1 answer:

agasfer [191]2 years ago

3 0

As we know by Doppler's effect of light we have

$\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$

here we will have

[tex}\frac{398.1 nm - 396.9 nm}{398.1 nm} = \frac{v}{c}[/tex]

here by solving above we have

$3.01 \times 10^{-3} = \frac{v}{c}$

here we have

$v = 904.3 km/s$

since wavelength is increased so we can say that it is moving away

so correct answer is

1- 904.3 km/s away from the Earth

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Answer:

F=UR

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2 years ago

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evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7

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Given :

Displacement , y = 0.75 m .

Angular acceleration , $\alpha=0.95\ s^{-2}$ .

Initial angular velocity , $\omega_o=6.3\ s^{-1}$ .

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The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

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$v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s$

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2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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2 years ago

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Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

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The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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2 years ago

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m_a_m_a [10]

Answer:

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Explanation:

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4 0

2 years ago