Answer:
At the closest point
Explanation:
We can simply answer this question by applying Kepler's 2nd law of planetary motion.
It states that:
"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"
In this problem, we have a comet orbiting around the Sun:
- Its closest distance from the Sun is 0.6 AU
- Its farthest distance from the Sun is 35 AU
In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: [tex]A\propto vr[/tex], therefore if r is larger, then v (velocity) must be lower).
On the other hand, when the the comet is closer to the Sun the line must move faster ([tex]A\propto vr[/tex], if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.
For the environment, why is the characteristic of regularity important?
Please use your own words.
Answer:
so we have a good place to live at.
Explanation:
Environment is understood to be the set of natural and human factors that surround man in his daily life. Thus, landforms, natural resources, buildings, etc., are part of it.
Taking these definitions into account, a regular environment is one in which all its conditions and components are found with the fewest possible alterations, or with human alterations that do not negatively affect its natural conditions.
In this way, an environment that has not definitively consumed its resources or that has not significantly affected the natural status of the region is considered regular. This characteristic is important because it allows the environment to not be negatively affected, allowing a normal development of human life.
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Can someone help? Thanks.
Franks car rolled 300 cm in 1.5 s. Noahs car rolled 360 cm in 2 s. What is the average speed for each car?
The average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.
Explanation:
Average speed is the measure of total distance covered at different time period. Since, the formula used for calculating the average speed is the ratio of total distance covered by each car to the time taken to cover that distance.
As there is only one set of data i.e., distance and time, the average speed will be equal to the speed of the car.
So in this case, the total distance covered by franks car is 300 cm = 3 m in 1.5 s. Then, the average speed will be
[tex]Avg. S_{frank} = \frac{Total\ distance\ rolled\ by\ his\ car}{Time\ taken}[/tex]
[tex]Avg.S_{frank} =\frac{300 \times 10^{-2} }{1.5}=2\ m/s[/tex]
Similarly, the average speed of Noahs car which rolled a distance of 360 cm = 3.6 m in time 2 s, will be
[tex]Avg.S_{Noah} =\frac{360 \times 10^{-2} }{2}=1.8\ m/s[/tex]
Thus, the average speed of Frank's car is 2 m/s and the average speed of Noah's car is 1.8 m/s.
The average speed for Frank's car and Noah's car is 200 cm/s and 180 cm/s respectively calculated using the formula speed = distance / time.
Explanation:Average speed is a measure of how quickly an object moves over a certain distance in a specific amount of time. It's calculated using the formula:
Average Speed (v) = Total Distance Traveled (d) / Total Time Taken (t)
To calculate the average speed of any moving object, use the formula: speed = distance / time. For Frank's car which rolled 300 cm in 1.5 seconds, the speed would be 300 cm / 1.5 s = 200 cm/s. Similarly, for Noah's car which rolled 360 cm in 2 seconds, the speed would be 360 cm / 2 s = 180 cm/s.
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In general it is expected that ________. A) osmotic pressure will remain relatively consistent throughout the capillary bed B) osmotic pressure will be lower in the arteriole end of the capillary bed compared to the venous end C) osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end D) hydrostatic pressure will remain constant throughout the capillary bed
Answer:
B. OSMOTIC PRESSURE WILL BE LOWER IN THE ARTERIOLE END OF THE CAPPILLARY BED COMPARED TO THE VENOUS END.
Explanation:
This is true for filtration to take place in the cappillary bed. Osmotic pressure is the net pressure that drives movement of fluid from the interstitial fluid back into the capillaries. Osmotic pressure increase favors reabsorption as water moves from region of higher water concentration in the interstitial fluid to the lower region of water concentration in the capillaries.
At the ends of a capillary bed, the difference in the hydrostatic and osmotic pressures provides a net filtration or reabsorption ratio. At the arteriole end of the capillary bed, hydrostatic pressure is greater than the osmotic pressure allowing movements of fluid to the interstitial fluid (filtration) while as the blood moves to the venous end, the osmotic pressure becomes greater than than hydrostatic pressure.
Osmotic pressure is usually higher at the arteriole end of the capillary bed than at the venous end (Option C). This happens because plasma proteins remain in the capillary causing water to move back into the capillary.
Explanation:In general, it is expected that osmotic pressure will be higher in the arteriole end of the capillary bed compared to the venous end (option C). This is because during capillary exchange, fluids and solutes are filtered out at the arteriole end of capillaries due to higher blood pressure, and then reabsorbed at the venous end due to higher osmotic pressure. This helps maintain fluid balance and prevent edema.
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Is it possible to have total internal reflection for light incident from air on water? Explain your answer.
Final answer:
Total internal reflection is possible for light incident from air on water.
Explanation:
Total internal reflection occurs when light traveling from a medium with a higher refractive index to a medium with a lower refractive index is incident at an angle greater than the critical angle.
In the case of light incident from air on water, the critical angle is 48.6°. If the angle of incidence is greater than this critical angle, the light will undergo total internal reflection and not refract out of the water. Therefore, total internal reflection is possible for light incident from air on water.
A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travels in the +x-direction. The force exerted by the expanding gas on the bullet is (16,000 + 10,000x − 26,000x2) N, where x is in meters. A) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.
B) If the barrel is 1.05 m long, how much work is done?
Answer:
Explanation:
Given that,
Length of barrel =0.54m
Mass of bullet=125g=0.125kg
Force extend
F=16,000+10,000x-26,000x²
a. Work done is given as
W= ∫Fdx
W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m
W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m
W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m
W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0
W=8640+1458-1364.69
W=8733.31J
The workdone by the gas on the bullet is 8733.31J
b. Work done is given as
Work done when the length=1.05m
W= ∫Fdx
W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m
W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m
W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm
W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0
W=16800+5512.5-10032.75
W=12,279.75J
The workdone by the gas on the bullet is 12,279.75J
Spud Webb was, at 5 ft 8 in, one of the shortest basketball players to play in the NBA. But he had an amazing vertical leap; he could jump to a height of 1.1 m off the ground, so he could easily dunk a basketball. For such a leap, what was his "hang time" - the time spent in the air after leaving the ground and before touching down again
Answer:
The hang time is 0.95 s
Explanation:
Spud Webb has a height of 5 feet 8 inches and is one of the shortest basketball players to play in the NBA.
He could make an amazing vertical leap of 1.1 m off the ground. For such a leap to calculate the hang time(the time spent in the air after leaving the ground and before touching down again), we use the formula:
[tex]S=ut -\frac{1}{2}gt^{2}[/tex]
Where S is the distance traveled, u is the initial velocity, t is the time taken and g is the acceleration due to gravity.
Given that:
g = 9.8 m/s²
S = [tex]y_{f} -y_{i} =0-1.1=-1.1[/tex]
u = 0
[tex]S=ut -\frac{1}{2}gt^{2}[/tex]
Substituting values:
[tex]-1.1=(0)t -\frac{1}{2}(9.8)t^{2}\\-1.1=0-4.9t^{2} \\-1.1=-4.9t^{2}[/tex]
Dividing through by -4.9 we get:
[tex]\frac{-1.1}{-4.9} =\frac{-4.9}{-4.9}t^{2}[/tex]
[tex]t^{2}=0.2245\\ t=\sqrt{0.2245}=0.474[/tex]
t = 0.474 s
The hang time = 2t = 2 × 0.474 = 0.95 s
The hang time is 0.95 s
A child on a merry-go-round takes 3.9 s to go around once. What is his angular displacement during a 1.0 s time interval?
Answer:
Angular displacement=2π/3.9 rad
Explanation:
Given data
Time t=3.9s
Required
The angular displacement during a 1.0 s time interval
Solution
In 3.9 second the child covers a full circle=2π rad
Angular displacement after 1.0 second is given as:
[tex]=\frac{2\pi }{3.9} rad[/tex]
Angular displacement=2π/3.9 rad
The child's angular displacement on the merry-go-round during a 1.0s time interval would be approximately 1.609 radians.
Explanation:To find the child's angular displacement on the merry-go-round, we first need to know the rate at which the merry-go-round is turning. This is called the angular velocity and is measured in radians per seconds (rad/s). If it takes 3.9 seconds for the merry-go-round to make one full revolution, this equals 2π radians. Therefore, the angular speed of the merry-go-round is 2π/3.9 rad/s.
Now, if we want to know how much the child displaces in 1.0 second, we simply multiply the angular speed by the time interval. So the angular displacement is (2π/3.9 rad/s)*1.0s = 1.609 rad.
Therefore, the child's angular displacement during a 1.0s time interval would be approximately 1.609 radians.
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When you float in fresh water, the buoyant force that acts on you is equal to your weight. When you float higher in the denser water of the Dead Sea, the buoyant force that acts on you is:_______.a) greater than your weight.
b) less than your weight.
c) equal to your weight.
Explanation:
On the increase in the density of the water, the buoyant force acting on the body weight is less than the body weight. In the dead sea, the water is saline due to the presence of salt in the water which increases the density of the water. Thus, the force acting on the body is less than the bodyweight which results in the sinking of the body.Hence, the correct option is (b) that is "less than your weight".A person ate 0.50 pound of cheese (an energy intake of 4000 kJ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes 44.0 kJ to vaporize 1 mole of water.)
Answer:
The answer to the question is;
1637.769 grams of water will need to be perspired in order to maintain his original temperature.
Explanation:
Energy intake of the person = 4000 kJ
Energy required to vaporize 1 mole of water = 44.0 kJ
That is 44.0 kJ/mole
Therefore
The number of moles of water that can be vaporized by 4000 kJ is given by
(4000 kJ)/ (44.0 kJ/mole) = 90.91 moles.
Mass of one mole of water = Molar mass of water = 18.01528 g/mol
Since number of moles of water = ([tex]\frac{Mass .of. water}{Molar. mass. of. water}[/tex])
We therefore have
Mass of water = (Number of moles of water)× (Molar mass of water)
Mass of water = 90.91 moles× 18.01528 g/mol = 1637.769 g
The mass (in grams) of water that he would need to perspire in order to maintain his original temperature is 1637.769 g.
The person would need to perspire approximately [tex]\( 1638.86 \, \text{g} \)[/tex] (or about 1.64 kg) of water to maintain their original temperature after consuming 0.50 pounds of cheese with an energy intake of 4000 kJ.
To determine the mass of water the person needs to perspire to maintain their original temperature, we need to calculate the amount of water required to dissipate the energy intake through perspiration. Given:
Energy intake: [tex]\( 4000 \, \text{kJ} \)[/tex]
Heat of vaporization of water: [tex]\( 44.0 \, \text{kJ/mol} \)[/tex]
First, we need to find out how many moles of water need to be vaporized to dissipate 4000 kJ.
[tex]\[ \text{Moles of water} = \frac{\text{Total energy intake}}{\text{Heat of vaporization per mole}} = \frac{4000 \, \text{kJ}}{44.0 \, \text{kJ/mol}} \][/tex]
[tex]\[ \text{Moles of water} = \frac{4000}{44.0} \approx 90.91 \, \text{moles} \][/tex]
Next, we convert the number of moles to mass. The molar mass of water [tex](\( \text{H}_2\text{O} \))[/tex] is approximately [tex]\( 18.015 \, \text{g/mol} \).[/tex]
[tex]\[ \text{Mass of water} = \text{Moles of water} \times \text{Molar mass of water} = 90.91 \, \text{moles} \times 18.015 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of water} = 1638.86 \, \text{g} \][/tex]
A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of the cable is 10×1010 N/m2. When the cable is fully extended, the end of the cable is 700 m below the support.
Answer:
The cable would stretch 14 cm when loaded with 1000 kg ore.
Explanation:
The question is incomplete.
The complete question would be.
A mine shaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex]. When the cable is fully extended, the end of the cable is 700 m below the support.
How much does the fully extended cable stretch when 1000 kg of ore is loaded into the elevator?
Given the diameter of the cable is [tex]2.5\ cm[/tex]. The length of the cable is [tex]700\ m[/tex].
And the mass of the ore is [tex]1000\ kg[/tex]. Also the Young's modulus of elasticity of the cable is [tex]10\times 10^{10}\ N/m^2[/tex].
We will use Hook's law
[tex]\sigma=E\epsilon[/tex]
Where [tex]\sigma[/tex] is stress. E is Young's modulus of elasticity. And [tex]\epsilon[/tex] is strain.
We can rewrite .
[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]
Where [tex]P[/tex] is the applied force, [tex]A[/tex] is the area of the cross-section. [tex]\delta l[/tex] is the change in length. [tex]l[/tex] is the initial length of the cable.
Also, the applied force [tex]P[/tex] is due to mass of the ore. That would be [tex]P=mg\\P=1000\times 9.81\ N[/tex]
Given diameter of the cable [tex](d)[/tex] [tex]2.5\ cm[/tex].
[tex]d=\frac{2.5}{100}=0.025\ m\\ A=\frac{\pi}{4}d^2\\ \\A=\frac{\pi}{4}(0.025)^2=4.91\times 10^{-4}\ m^2[/tex]
[tex]E=10\times 10^{10}\ N/m^2[/tex]
[tex]l=700\ m[/tex]
Plugging these values
[tex]\frac{P}{A}=E\times \frac{\delta l}{l}[/tex]
[tex]\frac{P}{A}\times \frac{l}{E}=\delta l \\\\ \delta l =\frac{1000\times 9.81\times 700}{4.91\times 10^{-4}\times 10\times 10^{10}} \\\delta l=.139\ m\\\delta l=14\ cm[/tex].
So, the cable would stretch 14 cm when loaded with 1000 kg ore.
________ is a type of polyploidy that arises from the hybridization between two different species.
Explanation:
The polyploids that arise within a species are known as autopolyploids, and those that arise from the hybridization of two different species are known as allopolyploids. Homoploid hybridization is also known as recombinational speciation. Allopolyploidy is a type of polyploidy in which two individuals that belong to two different species produce a hybrid individual. This hybrid individual which is formed undergoes duplication of the genome and forms a new species or individuals.Olivia places her pet frog on a line to observe the frog’s motion. The line is divided into sections that measure 1 centimeter each. The frog begins at 0, moves 18 centimeters forward, moves 6 centimeters backward, and then 12 centimeters backward. What is the frog’s displacement? 0 6 18 36
Answer:
A=0
Explanation:
I got 100% on this assignment
This question involves the basic concept of displacement.
The frog's displacement is "0 cm".
In order to calculate the displacement of the frog, we must consider both the magnitude and direction of the movements made by the frog. Here, we will take the forward direction as positive and the backward direction as negative. Hence, the displacement of the frog will be as follows:
Displacement = 18 cm - 6 cm - 12 cm
Displacement = 0 cm
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The attached picture shows the difference between the displacement and the distance.
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Calculate the energy required. 3) Calculate the average power required.
Answer:
a) τ = 4.47746 * 10^25 N-m
b) E = 2.06301 * 10^13 J
c) P = 3.25511*10^21 W
Explanation:
Given that,
The radius of earth r = 6.3781×10^6 m
The angular speed of earth w = 7.27*10^-5 rad/s
The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s
The mass of earth m = 5.972 × 10^24 kg
The inertia of sphere I = 2/5 * m* r^2
Solution:
angular acceleration of the earth from rest to w is given by α:
α = w / t
α = (7.27*10^-5) / (1.57784760 * 10^8)
α = 4.60754*10^-13 rad/s^2
The required torque τ is given by:
τ = I*α
τ = 2/5 * m* r^2 * α
τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)
τ = 4.47746 * 10^25 N-m
Power required P to turn the earth to the speed w is:
P = τ*w
P = (4.47746 * 10^25)*(7.27*10^-5)
P = 3.25511*10^21 W
Energy E required is :
E = P / t
E = (3.25511*10^21) / (1.57784760 * 10^8)
E = 2.06301 * 10^13 J
______ Was once widely used in the united states as a gasoline additive
Answer:
lead
Explanation:
Lead was once widely used in the United States as a gasoline additive.
Addition of lead is in the form of tetra ethyl lead(II).
It helps to improve the octane rating of gasoline and to produce more useful energy via each combustion step.
The compound containing lead was banned due to the huge health risk it poses to people. Lead poisoning is a known defect that affects people. When the gasoline is combusted, it releases lead as a by-product. Exposure to a high level of lead can cause brain damage and kidney failure.An rv travels 45 km east and stays the night at a KOA. The next day it travels for 3 hours to the north l, traveling 110 km. What is the displacement over the two days for the RV?
Answer:
The displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.
Explanation:
Given:
Distance moved in the East direction (d) = 45 km
Distance moved in the North direction (D) = 110 km
Displacement is defined as the difference of final position and initial position.
Let us draw a diagram representing the above situation.
Point A is the starting point and point C is the final position of RV.
So, the displacement of RV in two days is given as:
Displacement = Final position - Initial position = AC
Now, triangle ABC is a right angled triangle with AB = 45 km, BC = 110 km, and AC being the hypotenuse.
Using Pythagoras theorem, we have:
[tex]AC^2=AB^2+BC^2\\\\AC=\sqrt{AB^2+BC^2}[/tex]
Plug in the given values and solve for AC. This gives,
[tex]AC=\sqrt{(45\ km)^2+(110\ km)^2}\\\\AC=\sqrt{2025+12100}\ km\\\\AC=\sqrt{14125}\ km\\\\AC=118.85\ km[/tex]
Now, the direction of displacement with the east direction is given as:
[tex]\theta=\tan^{-1}(\frac{BC}{AB})\\\\\theta =\tan^{-1}(\frac{110}{45})=67.75^\circ[/tex]
Therefore, the displacement of RV for the two days is 118.85 km at an angle of 67.75° with the east direction.
The displacement of the rv over the two days is 118.85 km.
The given parameters;
initial displacement of the rv = 45 km eastfinal displacement of the rv, = 110 km northThe displacement of the rv over the two days is calculated by applying Pythagoras theorem as follows;
[tex]c^2 = a^2 + b^2\\\\c = \sqrt{a^2 + b^2} \\\\c = \sqrt{(45)^2 + (110)^2} \\\\c = 118.85 \ km[/tex]
Thus, the displacement of the rv over the two days is 118.85 km.
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The initial temperature of a bomb calorimeter is 28.50°C. When a chemist carries out a reaction in this calorimeter, its temperature decreases to 27.45°C. If the calorimeter has a mass of 1.400 kg and a specific heat of 3.52 J/(gi°C), how much heat is absorbed by the reaction? Use .
Answer:
5174.4 J
Explanation:
Parameters given:
Mass of calorimeter, m = 1.4 kg = 1400 g
Specific heat capacity, c = 3.52 J/g°C
Temperature difference, ΔT = 28.5 - 27.45 = 1.05 °C
Heat absorbed by reaction, Q = m * c * ΔT
Q = 1400 * 3.52 * 1.05
Q = 5174.4 J
Answer:
5174.4Joules
Explanation:
Heat capacity is defined as the quantity of heat required to raise the temperature of total mass of a substance by 1Kelvin. Mathematically,
Q = mc∆t where;
Q is the amount of heat absorbed (in Joules)
m is the mass of the substance (bomb calorimeter) in g or kg
c is the specific heat capacity of the bomb calorimeter in J/g°C
∆t is the change in temperature in °C
Given m = 1.4kg
Since 1kg = 1000g
1.4kg = (1.4×1000)g
m = 1.4kg = 1400g
c = 3.52J/g°C
∆t = final temperature - initial temperature
Since heat was absorb (heat gained), final temperature will be 28.5°C
Initial temperature = 27.45°C
Substituting the data given into the heat capacity formula will give us;
Q = 1400×3.52×(28.5-27.45)
Q = 1400×3.52× 1.05
Q = 5174.4Joules
Amount of heat absorbed by the reaction is 5174.4Joules
How much charge has been transferred from the negative to the positive terminal?
Answer:
45 C.
Explanation:
Given:
Time, t = 5.0 h
= 18000 s
Current, I = 2.5 mA
= 0.0025 A
Potential difference, V = 9 V
Q = I × t
= 18000 × 0.0025
= 45 C.
An ideal gas is kept at constant volume of 2.00 L as its temperature is increased, raising the pressure from 15.0 kPa to 30.0 kPa. What work is performed by or on the gas during this process?
30.0 J performed by the gas
4.00 J performed on the gas
0 J
30.0 J performed on the gas
Answer:
0 J
Explanation:
The work done by an ideal gas is given by the equation:
[tex]W=p\Delta V[/tex]
where
W is the work done by the gas
p is the pressure of the gas
[tex]\Delta V[/tex] is the change in volume of the gas
In this problem, we have a gas kept at a constant volume of
V = 2.00 L
This is an isochoric process (constant volume). Since the volume is constant, the change in volume is zero:
[tex]\Delta V=0[/tex]
And therefore, this means that the work done by the gas is zero:
W = 0
Radioactivity is a tendency for an element or a material to
Which motor and body should Devon use to build the car with the greatest acceleration?
Complete Question:
Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.
The bodies have the masses shown in this table (refer attached figure).
Which motor and body should Devon use to build the car with the greatest acceleration?
motor 1, with body 1
motor 1, with body 2
motor 2, with body 1
motor 2, with body 2
Answer:
Devon should build the car with motor 2 and body 1 for having the greatest acceleration.
Explanation:
As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.
It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,
[tex]acceleration =\frac{\text { Force }}{\text { mass }}[/tex]
It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.
The car with the greatest acceleration will be one that has a higher power-to-weight ratio, provided by a light yet powerful motor, and a lighter, aerodynamically efficient body. Both factors--lightweight and power-- are crucial for achieving high acceleration.
Explanation:In deciding what motor and body to use, Devon must consider factors like the power to weight ratio, the torque of the motor, and the aerodynamics of the body. A higher power-to-weight ratio generally translates to greater acceleration. Therefore, Devon should choose a motor that is powerful yet light. When considering the body, Devon should go for a lighter body as heavy bodies slows down a car's acceleration. Besides weight, a body whose design is aerodynamically efficient will enhance acceleration because it reduces air resistance.
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A particular heat engine has a mechanical power output of 5.00 kW and an efficiency of 25.0%. The engine expels 8.00 3 103 J of exhaust energy in each cycle. Find (a) the energy taken in during each cycle and (b) the time interval for each cycle.
Answer:
Explanation:
Given
Power output [tex]P=5\ kW[/tex]
efficiency [tex]\eta =25\ \%[/tex]
Engine expels [tex]Q_r=8\times 10^3\ J[/tex]
Efficiency is given by
[tex]\eta =1-\dfrac{Q_r}{Q_s}[/tex]
where [tex]Q_s[/tex]=Heat supplied
[tex]0.25=1-\dfrac{8\times 10^3}{Q_s}[/tex]
[tex]0.75=\dfrac{8\times 10^3}{Q_s}[/tex]
[tex]Q_s=\dfrac{8\times 10^3}{0.75}[/tex]
[tex]Q_s=10.667\ kJ[/tex]
Work Produced by cycle
[tex]W=Q_s-Q_r[/tex]
[tex]W=10.667-8[/tex]
[tex]W=2.667\ kJ[/tex]
Time interval for which power is supplied
[tex]P\times t=W[/tex]
[tex]t=\dfrac{W}{P}[/tex]
[tex]t=\dfrac{2.667}{5}[/tex]
[tex]t=0.5334\ s[/tex]
The energy taken in each cycle of this particular heat engine is 2.0 x 10^4 J, and the time interval for each cycle is four seconds.
Explanation:The efficiency of a heat engine (e) is derived from work output (W) divided by the energy input (Qin). Given the mechanical power output of the engine and its efficiency, we can use this formula to determine the energy input and the time interval for each cycle.
(a) Energy taken in during each cycle: Since efficiency e = Wout/Qin, then Qin = Wout / e = 5,000W (or 5 x 103 J/s) / 0.25 = 2.0 x 104 J. Where Wout is 5000W converted to Joules per second.
(b) Time interval for each cycle: The energy balance for one cycle is given by Qin = Wout + Qexhaust, where Wout is work output and Qexhaust is exhaust energy. The time for one cycle t = Qin / W = 2.0 x 104 J / 5,000 J/s = 4 seconds. The time interval for each cycle is four seconds.
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Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Incomplete question.The complete question is attached below as screenshot along with figure
Answer:
[tex]F=6.00*10^{-6}N[/tex]
Force is repulsive
Explanation:
Given data
Current I₁=5.00A
Current I₂=2.00A
Length L=1.20 m
Radius r=0.400m
To find
Force F
Solution
As the force is repulsive because currents are in opposite direction
From repulsive force we know that:
[tex]F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}[/tex]
Substitute the given values
[tex]F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N[/tex]
A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has vx =− 35.0 cm/s . Determine The phase constant.
Answer:
The phase constant is 7.25 degree
Explanation:
given data
mass = 265 g
frequency = 3.40 Hz
time t = 0 s
x = 6.20 cm
vx = - 35.0 cm/s
solution
as phase constant is express as
y = A cosФ ..............1
here A is amplitude that is = [tex]\sqrt{(\frac{v_x}{\omega })^2+y^2 }[/tex] = [tex]\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }[/tex] = 6.25 cm
put value in equation 1
6.20 = 6.25 cosФ
cosФ = 0.992
Ф = 7.25 degree
so the phase constant is 7.25 degree
What conclusion can be drawn from the statement that an element has high electron affinity, high electronegativity, and a high ionization energy?
A) The element is most likely from Group 1A or 2A and in period 1 or 2.
B) The element is most likely from Group 2A or 3A and in period 6 or 7.
C) The element is most likely from Group 4A or 5A and in period 4 or 5.
D) The element is most likely from Group 6A or 7A and in period 2 or 3.
Answer:
D) The element is most likely from Group 6A or 7A and in period 2 or 3.
Explanation:
Electronegativity of an atom is the tendency of an atom to attract shared paired of electron to itself. Electronegativity increase across the period from left to right.The ability of an atom to attract electron to itself is electronegativity. Group 7A and 6A elements can easily attract atoms to itself so they are highly electronegative. The most electronegative element in the periodic table is fluorine.Group 6A and 7A is likely to have high electronegativity.
Electron affinity of an atom is the amount of energy release when an atom gains electron . Generally, when atom gains electron they become negatively charged. Group 6A and 7A elements have high electron affinity.
Ionization energy is the energy required to remove one or more electron from a neutral atom to form cations. ionization energy of group 7A and 6A are usually high because the energy required to remove these electron is usually very high . The elements in this groups usually gain electron easily so the energy to remove electron is very high.
We have seen that the heart produces a magnetic field, and that this can be used to diagnose problems with the heart. The magnetic field of the heart is a dipole field that is produced by a loop current in the outer layers of the heart. Suppose the field at the center of the heart is 90 pT (a pT is 10−12T ) and that the heart has a diameter of approximately 12 cm. What current circulates around the heart to produce this field?
Answer:
Current circulating is 8.59×10^-6A
Explanation:
Magnetic field at the center of the loop is given by, B= μI/2R
I = 2RB/μ
given that B is 90x 10^-12 T
radius is 0.12 m /2 = 0.06m
we know that μ is 4π x 10^-7 T.A/m
Substituting the given values we get,
I= (((2(0.06m)(90x 10^-12 T))/(4π x 10^-7 T.A/m))
I is 8.59 x 10^-6 A
The current circulating around the heart is 8.59×10⁻⁶A
Calculating the current:The magnetic field B at the center of the loop of radius R with current I is given by,
B= μ₀I/2R
Rearranging the terms we get:
I = 2RB/μ₀
Given that B is 90×10⁻¹² T
radius is R = 0.12 m /2 = 0.06m
Substituting the given values we get,
I = 2×0.06×90×10⁻¹² / 4π×10⁻⁷
I = 8.59 x 10⁻⁶ A
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A satellite of mass 230 kg is placed in Earth orbit at a height of 500 km above the surface. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit
Answer:
Orbital period of satellite is 5.83 x 10³ s
Explanation:
The orbital period of satellite revolving around Earth is given by the equation :
[tex]T=\sqrt{\frac{4\pi ^{2} (R+h)^{3} }{GM} }[/tex] .....(1)
Here R is radius of Earth, h is height of satellite from the Earth's surface, M is mass of Earth and G is gravitational constant.
In this problem,
Height of satellite, h = 500 km = 500 x 10³ m
Substitute 6378.1 x 10³ m for R, 500 x 10³ m for h, 5.972 x 10²⁴ kg for M and 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻² for G in equation (1).
[tex]T=\sqrt{\frac{4\pi ^{2} [(6378.1+500)\times10^{3} ]^{3} }{6.67\times10^{-11} \times5.972\times10^{24} } }[/tex]
T = 5.83 x 10³ s
A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizontal, when will the ball hit the ground?
Answer:
[tex]t_t=4.131\ s[/tex]
Explanation:
Given:
height above the horizontal form where the ball is hit, [tex]y=1\ m[/tex]
angle of projectile above the horizontal, [tex]\theta=30^{\circ}[/tex]
initial speed of the projectile, [tex]u=40\ m.s^{-1}[/tex]
Firstly we find the vertical component of the initial velocity:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=40\times \sin30^{\circ}[/tex]
[tex]u_y=20\ m.s^{-1}[/tex]
During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.
So final vertical velocity during the course of ascend:
[tex]v_y=0\ m.s^{-1}[/tex]Using eq. of motion:
[tex]v_y^2=u_y^2-2g.h[/tex] (-ve sign means that the direction of velocity is opposite to the direction of acceleration)
[tex]0^2=20^2-2\times 9.8\times h[/tex]
[tex]h=20.4082\ m[/tex] (from the height where it is thrown)
Now we find the time taken to ascend to this height:
[tex]v_y=u_y-g.t[/tex]
[tex]0=20-9.8t[/tex]
[tex]t=2.041\ s[/tex]
Time taken to descent the total height:
we've total height, [tex]h'=h+y[/tex] [tex]=20.4082+1[/tex][tex]h'=u_y'.t'+\frac{1}{2} g.t'^2[/tex]
during the course of descend its initial vertical velocity is zero because it is at the top height, so [tex]u_y'=0\ m.s^{-1}[/tex][tex]21.4082=0+4.9t'^2[/tex]
[tex]t'=2.09\ s[/tex]
Now the total time taken by the ball to hit the ground:
[tex]t_t=t'+t[/tex]
[tex]t_t=2.09+2.041[/tex]
[tex]t_t=4.131\ s[/tex]
Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the magnetic field created by the current in the wire?
1. The magnetic field encircles the wire in a counterclockwise direction
2. The magnetic field encircles the wire in a clockwise direction
Answer:
1. The magnetic field encircles the wire in a counterclockwise direction
Explanation:
When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.
Ever current carrying conductor produces a magnetic field around it.
The potential difference between a pair of oppositely charged parallel plates is 402 V. If the spacing between the plates is doubled without altering the charge on the plates, what is the new potential difference between the plates
Final answer:
Doubling the distance between a pair of oppositely charged parallel plates, without altering the charge on them, does not change the potential difference. The potential difference remains at 402 V because it is directly influenced by the electric field and the charge on the plates, both of which remain constant in this scenario.
Explanation:
The question involves understanding how the potential difference between a pair of oppositely charged parallel plates changes when the distance between them is altered, without changing the charge on the plates. The original potential difference is given as 402 V when the spacing between the plates is a certain value.
If the spacing between the plates is doubled, the new potential difference remains the same, 402 V, because the potential difference between two plates is determined by the equation V = Ed, where 'E' is the electric field strength and 'd' is the distance between the plates. Since the charge on the plates remains constant, and assuming the electric field remains uniform, the electric field (E) does not change.
Therefore, doubling the distance without altering the charge or the electric field does not affect the potential difference. This is a key concept in electrostatics, highlighting how changes in geometry and charge distribution affect electric fields and potential differences.
The heat of fusion for water is 80. cal/g. How many calories of heat are released when 20.0 g of water at 0°C is frozen to ice?
Answer:
1600 cal
Explanation:
The formula
q=mΔ[tex]_{fus}[/tex][tex]H^{0}[/tex]
is used to calculate the heat required to melt a solid where q=amount of heat, m=mass and Δ[tex]_{fus}[/tex][tex]H^{0}[/tex] is the enthalphy of fusion
now we substitute, m=20g, Δ[tex]_{fus}[/tex][tex]H^{0}[/tex]=80cal/g
Therefore, q=20g x 80cal/g =1600 cal
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The heat released when 20.0 g of water at 0°C is frozen to ice is 1600 calories.
Explanation:When water freezes, it releases heat energy. The quantity of heat energy needed to convert a substance from a solid to a liquid or vice versa without affecting its temperature is known as the heat of fusion.
In this case, we are asked to find the number of calories of heat released when moment 20 g of water freezes to ice at zero degrees.
To calculate this, we can use the formula:
Heat released = mass of water * heat of fusion
Substituting the given values:
Heat released = 20.0 g * 80. cal/g = 1600 cal
Therefore, 1600 calories of heat are released when 20.0 g of water freezes to ice at 0 °C.
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