Answer:
How: It will have a very different chemical composition.
Why: Because the interstellar medium at the beginning of universe were compounded only of Hydrogen and Helium.
Explanation:
Once the Big Bang passed and the universe began its expansion, the temperature started to decrease, what allowed the combination of hydrogen and helium. Therefore, the chemical composition of the molecular cloud in which the first stars were formed only held those elements.
If the Sun and the solar system were formed from a molecular cloud with that composition, it will both have an absence of metallic elements (elements heavier than helium), since all of them are formed through nuclear reactions in the core of the stars, and each time a star dies, the interstellar medium for the next generation is enriched.
Key elements for life as Carbon, Nitrogen, Oxygen, Phosphorus and Sulfur will be missing for that hypothetical scenario.
For example, the lack of O3 (ozone) in the atmosphere of earth will allow that ultraviolet light hits the earth surface.
If our solar system formed with the first stars, the planets would mostly comprise hydrogen and helium, as heavier elements had not yet formed. The conditions would be hotter and harsher, making life as we know it unlikely to evolve. The Sun, too, would've been exhausted already due to its short lifespan.
Explanation:If our Sun and the solar system had formed around the same time as the first stars in the universe, our solar system would look very different. The universe at the time of the formation of the first stars was composed of only hydrogen and helium, with trace amounts of lithium. Thus, the planets in our system, formed from the leftover dust and gasses surrounding our Sun, would mostly be made up of hydrogen and helium.
Believed to have occurred just about 200 million years after the big bang, the first stars were much larger and hotter than the current generation of stars, including our Sun. Hence, the planets in our system could have been significantly hotter and larger. It's also worth noting that these first-generation stars had relatively short lifetimes and ended their lives in explosive supernovae, catalyzing the formation of heavier elements. But this means if our Sun was a first-generation star, it would have exhausted its nuclear fuel billions of years ago.
Due to these circumstances, life as we know it might not have been possible in our solar system. Life on Earth relies on heavier elements like carbon, nitrogen, and oxygen, which were only formed in the universe after generations of stellar evolution and supernova explosions. If our solar system was created at the same time as the first stars, these heavier elements wouldn't have existed yet. Therefore, life, at least life as we understand it, wouldn't likely evolve in our solar system.
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Two point charges are brought closer together, increasing the force between them by a factor of 22. By what factor did their separation decrease? Hint: If the force increases, distance between charges must decrease. Force is indirectly proportional to the distance squared.
Answer:0.21
Explanation:
Given
Two point charges are brought closer together, increasing the force by a factor of 22
Let the original force be
[tex]F=\frac{kq_1q_2}{r^2}---1[/tex]
where [tex]q_1,q_2[/tex] are charges and r is the distance between them
new force [tex]F'=\frac{kq_1q_2}{r'^2}----2[/tex]
divide 1 & 2
[tex]\frac{F'}{F}=\frac{\frac{kq_1q_2}{r'^2}}{\frac{kq_1q_2}{r^2}}[/tex]
[tex]22=\frac{r^2}{r'^2}[/tex]
[tex]r'=\frac{r}{\sqrt{22}}\approx 0.213 r[/tex]
Distance between them is decrease by a factor of 0.21
The separation between two point charges decreased by a factor of 5.
Explanation:To find the factor by which the separation between two point charges decreased when the force between them increased by a factor of 25, we need to understand the relationship between force and distance. According to Coulomb's law, the force between two point charges is inversely proportional to the square of the distance between them. This means that when the distance decreases, the force increases, and vice versa.
Since the force increased by a factor of 25, we can find the factor by which the distance decreased by taking the square root of 25, which is 5.
Therefore, the separation between the two point charges decreased by a factor of 5.
LA 7.0- g bullet is fired into a 1.5- kg ballistic pendulum.The
bullet emerges from the block with a speed of 200 m/s, and theblock
rises to a maximum height of 12 cm. Find the initial speed ofthe
bullet.
Final answer:
The problem involves a ballistic pendulum scenario to determine the initial speed of a bullet, requiring an analysis that combines conservation of momentum and energy. However, the twist in this problem is that the bullet exits the block, requiring indirect methods to deduce the initial conditions.
Explanation:
The question involves finding the initial speed of a bullet by analyzing a ballistic pendulum scenario. This is a physics problem that integrates concepts of momentum and energy conservation. The solution requires two steps: first, finding the velocity of the bullet-block system right after the bullet exits, and second, using energy conservation to link this velocity to the height reached by the block.
Use the conservation of momentum to find the system's velocity right after the bullet exits the block. The exiting speed of the bullet and the mass details are essential for this step. However, an important observation is that the information given does not allow for the direct application of momentum conservation in the traditional sense, since the bullet exits the block, unlike the typical scenario where it remains lodged in.
Next, use the conservation of energy principle to relate the kinetic energy of the block right after the bullet exits to the potential energy at the maximum height. The height given allows calculation of the velocity of the block right after the bullet exits, which indirectly informs the system's dynamics at impact.
This problem has a twist as the bullet does not remain in the block, which complicates the direct application of the conservation of momentum to find the initial speed of the bullet. Instead, it involves a more nuanced approach using the given exit velocity of the bullet and the rise of the block to infer the initial conditions.
The initial speed of the bullet is approximately 23.68 m/s.
To solve this problem, we can use the principle of conservation of momentum and conservation of energy.
1. Conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
[tex]\[m_b v_b = (m_b + m_p)v'\][/tex]
Where:
[tex]\(m_b = 7.0 \, \text{g} = 0.007 \, \text{kg}\)[/tex] (mass of the bullet)
\(v_b\) = initial speed of the bullet (which we need to find)
[tex]\(m_p = 1.5 \, \text{kg}\)[/tex] (mass of the ballistic pendulum)
\(v'\) = final velocity of the bullet and pendulum system
2. Conservation of energy:
The initial kinetic energy of the bullet is equal to the sum of the final kinetic energy of the bullet and the kinetic energy of the pendulum, plus the gravitational potential energy gained by the pendulum.
[tex]\[ \frac{1}{2} m_b v_b^2 = \frac{1}{2} (m_b + m_p) v'^2 + m_p g h\][/tex]
Where:
[tex]\(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)[/tex]
[tex]\(h = 12 \, \text{cm} = 0.12 \, \text{m}\) (maximum height reached by the pendulum)[/tex]
Let's solve these equations step by step:
Step 1: Conservation of momentum:
[tex]\[0.007 \, \text{kg} \times v_b = (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'\][/tex]
[tex]\[0.007 \, \text{kg} \times v_b = 1.507 \, \text{kg} \times v'\][/tex]
[tex]\[v_b = \frac{1.507}{0.007} \times v' \][/tex]
[tex]\[v_b = 215.28 \times v'\][/tex] (Equation 1)
Step 2: Conservation of energy:
[tex]\[ \frac{1}{2} \times 0.007 \, \text{kg} \times v_b^2 = \frac{1}{2} \times (0.007 \, \text{kg} + 1.5 \, \text{kg}) \times v'^2 + 1.5 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.12 \, \text{m}\][/tex]
[tex]\[0.5 \times 0.007 \, \text{kg} \times v_b^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Now, substitute [tex]\(v_b = 215.28 \times v'\)[/tex] from Equation 1 into the above equation:
[tex]\[0.5 \times 0.007 \, \text{kg} \times (215.28 \times v')^2 = 0.5 \times 1.507 \, \text{kg} \times v'^2 + 1.47 \, \text{J}\][/tex]
Simplify and solve for \(v'\):
[tex]\[0.5 \times 0.007 \times (215.28)^2 \times v'^2 = 0.5 \times 1.507 \times v'^2 + 1.47\][/tex]
[tex]\[ 160.49 \times v'^2 = 0.7535 \times v'^2 + 1.47\][/tex]
[tex]\[159.7365 \times v'^2 = 1.47\][/tex]
[tex]\[v'^2 = \frac{1.47}{159.7365}\][/tex]
[tex]\[v' = \sqrt{\frac{1.47}{159.7365}}\][/tex]
[tex]\[v' \approx 0.11 \, \text{m/s}\][/tex]
Finally, we can use this value of \(v'\) to find \(v_b\) using Equation 1:
[tex]\[v_b = 215.28 \times 0.11\][/tex]
[tex]\[v_b \approx 23.68 \, \text{m/s}\][/tex]
So, the initial speed of the bullet is approximately [tex]\(23.68 \, \text{m/s}\).[/tex]
A power station delivers 690 kW of power at 12,000 V to a factory through wires with total resistance 5.0 Ω. How much less power is wasted if the electricity is delivered at 50,000 V rather than 12,000 V? Express your answer using two significant figures.
Answer:
15.579 kW
Explanation:
current passing through the wire
[tex]I = \dfrac{P}{V} = \dfrac{690 \times 10^3}{12000} = 57.5 A[/tex]
power loss
[tex]P_{loss} = I^2R = (57.5)^2\times 5 = 16531.25 W[/tex]
the current transmission for 50,000 V is
[tex]I' = \dfrac{P}{V'} = \dfrac{690 \times 10^3}{50000} = 13.8 A[/tex]
power loss
[tex]P_{loss} = I^2R = (13.8)^2\times 5 = 952.2 W[/tex]
wasted power
= 16531.25 W - 952.2 W
= 15579.05 W = 15.579 kW
Hence, the power wasted is equal to 15.579 kW
Electric power delivers from power station is the rate of energy transfer per unit time. The less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.
What is electric power?The electric power is the amount of electric energy transferred per unit time. It can be given as,
[tex]P=IV[/tex]
Here, (I) is the current and (V) is the electric potential difference.
Given information-
The power delivers by the power station is 690 kW.
The voltage of the power is 12,000 V.
The resistance of the wire is 5.0 Ω.
As the power delivers by the power station is 690 kW and the voltage of the power is 12,000 V. Thus the current flowing through the wire is,
[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{12000}\\I=57.5 \rm A[/tex]
The power loss due to the resistance of the wire is 5.0 Ω is,
[tex]P_{loss}=I^2R\\P_{loss}=(57.5)^2\times5\\P_{loss}=16531.25\rm W\\[/tex]
Now the electricity is delivered at 50,000 V rather than 12,000 V.
The power delivers by the power station is 690 kW and the The voltage of the power is 50,000 V. Thus the current flowing through the wire is,
[tex]I=\dfrac{P}{V}\\I=\dfrac{690\times10^3}{50000}\\I=13.8 \rm A[/tex]
The power loss due to the resistance of the wire is 5.0 Ω is,
[tex]P_{loss}=I^2R\\P_{loss}=(13.8)^2\times5\\P_{loss}=952.9\rm W\\[/tex]
As when the voltage of the power is 12,000 V the power loss is 16531.25 W and when the voltage of the power is 50,000 V the power loss is 952.9 W.
Thus less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is,
[tex]P_{saved}=16531.25-952.2\\P_{saved}=15579\rm W\\P_{saved}=15.579\rm kW[/tex]
Hence, the amount of less power is wasted, when electricity is delivered at 50,000 V rather than 12,000 V is 15.579 kW.
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A coin is dropped from a height and reaches the ground in 2 seconds. Neglecting air resistance, from what height (in meters) was the coin dropped?
Using the formula for distance covered in free fall, the coin was determined to have been dropped from a height of 19.6 meters.
Explanation:To determine the height from which a coin was dropped if it reaches the ground in 2 seconds, we can use the formula for the distance covered in free fall (neglecting air resistance), which is given by:
Distance (d) = ½ * g * t2
where g is the acceleration due to gravity (9.8 m/s2 on Earth), and t is the time in seconds. Plugging in the values, we get:
Distance (d) = ½ * 9.8 * (22) = ½ * 9.8 * 4 = 19.6 meters.
Thus, the coin was dropped from a height of 19.6 meters.
The coin was dropped from a height of 19.62 meters, using the equation [tex]\( s = \frac{1}{2}gt^2 \)[/tex].
To solve this problem, we can use the equation of motion for an object in free fall under gravity:
[tex]\[ s = ut + \frac{1}{2}gt^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the displacement (height) of the object
- [tex]\( u \)[/tex] is the initial velocity (which is 0 because the object is dropped)
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.81 \, \text{m/s}^2 \)[/tex] on Earth)
- [tex]\( t \)[/tex] is the time taken
Given that the initial velocity [tex]\( u = 0 \)[/tex], we can simplify the equation to:
[tex]\[ s = \frac{1}{2}gt^2 \][/tex]
Substituting the known values:
[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (2 \, \text{s})^2 \][/tex]
[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times 4 \, \text{s}^2 \][/tex]
[tex]\[ s = \frac{1}{2} \times 39.24 \, \text{m} \][/tex]
[tex]\[ s = 19.62 \, \text{m} \][/tex]
So, the height from which the coin was dropped is [tex]\( 19.62 \, \text{m} \)[/tex].
The position of a particle moving under uniform acceleration is some function of time and the acceleration. Suppose we write this position as x = kamtn, where k is a dimensionless constant. Show by dimensional analysis that this expression is satisfied if m = 1 and n = 2. (Submit a file with a maximum size of 1 MB.)
Answer:
Explanation:
[tex]x = k\times a^m\times t^n[/tex]
k is constant , it is dimension is zero. Using dimensional unit , we cal write the relation as follows
L = [tex](LT^{-2})^m(T)^n[/tex]
= [tex]L^mT^{-2m+n}[/tex]
Equating power of like items
m=1
-2m+n = 0
n = 2
In a thundercloud there may be electric charges of +44.0 C near the top of the cloud and -44.0 C near the bottom of the cloud. These charges are separated by 1.60 km. What is the electric force on the top charge?
Answer:6.806 MN
Explanation:
Given
Charge on clouds is +44 C and -44 C
They are separated by 1.6 km
and electrostatic Force is given by
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
[tex]F=\frac{9\times 10^9\times 44\times 44}{1600^2}[/tex]
[tex]F=\frac{17,424\times 10^9}{256\times 10^4}[/tex]
F=6.806 MN
An object is undergoing uniform circular motion. Which of the following changes would increase the period of the motion? I. Increase the radius of the circular motion. II. Decrease the radius of the circular motion. III. Increase the speed of the object. IV. Decrease the speed of the object.
Answer:
See it in the pic
Explanation:
See it in the pic
Final answer:
Increasing the radius of the circular motion or decreasing the speed of the object would both increase the period of uniform circular motion. This is due to the direct proportionality of period to radius and inverse proportionality to speed.
Explanation:
To determine which changes would increase the period of an object's uniform circular motion, let's recall the relationship between period (T), speed (v), and radius (r) of the circle. According to the formula for calculating the centripetal acceleration of an object in uniform circular motion (a = v²/r or a = 4π²r/T²), we can deduce that the period T is directly proportional to the radius and inversely proportional to the speed. Therefore, increasing the radius (Option I) would increase the period since T is proportional to the square root of the radius when speed is constant. Decreasing the speed of the object (Option IV) would also increase the period because the period is inversely proportional to speed.
The correct answer to the question of which changes would increase the period of motion in uniform circular motion are: I. Increase the radius of the circular motion and IV. Decrease the speed of the object. Options II and III would decrease the period, not increase it.
The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows at a rate of 1.35 x 10^8 L/Hr. The water enters the plant at 25.4°C and exits at 30.7°C. (a) What is the power plant's total thermal power? (MWT (Megawatt thermal)
(b) What is the efficiency of the power plant?
Answer:
(a) 83475 MW
(b) 85.8 %
Explanation:
Output power = 716 MW = 716 x 10^6 W
Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3
mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000
= 1.35 x 10^8 kg
Time, t = 1 hr = 3600 second
T1 = 25.4° C, T2 = 30.7° C
Specific heat of water, c = 4200 J/kg°C
(a) Total energy, Q = m x c x ΔT
Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J
Power = Energy / time
Power input = [tex]P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W[/tex]
Power input = 83475 MW
(b) The efficiency of the plant is defined as the ratio of output power to the input power.
[tex]\eta =\frac{Power output}{Power input}[/tex]
[tex]\eta =\frac{716}{83475}=0.858[/tex]
Thus, the efficiency is 85.8 %.
Material A has a specific heat of 100 J/kg-K and Material B has a specific heat of 200 J/kg-K. If the two materials start at the same temperature and have heat energy added at the same constant rate, which would you expect to reach a higher temperature first? Material A
Material B
Both the same
Cannot determine from the information given None of the above
Answer:
material A
Explanation:
Specific heat of material A = 100 J/kg - K
Specific heat of material B = 200 J/kg - K
The specific heat of a material is defined as the amount of heat required to raise the temperature of substance of mass 1 kg by 1 °C.
So, material A requires 100 J of heat to raise the temperature of mass 1 kg by 1°C.
So, material B requires 200 J of heat to raise the temperature of mass 1 kg by 1°C.
So, the temperature of material A is higher is more than the material B when same amount of heat is added.
What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a velocity of 30 meters/second?
A. 1.6 meters/second2
B. 3.0 meters/second2
C. 3.4 meters/second2
D. 5.0 meters/second2
Answer: d. 5 m/s^2
Explanation:
Acceleration is the change in velocity in a given time.
a = (30-20)/2 = 5
The ball is travelling in horizontal direction with an acceleration of [tex] \bold{5\ m/ s^{2}} [/tex]
Answer: Option D
Explanation:
The rate at which velocity of an object changes is acceleration i.e., if the velocity of an object is changing, the object is accelerating. The unit of measurement of acceleration is Meter per Seconds Square. Acceleration is the derivative of the changing velocity wrt to time,
[tex]a=\frac{\Delta v}{\Delta t}-----{Eqn} 1[/tex]
Given,
Initial velocity = 20 m/s [tex](v_{i})[/tex]
Final velocity = 30 m/s [tex](v_{f})[/tex]
Initial time = 0 s [tex](t_{i})[/tex]
Final time = 2 s [tex](t_{f})[/tex]
From the equation 1,
[tex]\begin{aligned} a &=\frac{\Delta v}{\Delta t}=\frac{v_{f}-v_{i}}{t_{f}-t_{i}} \\ a &=\frac{30 \mathrm{m} / \mathrm{s}-20 \mathrm{m} / \mathrm{s}}{2 \mathrm{s}}=5 \mathrm{m} / \mathrm{s}^{2} \end{aligned}[/tex]
Hence, the acceleration of ball moving horizontally is [tex] \bold{5\ m/s^{2}} [/tex]
An electrostatic paint sprayer has a 0.8 m diameter metal sphere at a potential of 39 kV that repels paint droplets onto a grounded object. What charge is on the sphere? Give answer in terms of 10^-7 C.
Answer:
The charge on the sphere is [tex]1.735\times 10^{- 7} C[/tex]
Solution:
The electric potential on the surface of a sphere of radius 'b' and charge 'Q' is given by:
[tex]V_{sphere} = \frac{Q}{4\pi\epsilon_{o}b}[/tex]
According to the question:
Diameter, b = 0.8 m
Potential of sphere, [tex]V_{sphere} = 39 kV = 39000 V[/tex]
[tex]{4\pi\epsilon_{o}\frac{0.8}{2}\times 39000 = Q[/tex]
Q = [tex]1.735\times 10^{- 7} C[/tex]
Describe the role of impurities in dielectrics. Why are sapphire and diamond transparent and how can impurities change this?
Answer:
Sapphire and diamond impurities change with their transparency and role of impurities in dielectrics is discussed below.
Explanation:
Temperature dependence of the dielectrics with different degrees of purity measured in the range of 233-313 K. This was observed that the impurities presence is strongly influences the dielectric constant, dipole moment and melting point.
SAPPHIRE
Sapphire is a very important gem stone. It's color is blue, but the naturally occurring sapphires have purple, yellow,orange and green. This variety in color is due to trace amount of presence of impurities like Iron,titanium, chromium, copper and Magnesium. It is the third hardest materials in the Mohs scale. The transparency of Sapphire is due to the impurities present in it.
DIAMOND
Diamond is the very precious stone. The transparency in the diamond is due to the band structure, that is the band gap is high in the diamonds, so they are transparent in nature. Diamond has many unequaled qualities and is very unique among the other minerals. It has highest refractive index of any natural minerals and is transparent over greatest number of wavelengths due to conduction present in it.
A 243 mL cup of whole milk contains about 45 mg of cholesterol. Express the cholesterol concentration of the milk in kilograms per cubic meter (kg/m^3).
Answer: [tex]0.1851 \frac{kg}{m^{3}}[/tex]
Explanation:
We have this concentration in units of [tex]mg/ml[/tex]:
[tex]\frac{45 mg}{243 ml}[/tex]
And we need to express it in [tex]kg/m^{3}[/tex], knowing:
[tex]1 ml= 1 cm^{3}[/tex]
[tex]1 m^{3}= (100 cm) ^{3}[/tex]
[tex]1g = 1000 mg[/tex]
[tex]1 kg = 1000 g[/tex]
Hence:
[tex]\frac{45 mg}{243 cm^{3}} \frac{1 g}{1000 mg} \frac{1 kg}{1000 g} \frac{(100 cm) ^{3}}{1 m^{3}}=0.1851 \frac{kg}{m^{3}}[/tex]
The cholesterol concentration of the milk in kilograms per cubic meter is 0.1851 kg/m³
The calculation is done as follows;
1 mg= 10⁻⁶ kg
1ml= 10⁻⁶ m³
= 45 ÷243
= 0.1851 kg/m³
Hence the cholesterol concentration of the milk is 0.1851 kg/m³
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A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the center of the electrode?
Express your answer to two significant figures and include the appropriate units.
Answer:
The electric field strength is [tex]4.5\times 10^{4} N/C[/tex]
Solution:
As per the question:
Area of the electrode, [tex]A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}[/tex]
Charge, q = 50 nC = [tex]50\times 10^{- 9} C[/etx]
Distance, x = 2 mm = [tex]2\times 10^{- 3} m[/tex]
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:
[tex]\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}[/tex]
Now, the electric field strength of the electrode is:
[tex]\vec{E} = \frac{\sigma}{2\epsilon_{o}}[/tex]
where
[tex]\epsilon_{o} = 8.85\times 10^{- 12} F/m[/tex]
[tex]\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}[/tex]
[tex]\vec{E} = 4.5\times 10^{4} N/C[/tex]
A flying hawk makes a noise of frequency 50 Hz that spreads evenly in every direction. The intensity at the ground, 60 m below, is measured at 70 dB. What is the power radiated by the hawk?
Answer:
The power radiated by the hawk is 0.452 Watt.
Explanation:
Given that,
Frequency = 50 Hz
Distance r=60 m
Level = 70 dB
We need to calculate the intensity
Using formula of intensity
[tex]dB=10 log(\dfrac{I}{I_{0}})[/tex]
Put the value into the formula
[tex]70=10 log(\dfrac{I}{10^{-12}})[/tex]
[tex]I=10^{7}\times10^{-12}[/tex]
[tex]I = 1\times10^{-5}\ W/m^2[/tex]
We need to calculate the power radiated by the hawk
Using formula of power
[tex]P = I\times 4\pi r^2[/tex]
Put the value into the formula
[tex]P=1\times10^{-5}\times4\pi\times(60)^2[/tex]
[tex]P=0.452\ W[/tex]
Hence, The power radiated by the hawk is 0.452 Watt.
A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the jet is moving with a speed of 400 m/s. What is the jet’s acceleration as it speeds up?
Answer:
8.75 m/s^2
Explanation:
initial speed, u = 300 m/s
final speed, v = 400 m/s
distance, h = 4 km = 4000 m
Let a be the acceleration of the plane.
Use third equation of motion
[tex]v^{2}=u^{2}- 2 gh[/tex]
[tex]400^{2}=300^{2}- 2 \times a \times 4000[/tex]
a = 8.75 m/s^2
Thus, the acceleration of the plane is 8.75 m/s^2.
A rabbit is traveling at a steady 6 m/s and (without knowing it) passes a wolf. The wolf takes off at the instant the rabbit passes it, accelerating at a steady rate of 2.0 m/s^2. How much time elapses before the wolf is moving as fast as the rabbit? How far is the wolf from the rabbit when it reaches this speed?
Answer:
It takes the wolf 3 s to move as fast as the rabbit.
The wolf is 9 m from the rabbit when it reaches a speed of 6 m/s
Explanation:
The equations for the position and velocity of objects moving in a straight line are as follows:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the object at time t
x0 = initial position of the object
t = time
v0 = initial speed
a = acceleration
v = velocity of the object at time t
For the first question, let´s use the equation of velocity of the wolf to find at which time its velocity is the same as the velocity of the rabbit ( 6 m/s):
v = v0 + a · t (v0 = 0 because the wolf starts at rest)
6 m/s = 0 + 2.0 m/s² · t
t = 3 s
Now, with this calculated time, let´s obtain the position of the wolf:
x = x0 + v0 · t + 1/2 · a · t² (Placing the center of the frame of reference at the point when the wolf starts running makes x0 = 0)
x = 1/2 · a · t²
x = 1/2 · 2.0 m/s² · (3 s)²
x = 9 m
Now, let´s calculate the position of the rabbit. Notice that a = 0. Then:
x = x0 + v · t x0 = 0
x = v · t
x = 6 m/s · 3 s = 18 m
The wolf is (18 m - 9 m) 9 m from the rabbit when it reaches a speed of 6 m/s
Metals are usually more dense than wood. Pick the statement below which accurately compares metals to wood. Metals have less mass per unit volume than wood.
Metals have more mass per unit volume than wood.
Metal and wood have the same mass per unit volume.
Answer:
Metals have more mass per unit of volume than wood.
Explanation:
Density is defined as the amount of matter contained in a unit of volume. A material that is denser than another will therefore have more mass per unit of volume.
The density of a body can affect buoyancy, bodies with low enough densities can float in fluids. For example wood can float in water while metals can't.
A car is driven east for a distance of 50 km, then north for 28 km, and then in a direction 28° east of north for 21 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.
Answer:
A ) displacement=75.69km
B) Angle[tex]= 28.92^o[/tex]
Explanation:
This is a trigonometric problem:
in order to answer A and B , we first need to know the total displacement on east and north.
the tricky part is when the car goes to the direction northeast, but we know that:
[tex]sin(\alpha )=\frac{opposite}{hypotenuse} \\where:\\opposite=north side\\hypotenuse=distance[/tex]
North'=9.86km
and we also know:
[tex]cos(\alpha )=\frac{adjacent}{hypotenuse} \\where:\\adjacent=east side\\hypotenuse=distance[/tex]
East'=18.54km
So know we have to total displacement
North=28km+North'=37.86km
East=50km+East'=68.54km
To calculate the total displacement, we have to find the hypotenuse, that is:
[tex]Td=\sqrt{North^2+East^2} =75.69km[/tex]
we can find the angle with:
[tex]\alpha = arctg (\frac {North} {East}) \\\\ \alpha = arctg (\frac {37.86} {68.54})= 28.92^o[/tex]
A grating with 400 lines per mm is illuminated with light of wavelength 600.0 nm. a Determine the angles at which maxima are observed b Determine the largest order that can be seen with this grating and this wavelength
Answer:
(a) angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side
b] largest order = 4
Explanation:
(a) for diffraction maxima,
[tex]sin \theta =m\times \lambda/d[/tex]
Here, m is the order, [tex]\lambda[/tex] is the wavelength, [tex]\theta[/tex] is the angle at which maxima occur, d is inter planar spacing.
And we know that lines per mm (N) is related with d as,
[tex]N=\frac{1}{d}[/tex]
Given that the wavelength is,
[tex]\lambda=600.0 nm=600\times 10^{-9}m[/tex]
And [tex]N=\frac{400 lines}{mm} \\N=\frac{400 lines}{10^{-3}m }[/tex]
Now,
[tex]sin \theta =m\times \lambda\times N[/tex]
Therefore,
[tex]sin \theta= m\times600\times 10^{-9} \times 400\times 10^{3}\\sin \theta=0.24m[/tex]
Here, m can be 1,2,3,4 as sin theta has to be less than 1.
[tex]\theta = arcsin 0.24 , arcsin 0.48 , arcsin 0.72 , arcsin 0.96[/tex]
Therefore, angles of maxima = 13.9°, 28.7° , 46°, 73.7° on either side
b] largest order = 4
Final answer:
The angles at which the maxima are observed can be determined using dsinθ = mλ. For a grating with 400 lines per mm and light of wavelength 600.0 nm, the largest order that can be seen occurs when the angle of the next order is equal to or greater than the angle of the first order maximum.
Explanation:
The angles at which the maxima are observed can be determined using the formula:
dsinθ = mλ
Where d is the grating spacing, θ is the angle, m is the order of the maximum, and λ is the wavelength of the light.
For a grating with 400 lines per mm and light of wavelength 600.0 nm:
d = 1/400 mm = 0.0025 mm = 0.0025 cm
θ = sin-1((mλ)/d)
To determine the largest order that can be seen, we need to find the angle at which the next order would overlap with the first order. This would be when the angle of the next order is equal to or greater than the angle of the first order maximum.
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.3) 10 units 110 units 50 units 30 units
Answer:
the correct answer is option C which is 50 units.
Explanation:
given,
two vector of magnitude = 30 units and of 70 units
to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}
cos θ value varies from -1 to 1
so, resultant vector
=[tex]\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}[/tex]
a = 30 units and b = 70 units
= [tex]\sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}[/tex]
= 40 units to 100 units
hence, the correct answer is option C which is 50 units.
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.520 of the escape speed from Earth and (b) its initial kinetic energy is 0.520 of the kinetic energy required to escape Earth
Answer:
a)[tex]r_1=1.36R[/tex]
b)[tex]r_2=2.083R[/tex]
Explanation:
Given:
a) when the initial velocity of the projectile is 0.520 times the escape velocity from the earth.
Let r be the radial distance from the earth's surface Let M be the mass of the Earth and R be the radius of the Earth
Now using conservation of Energy at earths surface and at distance r we have
[tex]\dfrac{-GMm}{R}+\dfrac{m(0.52V_e)^2}{2}=\dfrac{-GMm}{r_1}\\\dfrac{-GMm}{R}+\dfrac{m\times 0.52^2\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_1}\\r_2=1.36\ R[/tex]
b) when the Initial kinetic Energy of the projectile is 0.52 times the Kinetic Energy required to escape the Earth
Conservation of Energy we have
[tex]\dfrac{-GMm}{R}+0.52\times KE_{escape}=\dfrac{-GMm}{r_2}\\\dfrac{-GMm}{R}+0.52\times\dfrac{m\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_2}\\r_2=2.083\ R[/tex]
Low voltage power system: You want to "supercharge"1 your brand new Tesla Model 3 at a rate of 100 kW at home. For safety reasons, you chose a 5 V direct current (DC) voltage source. a) Calculate the required current. b) Assuming a cylindrical conductor with a current density of 10 A/mm2 , what diameter conductor would you need? c) Assuming copper with a resistivity of rhoCu = 1.67×10−8 Ωm and a cable length
Answer:
(a) 20 kA (B) [tex]diameter=5.0035\times 10^{-2}m[/tex] (c) length = 29.94 m
Explanation:
We have given power P = 100 KW
Voltage V = 5 V
Current density [tex]10A/m^2=10\times 10^{6}A/m^2[/tex]
(A) We have to calculate current
We know that power P = voltage × current
So [tex]100\times 1000=5\times i[/tex]
[tex]i=20kA[/tex]
(b) We know that current density is given by
Current density [tex]=\frac{current}{area}[/tex]
So [tex]10\times 10^{6}A/m^2=\frac{20\times 10^3}{area}[/tex]
[tex]area=2\times 10^{-3}m^2[/tex]
We know that [tex]a=\pi r^2[/tex]
[tex]2\times 10^{-3}=3.14\times r^2[/tex]
[tex]r^2=6.34\times 10^{-4}[/tex]
[tex]r=2.517\times 10^{-2}m[/tex]
[tex]d=2 r=2\times 2.517\times 10^{-2}=5.035\times 10^{-2}m[/tex]
(c) We have to find the length of the cable
We know that resistance [tex]R=\frac{V}{I}=\frac{5}{20\times 10^3}=0.25mohm[/tex]
Resistance is given by [tex]R=\rho \frac{l}{A}[/tex]
So [tex]0.25\times 10^{-3}=1.67\times 10^{-8}\frac{l}{2\times 10^{-3}}[/tex]
l = 29.94 m
Calculate capacitance of two parallel flat circular conducting plates with a space d between them and radius R>>d. How good an approximation is your result?
Answer:
[tex]C=\epsilon_{o}\frac{\pi R^_{2}}{d}[/tex]
This result is a very good approximation, because R>>d
Explanation:
The capacitance for two parallel flat plates, with surface S, space between them d, is:
[tex]C=\epsilon_{o}\frac{S}{d} =\epsilon_{o}\frac{\pi R^_{2}}{d}[/tex]
This calculation is based on the fact that the electric field is constant between the plates. This only happens if the area of the plates is much larger than the distance between them: S>>d, i.e. R>>d. If we assume these factors, the result has a very good approximation.
A skater is standing still on a frictionless rink. Herfriend
throws a Frisbee straight at her. In which of the followingcases is
the largest momentum transferred to the skater?
a) the skater catches the frisbee and hols onto it.
b) the sjater catches the frisbee momentarily, but then
dropsit vertically downward.
c) the skater catches the frisbee, holds it momentarily,
andthrows it back to her friend.
Answer:
option C
Explanation:
the correct answer is option C
Skater is standing on the frictionless surface when her friend through Frisbee at her he starts moving in the backward direction as there is no friction acting to stop him.
but when again the skater through Frisbee back to his friend there is backward force which will increase the velocity of the skater.
so, momentum is directly proportional to velocity so, velocity increases momentum also increases.
Final answer:
The largest momentum transfer to the skater happens when she catches and then throws the Frisbee back to her friend, utilizing the conservation of momentum.
Explanation:
The largest momentum transferred to the skater occurs in the scenario where the skater catches the Frisbee, holds it momentarily, and then throws it back to her friend. In this case, not only is the momentum of the incoming Frisbee transferred to the skater when caught, but additional momentum is transferred when she throws it back due to the conservation of momentum.
A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back past its original point, (b) When it is 15 m above the street, and (c) Just before it hits the street. A horse drags a 100 kg sled a distance of 4 km in 20 minutes. The horse exerts one horsepower, of course. What is the coefficient of sliding friction between the sled and the ground?
Answer:
a) 5 m/s downwards
b) 17.86 m/s
c) 24.82 m/s
d) 0.228
Explanation:
We can set the frame of reference with the origin on the top of the building and the X axis pointing down.
The rock will be subject to the acceleration of gravity. We can use the equation for position under constant acceleration and speed under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
V(t) = V0 + a * t
In this case
X0 = 0
V0 = -5 m/s
a = 9.81 m/s^2
To know the speed it will have when it falls back past the original point we need to know when it will do it. When it does X will be 0.
0 = -5 * t + 1/2 * 9.81 * t^2
0 = t * (-5 + 4.9 * t)
One of the solutions is t = 0, but this is when the rock was thrown.
0 = -5 + 4.69 * t
4.9 * t = 5
t = 5 / 4.9
t = 1.02 s
Replacing this in the speed equation:
V(1.02) = -5 + 9.81 * 1.02 = 5 m/s (this is speed downwards because the X axis points down)
When the rock is at 15 m above the street it is 15 m under the top of the building.
15 = -5 * t + 1/2 * 9.81 * t^2
4.9 * t^s -5 * t - 15 = 0
Solving electronically:
t = 2.33 s
At that time the speed will be:
V(2.33) = -5 + 9.81 * 2.33 = 17.86 m/s
When the rock is about to reach the ground it is at 30 m under the top of the building:
30 = -5 * t + 1/2 * 9.81 * t^2
4.9 * t^s -5 * t - 30 = 0
Solving electronically:
t = 3.04 s
At this time it has a speed of:
V(3.04) = -5 + 9.81 * 3.04 = 24.82 m/s
---------------------
Power is work done per unit of time.
The work in this case is:
L = Ff * d
With Ff being the friction force, this is related to weight
Ff = μ * m * g
μ: is the coefficient of friction
L = μ * m * g * d
P = L/Δt
P = (μ * m * g * d)/Δt
Rearranging:
μ = (P * Δt) / (m * g * d)
1 horsepower is 746 W
20 minutes is 1200 s
μ = (746 * 1200) / (100 * 9.81 * 4000) = 0.228
A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC is at the point x = 0.60 m , y = 0. A. Calculate the magnitude of the net electric field at the origin due to these two point charges.
B. Calculate the direction of the net electric field at the origin due to these two point charges.
Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].
Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.
To solve this problem, we need to calculate the net electric field at the origin (0, 0) due to the two-point charges q₁ and q₂. Here’s a step-by-step breakdown of the process:
Given Information:
Charge q₁ = -4.00 nC located at (0.600 m, 0.800 m)
Charge q₂ = +6.00 nC located at (0.600 m, 0.0 m)
Part A: Calculate the Magnitude of the Electric Field at the Origin
Calculate the Distance from Each Charge to the Origin:
For q₁:
[tex]r_1 = \sqrt{(0 - 0.600)^2 + (0 - 0.800)^2} \\= \sqrt{0.36 + 0.64} \\= \sqrt{1.00} \\= 1.00 \text{ m}[/tex]
For q₂:
[tex]r_2 = \sqrt{(0 - 0.600)^2 + (0 - 0.0)^2} \\= \sqrt{0.36} \\= 0.600 \text{ m}[/tex]
Calculate the Electric Field due to Each Charge:
The formula for the electric field due to a point charge is:
[tex]E = \frac{k |q|}{r^2}[/tex]
Where [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex]
For q₁:
[tex]E_1 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 4.00 \times 10^{-9} \text{ C}}{(1.00)^2} \\= 35.96 \text{ N/C}[/tex]
Direction of E₁ is towards q₁ (since it is negative) and therefore directed along the vector from (0.600, 0.800) to the origin. The components can be calculated as:
[tex]E_{1x} = -E_1 \cdot \frac{0.600}{1.00} = -35.96 \cdot 0.600 \\= -21.576 \text{ N/C}[/tex]
[tex]E_{1y} = -E_1 \cdot \frac{0.800}{1.00} \\= -35.96 \cdot 0.800 \\= -28.768 \text{ N/C}[/tex]
For q₂:
[tex]E_2 = \frac{8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \times 6.00 \times 10^{-9} \text{ C}}{(0.600)^2} \\= 24.99 \text{ N/C}[/tex]
Direction of E₂ is away from the charge because it is positive, directed along the negative y-axis:
[tex]E_{2x} = 0[/tex]
[tex]E_{2y} = -E_2 = -24.99 \text{ N/C}[/tex]
Calculate the Net Electric Field at the Origin:
Combine the x and y components of the electric fields:
[tex]E_{net,x} = E_{1x} + E_{2x} \\= -21.576 + 0 \\= -21.576 \text{ N/C}[/tex]
[tex]E_{net,y} = E_{1y} + E_{2y} \\= -28.768 - 24.99 \\= -53.758 \text{ N/C}[/tex]
The total electric field magnitude is:
[tex]E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2}[/tex]
[tex]E_{net} = \sqrt{(-21.576)^2 + (-53.758)^2} \\= \sqrt{465.33 + 2884.46} \\= \sqrt{3349.79} \\= 57.89 \text{ N/C}[/tex]
Final Answer for Part A: The magnitude of the electric field at the origin is approximately [tex]57.89 \text{ N/C}[/tex].
Part B: Calculate the Direction of the Net Electric Field
Calculate the Angle of the Electric Field:
Final Answer for Part B: The direction of the net electric field at the origin is approximately [tex]249.44°[/tex] relative to the negative x-axis.
Complete Question:
A point charge q1 = −4.00nC is at the point x = 0.600 meters, y = 0.800 meters, and a second point charge q2 = +6.00nC is at the point x = 0.600 meters, y = 0.
Part A
Calculate the magnitude E of the net electric field at the origin due to these two-point charges. (Express your answer in newtons per coulomb to three significant figures.)
Part B
What is the direction, relative to the negative x-axis, of the net electric field at the origin due to these two-point charges. (Express your answer in degrees to three significant figures).
The driver of a car traveling at a speed of 25.5 m/s slams on the brakes and comes to a stop in 3.4 s. If we assume that the car's speed changed at a constant rate (constant net force), find the following. What was the car's average speed during this 3.4 s interval? ______ m/s
How far did the car go in this 3.4 s interval? ____ m
Answer:
(a). The average speed of the car is 12.75 m/s
(b). The distance is 43.35 m.
Explanation:
Given that,
Initial speed = -25.5 m/s
Final speed = 0
Time = 3.4 s
(a). We need to calculate the average speed
Using formula of average speed
[tex]v_{avg}=\dfrac{v_{f}-v_{i}}{2}[/tex]
Put the value into the formula
[tex]v_{avg}=\dfrac{0-(-25.5)}{2}[/tex]
[tex]v_{avg}=12.75\ m/s[/tex]
(b). We need to calculate the acceleration
Using equation of motion
[tex]v_{f}=v_{i}+at[/tex]
[tex]a =\dfrac{v_{f}-v_{i}}{t}[/tex]
[tex]a=\dfrac{-25.5-0}{3.4}[/tex]
[tex]a=-7.5\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s=25.5\times3.4+\dfrac{1}{2}\times(-7.5)\times(3.4)^2[/tex]
[tex]s=43.35\ m[/tex]
Hence, (a). The average speed of the car is 12.75 m/s
(b). The distance is 43.35 m.
Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.
A. Wants, needs
B. Culture, heritage
C. Likes, dislikes
D. strengths, weaknesses
Answer:
D. strengths, weaknesses
Explanation:
Each individual has characteristics that makes them different from one another. Their innate ability or cannot be determined without testing them in various situations.
The various tests which determine a person's aptitude and intelligence. Another aspect to consider is the emotional intelligence of a person i.e., the capability a person has to recognize others' emotions as well as their own emotions.
Testing children in different types of tests which measures their ability in each of them is the only way to know their strength and weaknesses.
An electron with speed v0 = 5.00 x 10^6 m/s is traveling parallel to an electric field of magnitude E = 9900 N/C . 1)How far will the electron travel before it stops?
Express your answer to three significant figures and include the appropriate units.
2)How much time will elapse before it returns to its starting point?
Express your answer to three significant figures and include the appropriate units
Answer:
Explanation:
in the electric field electron will face a force which will create an acceleration
( here - ve ) as follows
Force on electron
= charge on electron x electric field = Q X E
acceleration= Force / mass = Q E / m
mass of electron = 9.1 x 10⁻³¹
acceleration a = [tex]\frac{1.6\times10^{-19}\times 9900}{1.67\times10^{-27}}[/tex]
= 17.4 x 10¹⁴ ms⁻² .
Now initial velocity u = 5 x 10⁶ m/s
Final velocity v = 0
acceleration a = 17.4 x 10¹⁴ ms⁻²
distance of travel = s
v² = u² - 2as
0 = (5 x 10⁶)² - 2 x 17.4 x 10¹⁴ s
s = 7.18 mm
2 ) v = u - at
0 = 5 x 10⁶ - 17.4 x 10¹⁴ t
t = .287 x 10⁻⁸ s
Total time elapsed = 2 x .287 x 10⁻⁸
= .57 x 10⁻⁸ s .