Answer:
265.9Hz
Explanation:
In an open pipe, both ends of the pipes are opened. The fundamental frequency in an open pipe is expressed as fo = V/2L where;
f is the frequency of the wave
V is the velocity of the wave = 343m/s
L is the length of the pipe = 1.29m
Substituting the value to get the fundamental frequency in the open pipe we have;
Fo = 343/2(1.29)
Fo = 343/2.58
Fo = 132.95Hz
Harmonics are integral multiples of the fundamental frequency e.g 2fo, 3fo, 4fo, 5fo...
The first harmonic in the open pipe will be f1 = 2fo
Since f1 =2(132.95)
f1 = 265.9Hz
The frequency of the first harmonic if the pipe is open at each end is 265.9Hz
Answer:
132.95 Hz.
Explanation:
Given:
v = 343 m/s
L = 1.29 m.
Since the pipe is open at both ends,
L = λ/2
λ = v/f = 2L
= 2 × 1.29
= 2.58 m
f = 343/2.58
= 132.95 Hz.
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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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.
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
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
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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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
Two carts are involved in an elastic collision. Cart A with mass 0.550 kg is moving towards Cart B with mass 0.550 kg, which is initially at rest. Cart A stops after the collision and Cart B begins to move.A)If cart A has an initial velocity of 0.8 m/s , what is the velocity of Cart B after the collision?B)What is the initial kinetic energy of Cart A?C)What is the initial kinetic energy of Cart B?D)What is the final kinetic energy of Cart A?E)What is the final kinetic energy of Cart B?F)Is the kinetic energy conserved for elastic collisions?G)Is the momentum conserved for elastic collisions?
Answer:
A) [tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B) [tex]KE_A=0.176\ J[/tex]
C) [tex]KE_B=0\ J[/tex]
D) [tex]ke_a=0\ J[/tex]
E) [tex]ke_B=0.176\ J[/tex]
F) Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G) Yes, momentum is always conserved for an elastic collision.
Explanation:
Given:
mass of car A, [tex]m_a=0.55\ kg[/tex]mass of car B, [tex]m_b=0.55\ kg[/tex]initial velocity of car A, [tex]u_a=0.8\ m.s^{-1}[/tex]final velocity of the car A, [tex]v_a=0\ m.s^{-1}[/tex]A)
As given in the question that the cars undergo an elastic collision:
According to the conservation of momentum:
[tex]m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b[/tex]
[tex]0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b[/tex]
[tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B)
Initial kinetic energy of cart A:
[tex]KE_A=\frac{1}{2} m_a.u_a^2[/tex]
[tex]KE_A=0.5\times 0.55\times 0.8^2[/tex]
[tex]KE_A=0.176\ J[/tex]
C)
Initial kinetic energy of cart A:
[tex]KE_B=\frac{1}{2} \times m_b.u_b^2[/tex]
[tex]KE_B=0.5\times 0.55\times 0^2[/tex]
[tex]KE_B=0\ J[/tex]
D)
The final kinetic energy of cart A:
[tex]ke_A=\frac{1}{2} m_a.v_a^2[/tex]
[tex]ke_a=0.5\times 0.55\times 0^2[/tex]
[tex]ke_a=0\ J[/tex]
E)
The final kinetic energy of cart B:
[tex]ke_B=\frac{1}{2} m_b.v_b^2[/tex]
[tex]ke_B=0.5\times 0.55\times 0.8^2[/tex]
[tex]ke_B=0.176\ J[/tex]
F)
Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G)
Yes, momentum is always conserved for an elastic collision.
(a) The velocity of cart B is [tex]0.8m/s[/tex]
(b) Initial kinetic energy of cart A is [tex]0.176J[/tex]
(c) Initial kinetic energy of cart B is 0
(d) The final kinetic energy of cart A is 0
(e) The final kinetic energy of cart B is [tex]0.176J[/tex]
(f) The kinetic energy is conserved
(g) The momentum is conserved
Elastic collision:(a) The momentum of the system must be conserved, which means momentum before the collision must be equal to momentum after the collision.
Momentum before collision = [tex]m_Au_A+m_bu_B=0.55\times0.8+0.55\times0=0.44kgm/s[/tex]
Momentum after collision:
[tex]m_A+m_A+m_Bv_B=0.55\times0+0.55v_B=0.44\\\\v_B=0.8m/s[/tex]
(b) Initial kinetic energy of cart A:
[tex]K_{iA}=\frac{1}{2}m_Au_A^2=0.5\times0.55\times0.64=0.176J[/tex]
(c) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_Bu_B^2=0.5\times0.55\times0=0J[/tex]
(d) The final kinetic energy of cart A:
[tex]K_{fA}=\frac{1}{2}m_Av_A^2=0.5\times0.55\times0=0J[/tex]
(e) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_B_v_B^2=0.5\times0.55\times0.64=0.176J[/tex]
(f) yes the kinetic energy of the system is conserved for elastic collision
(g) The momentum is conserved in any case including the elastic collision
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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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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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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 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
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.
Radioactivity is a tendency for an element or a material to
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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The speed of a moving sidewalk at an airport is 7 ft/sec. A person can walk 74 ft forward on the moving sidewalk in the same time it takes to walk 15 ft on a non moving sidewalk in the opposite direction. At what rate would a person walk on a non moving sidewalk
Answer:
10.56 ft/s.
Explanation:
given,
Speed of side walk = 7 ft/s
time taken to travel 74 ft on the moving sidewalk = time taken to walk 15 ft on non moving side walk.
We know,
[tex]time = \dfrac{distance}{speed}[/tex]
[tex]\dfrac{74}{w+7}=\dfrac{15}{w-7}[/tex]
[tex]59 w = 7\times 74 + 15\times 7[/tex]
[tex] 59 w = 623[/tex]
[tex]w = 10.56\ ft/s[/tex]
Rate at which person walk on non moving side walk is equal to 10.56 ft/s.
The person's walking speed on a non-moving sidewalk is approximately 1.78 ft/sec, determined by setting up and solving the equation based on equal times on the moving and non-moving sidewalks.
To find the rate at which a person would walk on a non-moving sidewalk, we need to set up an equation based on the given information.
We know the speed of the moving sidewalk is 7 ft/sec. Let's denote the walking speed of the person (on a non-moving sidewalk) as v ft/sec.
From the problem, the time taken by the person to walk 74 ft on the moving sidewalk should be equal to the time taken to walk 15 ft on a non-moving sidewalk in the opposite direction:
Time on the moving sidewalk:[tex]\frac{74}{7 + v} = \frac{15}{v}[/tex]
To solve for v, cross-multiply to get: 74v = 15(v+7)
74v = 15v + 105
74v − 15v = 105
59v = 105
[tex]v = \frac{105}{59}[/tex]
v ≈ 1.78 ft/sec
Therefore, the person's walking speed on a non-moving sidewalk is approximately 1.78 ft/sec.
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]
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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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.
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]
Estimate how long the sun would last if it were merely a huge fire that was releasing chemical energy. Assume that the sun begins with roughly 108 joules per kilogram
Answer:
≅ 17000 years or 1.7 x 10⁴ years
Explanation:
time= total energy/power
= (10⁸J/kg)(2x10³⁰ kg) / 3.8 x 10²⁶ J/s
= 526,315,789,473 s
= 16689 years
≅ 17000 years or 1.7 x 10⁴ years
Final answer:
If the Sun were simply a large fire releasing chemical energy, it would have a lifetime of approximately 16,438 years, considering a mass of 2.0 x 10³⁰ kilograms and an energy release of 10^8 joules per kilogram. This is significantly shorter than the Sun's actual lifespan, confirming that nuclear reactions, not chemical combustion, power the Sun.
Explanation:
If the Sun were simply a large fire releasing chemical energy, we can estimate its lifetime using certain assumptions. Firstly, if the Sun's mass is predominantly hydrogen and its energy output is due to combustion with oxygen, the amount of energy released per kilogram would be about 108 joules. With this in mind, we recall that the Sun's mass is approximately 2.0 × 1030 kilograms. The resource provided illustrates that combustion produces about 45,000,000 calories, or approx. 1.88 × 108 joules, per second for a combusting mass of one kilogram (previously cited as 1.6 × 10-19 joules for a typical energy in a chemical bond). With a simple energy balance calculation, we find that the theoretical Sun-as-a-fire would consume its mass in:
Total energy available = mass × energy per kilogram = 2.0 × 1030 kg × 108 joules/kg = 2.0 × 1038 joules.
If the Sun's luminosity is 3.8 × 1026 watts and 1 watt is equivalent to 1 joule per second, the time in seconds for the Sun to deplete this energy at its current luminosity is:
Lifetime in seconds = Total energy / Luminosity = 2.0 × 1038 joules / 3.8 × 1026 watts = 5.26 × 1011 seconds.
Since there are roughly 32 million seconds in a year, the lifetime of the Sun, if it were burning chemically, can be estimated as:
Lifetime in years = Lifetime in seconds / seconds per year = 5.26 × 1011 seconds / 3.2 × 107 seconds/year = approximately 16,438 years.
This is a very short lifetime compared to the Sun's actual age of approximately 4.6 billion years, indicating the real Sun is sustained by nuclear reactions and not chemical combustion.
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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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.
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.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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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]
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.
Torque, a car manufacturer, aims to make its slogan "Burn Rubber" known to every individual in a way that they immediately associate it with the firm. The company has created a social media page where almost a thousand fans have registered. However, there are two competitors who advertise the same features as those of Torque cars. The firm has a large number of customers who think it manufactures the best cars and there is no other manufacturer better than Torque.
Answer:
A customer is about to buy a limited edition sports car from Torque. It is most likely that the customer will have d. limited problem solving, in which consumers decision rules to purchase are simple, and are not motivated to search for information about other optons, deciding to buy the car.
Explanation:
a. advertising clutter is the big amount of ad-messages that the consumer is exposed to everyday.
b. high involvement product or purchases features many variables to be considered by the customer before getting to a decision.
c. cognitive dissonance or conflicting attitudes causing mental discomfort.
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.