The end of Hubbard Glacier in Alaska advances by an average of 105 feet per year. What is the speed of advance of the glacier in m/s? ( 1 ft = 0.305 m, 1 yr = 3.156 × 10's )

Answer:

A. iron.

In the process of nuclear fusion, lighter elements are fused together to form heavier elements. This process releases energy and is what powers the star. However, the fusion of elements heavier than iron requires energy, rather than releasing it. Therefore, once a star has produced iron in its core, it is no longer able to sustain nuclear fusion and will eventually undergo a supernova explosion.

The force is 400.2 N

The work done is 120 J

The power is 48W

Force is a physical concept that describes the influence that one object has on another object, causing it to accelerate or deform. Force can be defined as any influence that changes the motion of an object, such as a push or a pull.

How to solve:

under equilibrium condition

F * 1.45 m =68 kg * 9.81 m/s^2 *0.87 m

F =400.2 N

b)

work done = m*g*h =68 kg*9.81 m/s^2*0.180 m =120.0744 J =120 J

c)

power =120.0744 J *(24 /60 s) =48.02976 W = 48 W

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Okay, let's break this down step-by-step:

1) A neutron has a mass of 1.7 x 10-27 kg and an approximate radius of 1.2 x 10-15 m.

So we know the mass and radius of a single neutron.

2) The mass of the sun is 2.0 x 1030 kg.

So we know the total mass of the sun, which is much greater than a neutron.

3) When a massive star reaches the end of its life, it can explode as a supernova.

This supernova can form a neutron star.

4) A neutron star has a density about the same as a neutron.

So we can conclude that a neutron star has a density of:

Density = Mass / Volume

= (1.7 x 10-27 kg) / (4/3 * pi * (1.2 x 10-15 m)3)

= 1.6 x 1017 kg/m3

5) A neutron star forms from the core collapse of a massive star during supernova.

So it has a mass on the order of 1-2 times that of the sun (2 x 1030 kg),

but compressed into a sphere only about 10-20 km in radius.

So its mass would be huge, around 2 x 1030 kg, but confined to a tiny volume,

giving it an immense density, around 1.6 x 1017 kg/m3, the same as a neutron.

Does this help explain the concepts and walk through the calculations? Let me know if you have any other questions!

Answer:

The formula to calculate average speed is:

average speed = total distance / total time

In this case, the distance covered is 500 meters and the time taken is 14.5 minutes. However, we need to convert the time to seconds to get the answer in meters per second.

14.5 minutes = 14.5 x 60 seconds = 870 seconds

So, the average speed is:

average speed = total distance / total time

average speed = 500 meters / 870 seconds

average speed = 0.57 meters per second

Therefore, the average speed of the runner is 0.57 meters per second.

Answer:

19.2 centimeters or 0.192 meters. Both are the same.

Explanation:

T = 2π√(L/g)

T = 60.0 s / 14 ≈ 4.29 s

L = (gT²) / (4π²) ≈ 0.384 m

h + 0.384 m = height above the field

H = h + 0.192 m

H = h

h ≈ 0.192 m or 19.2 cm.

The magnifying glass should be held 16.4 cm away from the nickel to obtain an image with a magnification of +1.53.

Using the thin lens equation, 1/f = 1/o + 1/i, where f is the focal length, o is the object distance, and i is the image distance, and the magnification equation, M = -i/o, where M is the magnification, we can solve for the object distance.

First, solve for the image distance i:

M = -i/o

1.53 = -i/o

i = -1.53o

Then, substitute i into the thin lens equation:

1/0.103 = 1/o + 1/(-1.53o)

Solving for o, we get:

o = 16.4 cm

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Answer:m 1= 36 g=0.036 kg - the mass of ice T1= −77 ◦C. - temperature of ice m2=589 g=0.589 kg - mass of water C1=2090 J/kg · ◦C - specific heat of ice  λ = 3.33 × 10^5 J/kg - latent heat of fusion of water T2=26◦C. - temperature of water C2= 4186 J/kg · ◦C . - - specific heat of water m3=74 g=0.074kg - mass of copper T3=26◦C. - temperature of copper C3=387 J/kg ·◦C - specific heat of copper is and of ice is T - ? - final temperature  1  1 ( 0 −  1 ) −  ℎ  0  1  −  1  2 (  − 0 ) −  ℎ   2  2 (  −  2 ) −   3  3 (  −  3 ) −  m 1 ​ C 1 ​ (0−T 1 ​ )−ice heating to 0 o C m1λ−ice melting m 1 ​ C 2 ​ (T−0)−melted water heating m 2 ​ C 2 ​ (T−T 2 ​ )−water cooling m 3 ​ C 3 ​ (T−T 3 ​ )−copper cooling  1  1 ( 0 −  1 ) +  1  +  1  2 (  − 0 ) +  2  2 (  −  2 ) +  3  3 (  −  3 ) = 0 m 1 ​ C 1 ​ (0−T 1 ​ )+m1λ+m 1 ​ C 2 ​ (T−0)+m 2 ​ C 2 ​ (T−T 2 ​ )+m 3 ​ C 3 ​ (T−T 3 ​ )=0 0.036 ⋅ 2090 ⋅ 77 + 0.036 ⋅ 3.33 ⋅ 1 0 5 + 0.036 ⋅ 4186 ⋅  + 0.589 ⋅ 2090 ⋅ (  − 26 ) + 0.074 ⋅ 387 ⋅ (  − 26 ) = 0 0.036⋅2090⋅77+0.036⋅3.33⋅10 5 +0.036⋅4186⋅T+0.589⋅2090⋅(T−26)+0.074⋅387⋅(T−26)=0 1410.344 ⋅ = 14969.368  = 10.6 1  1410.344⋅T=14969.368 T=10.61 o C Answer:  = 10.6 1  T=10.61 o C

Explanation:

D. The universe is expanding. Hubble's observation of red shift in the light from distant galaxies led him to conclude that these galaxies were moving away from us and that the universe was expanding.

Hubble's most famous observation was his discovery of the relationship between the redshift of light from distant galaxies and their distance from Earth. He observed that the light from distant galaxies was shifted toward longer, redder wavelengths, which indicated that the galaxies were moving away from us. By analyzing the degree of redshift, Hubble was able to calculate the distance of these galaxies from Earth and found that they were much farther away than previously thought. This led him to conclude that the universe was expanding and laid the foundation for the Big Bang theory.

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(a) The impedance of the circuit at 485 Hz is approximately 1253.53 Ω.

(b) The impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.

(c) The current (Irms) at 485 Hz is approximately 326 mA, and the current at 7.50 kHz is approximately 325 mA.

(d)  The resonant frequency of the circuit is approximately 25.74 kHz.

(e) The current (Irms) at resonance is approximately 408 mA.

a) The impedance (Z) of an RLC series circuit can be calculated using the formula:

Z = √(R^2 + (ωL - 1/ωC)^2)

where;

Given:

R = 1.00 kΩ = 1000 Ω

L = 155 mH = 0.155 H

C = 25.0 nF = 25.0 × 10^(-9) F

f = 485 Hz

First, we need to convert the frequency from Hz to radians per second:

ω = 2πf

Substituting the given values into the impedance formula:

Z = √((1000)^2 + ((2π × 485) × 0.155 - 1/(2π × 485 × 25.0 × 10^(-9)))^2)

Calculating Z:

Z = √((1000)^2 + (481.663)^2) ≈ 1253.53 Ω

(b) Given:

f = 7.50 kHz = 7500 Hz

ω = 2πf

Substituting the given values into the impedance formula:

Z = √((1000)^2 + ((2π × 7500) × 0.155 - 1/(2π × 7500 × 25.0 × 10^(-9)))^2)

Calculating Z:

Z = √((1000)^2 + (568.28)^2) ≈ 1256.04 Ω

So, the impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.

(c) Given:

Vrms = 408 V

At 485 Hz:

Irms = Vrms / Z (using the impedance calculated in part (a))

Irms = 408 / 1253.53 ≈ 0.326 A = 326 mA

At 7.50 kHz:

Irms = Vrms / Z (using the impedance calculated in part (b))

Irms = 408 / 1256.04 ≈ 0.325 A = 325 mA

(d) The resonant frequency of an RLC circuit can be calculated using the formula:

f_resonant = 1 / (2π√(LC))

Given:

L = 155 mH = 0.155 H

C = 25.0 nF = 25.0 × 10^(-9) F

Substituting the given values into the resonant frequency formula:

f_resonant = 1 / (2π√(0.155 × 25.0 × 10^(-9)))

Calculating f_resonant:

f_resonant ≈ 25.74 kHz

(e) At resonance, the impedance of the inductor (ωL) and the capacitor (1/(ωC)) cancel each other out, resulting in the minimum impedance of the circuit.

Therefore, at resonance, the impedance (Z) of the circuit is equal to the resistance (R) only.

Given:

R = 1.00 kΩ = 1000 Ω

Vrms = 408 V

Using Ohm's law, we can calculate the current (Irms) at resonance:

Irms = Vrms / R

Irms = 408 / 1000 = 0.408 A = 408 mA

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The magnitude of the force is 3.6 x 10^-6 N, rounded to two significant figures.

The magnitude of the force between two point charges can be calculated using Coulomb's law:

F = k * (q1 * q2) /[tex]r^2[/tex]

where F is the magnitude of the force in Newtons (N), k is Coulomb's constant (9 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs (C), and r is the distance between the charges in meters (m).

Plugging in the given values, we get:

F =[tex](9 * 10^{9} N*m^{2}/C^2) * (-0.0004\ C) * (0.0003 C) / (3 m)^{2}[/tex]

Simplifying the expression, we get:

F = [tex]-3.6 * 10^{-6} N[/tex]

Note that the negative sign in the result indicates that the force is attractive. The magnitude of the force is [tex]3.6 * 10^{-6} N[/tex], rounded to two significant figures.

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Ammeter is always connected in series with the circuit in which the current is to be measured.

An ammeter is connected, to gauge the current flowing through a part or circuit. Since current in a series connection stays constant and an ammeter's resistance is relatively low, the current being measured is unaffected. For the purpose of measuring current, an ammeter is linked in series.

A shunt running parallel to the metre carries the majority of the current at high current values, hence an ammeter can measure a wide range of current values. An ammeter is represented by a circle with a capital A inside it on circuit diagrams.

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