Answer:
Explanation:
Let the velocity of rocket case and payload after the separation be v₁ and v₂ respectively. v₂ will be greater because payload has less mass so it will be fired with greater speed .
v₂ - v₁ = 910
Applying law of conservation of momentum
( 250 + 100 ) x 7700 = 250 v₁ + 100 v₂
2695000 = 250 v₁ + 100 v₂
2695000 = 250 v₁ + 100 ( 910 +v₁ )
v₁ = 7440 m /s
v₂ = 8350 m /s
Total kinetic energy before firing
= 1/2 ( 250 + 100 ) x 7700²
= 1.037575 x 10¹⁰ J
Total kinetic energy after firing
= 1/2 ( 250 x 7440² + 100 x 8350² )
= 1.0405325 x 10¹⁰ J
The kinetic energy has been increased due to addition of energy generated in firing or explosion which separated the parts or due to release of energy from compressed spring.
The diameters of the main rotor and tail rotor of a single-engine helicopter are 7.56 m and 1.00 m, respectively. The respective rotational speeds are 453 rev/min and 4137 rev/min. Calculate the speeds of the tips of both rotors.
Answer:
Explanation:
Given
Diameter of main rotor [tex]d_1=7.56 m[/tex]
Tail rotor [tex]d_2=1 m[/tex]
[tex]N_1=453 rev/min[/tex]
[tex]N_2=4137 rev/min[/tex]
[tex]\omega =\frac{2\pi N}{60}[/tex]
[tex]\omega _1=\frac{2\pi 453}{60}=47.44 rad/s[/tex]
[tex]\omega _2=\frac{2\pi 4137}{60}=433.28 rad/s[/tex]
Speed of the tip of main rotor[tex]=\omega _1\times r_1=47.44\times \frac{7.56}{2}=179.32 m/s[/tex]
Speed of tail rotor[tex]=\omega _2\times r_2=433.28\times \frac{1}{2}=216.64 m/s[/tex]
A gas, behaving ideally, fills a fixed volume container at a pressure P₁ and at a temperature T₁. The temperature of the container is changed to T₂. Using Avogadro’s, Charles’, or Boyle’s law, develop an expression that would solve for the new pressure P₂.a. V₁/T₁=V₂/T₂b. P₁*V₁=P₂*V₂c. P₁/T₁=P₂/T₂d. P₁*T₁=P₂*T₂
Answer:
c. P₁/T₁=P₂/T₂
Explanation:
neither Avogadro’s, Charles’, or Boyle’s law formula can be used, since some parameters like volume is not given,
to find P₂, given P₁, T₁, and T₂ we will therefore use Gay-lussac's law.
gay lussacs law state that, provided volume is kept constant, pressure is directly proportional to temperature.
the volume volume is said to be filled, i.e its is kept constants when temperature is change
In order from lowest urgency to highest, which sequence properly ranks the product categories issued by the National Weather Service? Note that not all product categories are issued for all hazard types.
A. Watch, Advisory, Outlook, Warning
B. Outlook, Watch, Advisory, Warning
C. Outlook, Advisory, Watch, Warning
D. Advisory, Outlook, Watch, Warning
Answer:
In order from lowest urgency to highest, the sequence which properly ranks the product categories issued by the National Weather Service are as follows:
B- Outlook, Watch, Advisory, Warning.
Explanation:
An outlook for a hazardous weather describes the potential hazardous weather of concern in day 1 through 7. There are total two segments of the outlook, one for the marine zones and the second for the land based zones.
A watch is issued when there is the possibility of hazardous weather within 48 hours, it does not guarantee that a hazardous weather is going to come, it just reminds the possibilities of any such weather to come.
Advisory comes third in the urgency ranking, we can explain this through an example: a winter weather advisory may be issued for amount of freezing rain or when there are chance of 2 to 4 inches of snow. And, in winter weather, an warning may be issued when there is one fourth inch or more of ice accumulation.
The National Weather Service orders their product categories based on urgency: Outlook, Watch, Advisory, and finally Warning. An outlook is prospective, a watch means conditions are favorable, an advisory means the event is likely, and a warning means the event is expected within 24 hours.
Explanation:The National Weather Service issues their product categories in a sequence based on the impending urgency of the weather condition. The correct order from lowest to highest urgency is: Outlook, Watch, Advisory, and then Warning.
An Outlook is used when hazardous weather is possible over the next week. A Watch is used when conditions are favorable for a hazard to occur; it typically means there is a possibility of the occurrence within the next 48 hours. An Advisory is used when a hazardous event is occurring, imminent or likely, but it is considered less severe than a warning.
A Warning is issued when a hazardous weather event is occurring, imminent or has a very high probability of occurring; it typically means the event is expected within the next 24 hours. Therefore, the correct answer is B. Outlook, Watch, Advisory, Warning.
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A molecule with a seesaw molecular geometry has a bond angle of
Answer
A molecule with seesaw molecular geometry has a bond angle of 90° and 120°.
Seesaw is a type of molecular geometry where there are four bonds attached to a central atom.
Bond is observed in the shape of playground seesaw that is why it is known as a seesaw bond.
Four bonds attached to the central bond formed results as tetrahedral or square planar geometry.
In see-saw, shape maximizes the bond angle of the lone pair and the other atoms in the molecule.
lone pair in the equatorial position offer 120 and 90-degree bond angles whereas bond angle will be 90-degree bond angles if placed at axial position.
A molecule with seesaw molecular geometry has bond angles that are less than 120 degrees for equatorial positions and less than 90 degrees for axial positions, due to lone pair repulsion. Hence the correct option is 1.
Bond Angles in Seesaw Molecular Geometry
A molecule with a seesaw molecular geometry has four nuclei and one lone pair of electrons. This molecular structure is based on a trigonal bipyramidal geometry but with one equatorial position occupied by a lone pair. As a result, the bond angles are adjusted due to electron pair repulsion.
The bond angles for a seesaw geometry are:
Less than 120 degrees for equatorial bonds, as the lone pair causes slight compression of these angles.Less than 90 degrees for axial bonds due to the same repulsive effects from the lone pair.Approximately 173 degrees between the two axial positions, slightly less than the ideal 180 degrees due to lone pair repulsion.The correct question is:
A molecule with a seesaw molecular geometry has a bond angle of
1. <120 for equatorial bonds and <90 for axial bonds.
2.180
3. <90
4. 120 for equatorial bonds and 90 for axial bonds.
5.120
A mover pushes a 30.0 kg crate across a wooden floor at a constant speed of 0.75 m/s. If the coefficient of static friction for wood-on-wood is 0.20, what is the normal force exerted by the floor on the crate?
Answer:
294.3 N
Explanation:
In this situation, we are told that the crate is not accelerating in the horizontal plane. But also it is not accelerating in the vertical plane. Meaning that the sum of all vertical forces add up to zero.
Fnet = ma
Weight + Normal force = mass * acceleration
-(30 kg * 9.81 m/s²) + Normal force = 30.0 kg * 0 m/s²
Normal force = 294.3 N
The normal force exerted by the floor on the crate is equal to the weight of the crate when it's at rest. It can be calculated by multiplying mass and acceleration due to gravity. For a crate with a mass of 30.0 kg, the normal force would be 294 N.
Explanation:The key to answering this question involves understanding how friction and forces work. In this scenario, the crate is not accelerating, so the forces acting on it are balanced, meaning the upward normal force equals the downward gravitational force. The normal force exerted by the floor on the crate is determined by the weight of the crate, which can be calculated using the formula for weight, W = mg (mass times gravity). Here, the mass m is 30.0 kg and the acceleration due to gravity g is 9.8 m/s².
So, we can substitute these values into the formula to find the weight: W = (30.0 kg) * (9.8 m/s²) = 294 N. This weight also equals the normal force since the crate is not moving vertically, so the normal force exerted by the floor on the crate is 294 N.
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A 7.29 kg bowling ball moves at 1.73 m/s. How fast must a 2.28 g Ping-Pong ball move so that the two balls have the same kinetic energy? Answer in units of m/s.
The speed of the ping-pong ball must be 97.8 m/s
Explanation:
The kinetic energy of an object is the energy possessed by an object due to its motion. It is calculated as
[tex]E_k = \frac{1}{2}mv^2[/tex]
where
m is the mass of the object
v is its speed
Here we have a bowling ball that has
Mass: m = 7.29 kg
Speed: v = 1.73 m/s
So its kinetic energy is
[tex]E_k = \frac{1}{2}(7.29)(1.73)^2=10.9 J[/tex]
Now we want the ping-pong ball to have the same kinetic energy, so
[tex]E_k = 10.9 J[/tex]
its mass is
[tex]m=2.28 g = 2.28\cdot 10^{-3} kg[/tex]
So, we can find the speed it should have by re-arranging the equation:
[tex]v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2(10.9)}{2.28\cdot 10^{-3}}}=97.8 m/s[/tex]
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2. Which of the following is an example of a passive immunization? A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog B. Heat-killed flu antigen grown in chicken eggs C. Live viral antigen given on a sugar cube to protect against polio D. Both A and B. E. Both B and C.
Answer:
A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog
Explanation:
passive immunization is the immunity or protection against certain infectious diseases by injecting antigens or antibodies to the blood stream or certain antibodies can be gotten through colostrum or breast milk.
for option A. Antibodies against rabies given to someone who was bitten by a potentially rabid dog. the immunity against rabies infection is administer to the person
For option B:is only telling us the type of infection but not stating the immunity administered
option c can be fairly correct, because its another method of immunization but its not passive immunization
Suppose a force of 1.00 x 104 n is put on a 10.0 cm by 10.0 cm square. What is the pressure?
Answer:
[tex]P=1000000Pa=1MPa=1000kPa[/tex]
Explanation:
When the force that is applied is normally and uniformly distributed on a surface, the magnitude of the pressure is obtained by dividing the force applied on the corresponding area:
[tex]P=\frac{F}{A}[/tex]
Where:
[tex]P=Pressure\hspace{3}in\hspace{3}Pa.\\F=Force\hspace{3}in\hspace{3}N\\A=Area\hspace{3}in\hspace{3}m^2[/tex]
Converting cm to m:
[tex]10cm*\frac{1m}{100cm} = 0.1m[/tex]
The area of a square is:
[tex]A=l^2=(0.1)^2=0.01m^2[/tex]
Therefore:
[tex]P=\frac{1.00\times10^4}{0.01} =1000000Pa=1MPa=1000kPa[/tex]
Which type of intermolecular attractive force operates between all molecules?
Answer:
Hydrogen bonding
Ionic bonding
Non- induced dipole forces
Ion dipole forces
Van-der wall forces
Explanation:
Intermolecular forces are the forces which operates at molecular level they may be attractive or repulsive in nature. But the given question ask only about attractive molecular forces. These are of basically five types:-
1. Hydrogen bonding- It is the force of attraction between hydrogen atom and an electro-negative atom.
2 .Ionic bonding- It is the force of attraction between cationic and anionic pair.
3.Non - Induced dipole forces- It is the dipole- dipole interaction between permanent electromagnetic materials.
4.Ion-dipole forces- These bonds are stronger than hydrogen bond and also stronger than dipole-dipole bond.
Van-der walls forces- It is the force of attraction between uncharged molecules.
Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 5.0 m/s. Once free of this area, it speeds up to 12 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final speed?
Answer:
26m/s
Explanation:
Assuming that the acceleration is constant, we can start by calculating the train speed when it's free of the congested area:
[tex]a = \frac{\deltav}{\deltat} = \frac{12 - 5}{8} = \frac{7}{8} = 0.875 m/s^2[/tex]
Then with the same acceleration we can find out the final speed:
[tex]v = v_0 + at = 12 + 0.875*16 = 26m/s[/tex]
This star map shows stars as we see them in our sky from Earth, centered around the constellation Canis Major. Larger dots represent brighter stars, and a few of the brightest stars are identified. From this view alone, what can you conclude about Sirius?
Answer:
It has the greatest apparent brightness of any star in this region of the sky.
Explanation:
We can conclude that Sirius is brighter than the other stars by the question saying "Larger dots represent brighter stars, and a few of the brightest stars are identified."
It can be viewed that it has the greatest apparent brightness of any star in this region of the sky.
What is Sirius?The brightest star in the sky at night is Sirius. It gets its name from the Greek word for 'glowing' or 'scorching.'
Sirius is known as the Dog Star because it is the brightest star in the Canis Major constellation. It's extremely bright because it's one of the stars closest to our sun. The name could be derived from ancient Egypt.
This star map depicts the stars as seen from Earth, centered on the constellation Canis Major. The brighter stars are represented by larger dots, and a few of the brightest stars are identified.
It has the highest apparent brightness of any star in this region of the sky, as can be seen.
Thus, this can be concluded regarding Sirius.
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Your question seems incomplete, the missing image is attached below:
A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 7.70 m2. The total power it absorbs from the star is 3800 W.
Assuming the surface is an ideal absorber and radiator, calculate the equilibrium temperature of the asteroid (in K)
Answer:
Temperature will be 305 K
Explanation:
We have given The asteroid has a surface area [tex]A=7.70m^2[/tex]
Power absorbed P = 3800 watt
Boltzmann constant [tex]\sigma =5.67\times 10^{-8}Wm/K^4[/tex]
According to Boltzmann rule power radiated is given by
[tex]P=\sigma AT^4[/tex]
[tex]3800=5.67\times 10^{-8}\times 7.70\times T^4[/tex]
[tex]T^4=87.0381\times 10^8[/tex]
[tex]T=305K[/tex]
So temperature will be 305 K
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.5 h. How long does it take for the water to fall a distance 0.250d from its highest level?Number__________Unit____________.
Answer:
2.08335 hours
Explanation:
T = Time period = 12.5 h
Angular frequency is given by
[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{12.5}\\\Rightarrow \omega=0.50265\ rad/h[/tex]
The distance moved from highest to lowest level is given by
[tex]d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d[/tex]
At the ocean surface [tex]x_m=0.25d[/tex]
[tex]x_0=x_mcos(\omega t+\phi)[/tex]
[tex]\phi[/tex] = Phase constant = 0 as clock is started at [tex]x_0=x_m[/tex]
[tex]0.25d=0.5dcos(0.50265t)\\\Rightarrow cos(0.50265t)=\frac{0.25}{0.5}\\\Rightarrow 0.50265t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.50265}\\\Rightarrow t=2.08335\ h[/tex]
The time taken for the water to fall the distance is 2.08335 hours
A reason that one typically does not notice a blind spot in the visual field is that
Explanation:
Both of our eyes have a blind spot, in the retina of eye where there are no rods (light vision) or cons (color vision) cells. Its about a size of pinhead. The blind spot is place where the optic nerves exit the eye and connect to our brain. The problem is that we cannot notice this blind as the brain fills in information for us.
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sleds move on an icy surface. The dogs exert a 240-N force on the rope attached to the front sled. The front rope pulls horizontally. Find the acceleration of the sleds.
Answer:
Acceleration, [tex]a=1.2\ m/s^2[/tex]
Explanation:
Given that,
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg
Force exerted by the doges on the rope attached to the front sled, F = 240 N
To find,
The acceleration of the sleds.
Solution,
Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :
F = ma
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{240\ N}{2\times 100\ kg}[/tex] (m = 2m)
[tex]a=1.2\ m/s^2[/tex]
So, the acceleration of the sleds is [tex]1.2\ m/s^2[/tex].
An aircraft is loaded 110 pounds over maximum certificated gross weight. If fuel (gasoline) is drained to bring the aircraft weight within limits, how much fuel should be drained?
Answer:
242 lts de fuel
Explanation:
We know density of fuel (is variable) but we can say that is close to 1 Kg/lts.
If we have 110 pounds over maximun certificated gross weight, we need to get rid of that 110 pounds or 110 * 2.2 = 242 Kgs
Now by rule of three
If 1 lt weight 1 kg.
x 242 kgs
x = 242 lts de fuel
To bring the aircraft within the maximum gross weight limits, approximately 18.33 gallons of fuel need to be drained, calculated by dividing the excess weight (110 pounds) by the weight of fuel per gallon (6 pounds/gallon).
To calculate how much fuel needs to be drained to bring an aircraft weight within the maximum certificated gross weight limits, we need to know two pieces of information: the weight by which the aircraft is over its maximum gross weight and the weight of the fuel that needs to be drained.
In this scenario, the aircraft is 110 pounds over the maximum certificated gross weight. In general, aviation gasoline (avgas) weighs about 6 pounds per gallon. To solve this problem, we will divide the excess weight by the weight of the fuel per gallon to find out how many gallons need to be drained.
So, the calculation is as follows:
Excess weight: 110 pounds.
Weight of fuel per gallon: 6 pounds/gallon.
Number of gallons to be drained = Excess weight / Weight of fuel per gallon.
Number of gallons to be drained = 110 pounds / 6 pounds/gallon.
Number of gallons to be drained = 18.33 gallons (approximately).
Therefore, to bring the aircraft back within weight limits, approximately 18.33 gallons of fuel need to be drained.
A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the well? Assume that the speed of the bucket remains constant as it is lifted. Answer in units of m.
The depth of the well can be calculated using the formula for gravitational potential energy, taking the work done to lift the bucket as equal to the gravitational potential energy gained by the bucket. This gives us the equation: Depth of the well (h) = Work done (W) / (mass (m) * acceleration due to gravity (g)), which we solve to find the depth.
Explanation:The physics concept involved in this question is called gravitational potential energy, which refers to the energy an object possesses because of its higher position in a gravity field.
In such cases, the energy can be calculated using the formula for work, which is force times distance. When you're lifting an object, the force you're countering is the force of gravity on the object, which is its mass multiplied by the acceleration due to gravity (9.8 m/s^2). This amount of work is equivalent to the gravitational potential energy gained by the bucket when it is lifted.
So, the formula we'll use is: Work = mgh (mass * gravity * height)
We know the work done (5.92 kJ or 5920 J), the mass (25.9 kg) and the acceleration due to gravity (9.8 m/s^2). We want to find height (h), which in this context, means the depth of the well.
Rearranging the formula and solving for h gives us: h = Work / (m*g). By substituting the known values, h = 5920 J / (25.9 kg * 9.8 m/s^2).
By solving this, we get the depth of the well.
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When output is below potential and the policy rate has hit the floor of zero, the resulting fall in inflation leads to ________ real interest rates, which ________ output further, which causes inflation to fall further.
Answer:
high real rate of interest, depressing
Explanation:
If output is less than potential and the rate of policy has fall to the ground of zero, the subsequent decline in inflation results in high real rate of interest which further depressing output, causing inflation to even further decline.
If the actual output is less than potential output, demand or supply has fallen, resulting in job and output declines
A 100 kg cart goes around the inside of a vertical loop of a roller coaster. The radius of the loop is 3 m and the cart moves at a speed of 6 m/s at the top. The force exerted by the track on the cart at the top of the loop is half the cart's mass, how fast is cart moving in m/s?
Answer:
The speed of cart is 5.5 m/s.
Explanation:
Given that,
Mass of cart = 100 kg
Distance = 3 m
Speed = 6 m/s
We need to calculate the speed of cart
Using relation of centripetal force and normal force
[tex]N+mg=\dfrac{mv^2}{r}[/tex]
[tex]\dfrac{m}{2}+mg=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}+9.8=\dfrac{v^2}{3}[/tex]
[tex]v=\sqrt{3(\dfrac{1}{2}+9.8)}[/tex]
[tex]v=5.5\ m/s[/tex]
Hence, The speed of cart is 5.5 m/s.
Final answer:
The speed of the cart at the top of the loop is 4.24 m/s.
Explanation:
In order to find the speed of the cart at the top of the loop, we can use the principle of conservation of mechanical energy. The total mechanical energy of the cart is the sum of its gravitational potential energy and its kinetic energy. At the top of the loop, the gravitational potential energy is zero, so the total mechanical energy is equal to the kinetic energy. The kinetic energy of the cart is given by the formula KE = 1/2 * m * v^2, where m is the mass of the cart and v is its velocity.
Given that the mass of the cart is 100 kg and the radius of the loop is 3 m, we can first find the centripetal acceleration at the top of the loop using the formula a = v^2 / r. Plugging in the values, we get a = (6 m/s)^2 / 3 m = 12 m/s^2. Since the force exerted by the track on the cart at the top of the loop is half the cart's mass, we can use this force to find the net force acting on the cart. The net force is equal to the centripetal force, which is given by F = m * a. Plugging in the values, we get F = (1/2 * m) * a = (1/2 * 100 kg) * 12 m/s^2 = 600 N.
Since the net force equals the centripetal force, we can equate it to the formula F = m * v^2 / r. Plugging in the values, we get 600 N = 100 kg * v^2 / 3 m. Solving for v, we get v^2 = 600 N * 3 m / 100 kg = 18 m^2/s^2. Taking the square root of both sides, we find that v = sqrt(18 m^2/s^2) = 4.24 m/s.
A 0.90-kg ball is thrown with a speed of 9.0 m/s at an upward angle of 27 ∘.A) what is its speed at its highest point, andb)how high does it go?
Answer:a)8.01 m/s
Explanation:
Given
mass of ball [tex]m=0.9 kg[/tex]
initial speed [tex]u=9 m/s[/tex]
launch angle [tex]\theta =27 [/tex]
As the projectile reaches its highest point its vertical velocity component becomes zero and there will only be horizontal component
velocity at highest Point [tex]v=u\cos \theta [/tex]
[tex]v=9\cos 27[/tex]
[tex]v=8.01 m/s[/tex]
(b)maximum height h
Maximum height is given by
[tex]H_{max}=\frac{u^2\sin^2 \theta }{2g}[/tex]
[tex]H_{max}=\frac{9^2\sin^2 (27)}{2\times 9.8}[/tex]
[tex]H_{max}=0.85 m[/tex]
Final answer:
The speed of the ball at its highest point is its horizontal velocity component, calculated using the cosine of the angle. The height reached by the ball is found using the kinematic equation for vertical motion, considering the initial vertical velocity and the acceleration due to gravity.
Explanation:
When addressing the question of a 0.90-kg ball thrown at an upward angle, we are dealing with a two-dimensional projectile motion problem in physics. This type of problem can be broken down into horizontal and vertical components. Since gravity acts only in the vertical direction, it does not affect the horizontal velocity of the projectile.
(A) Speed at the highest point: The speed of the ball at its highest point is solely the horizontal component of its initial velocity because the vertical component will be zero at that point. The horizontal velocity (
vx) can be calculated using
vx =
v cos(
θ). So,
vx = 9.0 m/s cos(27°).
(B) Maximum height: To find the maximum height, we need to consider the vertical component of the velocity (
vy) and the acceleration due to gravity (
g), which is -9.81 m/s² (negative sign indicates the direction is opposite to the velocity). The initial vertical velocity (
vy) is
v sin(
θ), thus
vy = 9.0 m/s sin(27°). We can then use the kinematic equation
v2 = u2 + 2as to find the maximum height (
h), where u is the initial velocity, s is the displacement, and a is the acceleration. At the highest point, the final vertical velocity (
v) is 0 m/s. Rearranging the equation to solve for
s, we get
s = v2 / (2g).
What is the average power consumption in watts of an appliance that uses 5.00 kWh of energy per day? How many joules of energy does this appliance consume in a year?
Answer:
(A) power = 0.208 kW = 208 watts
(B) energy = 6.6 x 10^{9} joules
Explanation:
energy consumed per day = 5 kWh
(a) find the power consumed in a day
1 day = 24 hours
power = \frac{energy}{time}
power = \frac{5}{24}
power = 0.208 kW = 208 watts
(b) find the energy consumed in a year
assuming it is not a leap year and number of days = 365 days
1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds
energy = power x time
energy = 208 x 31,536,000
energy = 6.6 x 10^{9} joules
The average power consumption of an appliance that uses 5.00 kWh of energy per day is approximately 208.33 watts. In a year, this appliance consumes about 6.57 × 10^12 joules of energy.
Explanation:The average power consumption of an appliance that uses 5.00 kWh of energy per day can be calculated as follows:
To convert kilowatt-hours to watts, we need to know that 1 kilowatt-hour equals to 1,000 watt-hours, and there are 24 hours in a day. Hence, 5.00 kWh is equal to 5,000 watt-hours. Therefore, the power consumption in watts is 5,000 watt-hours divided by 24 hours, yielding an average power consumption of approximately 208.33 watts.
To calculate the energy consumed in a year, we know that 1 kWh equals 3.6 million joules. Therefore, 5.00 kWh equals 18 million joules. Since there are 365 days in a year, the appliance uses 18 million joules per day times 365 days which equals 6.57 × 10^12 joules per year.
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The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
The distance from the radio station is 0.28 light years away.
Solution:
As per the question:
Distance, d = 4 ly
Frequency of the radio station, f = 854 kHz = [tex]854\times 10^{3}\ Hz[/tex]
Power, P = 50 kW = [tex]50\times 10^{3}\ W[/tex]
[tex]I_{p} = 1\ photon/s/m^{2}[/tex]
Now,
From the relation:
P = nhf
where
n = no. of photons/second
h = Planck's constant
f = frequency
Now,
[tex]n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s[/tex]
Area of the sphere, A = [tex]4\pi r^{2}[/tex]
Now,
Suppose the distance from the radio station be 'r' from where the intensity of the photon is [tex]1\ photon/s/m^{2}[/tex]
[tex]I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}[/tex]
[tex]1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}[/tex]
[tex]r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m[/tex]
Now,
We know that:
1 ly = [tex]9.4607\times 10^{15}\ m[/tex]
Thus
[tex]r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly[/tex]
What is happening inside a star while it expands into a subgiant?
Answer:
What happens inside a star while it expands into a sunlight is that It is fusing hydrogen into helium in a shell outside the core after which the fused hydrogen in the shell outside the core generates enough thermal pressure to push the upper layers outward.
Calculate the angular velocity of the earth in its orbit around the sun.
Answer:
[tex]0.0172rad/day=1.99x10^{-7}rad/second[/tex]
Explanation:
The definition of angular velocity is as follows:
[tex]\omega=2\pi f[/tex]
where [tex]\omega[/tex] is the angular velocity, and [tex]f[/tex] is the frequency.
Frequency can also be represented as:
[tex]f=\frac{1}{T}[/tex]
where [tex]T[/tex] is the period, (the time it takes to conclude a cycle)
with this, the angular velocity is:
[tex]\omega=\frac{2\pi}{T}[/tex]
The period T of rotation around the sun 365 days, thus, the angular velocity:
[tex]\omega=\frac{2\pi}{365days}=0.0172rad/day[/tex]
if we want the angular velocity in rad/second, we need to convert the 365 days to seconds:
Firt conveting to hous
[tex]365days(\frac{24hours}{1day} )=8760hours[/tex]
then to minutes
[tex]8760hours(\frac{60minutes}{1hour} )=525,600minutes[/tex]
and finally to seconds
[tex]525,600minutes(\frac{60seconds}{1minute})=31,536x10^3seconds[/tex]
thus, angular velocity in rad/second is:
[tex]\omega=\frac{2\pi}{31,536x10^3seconds}=1.99x10^{-7}rad/second[/tex]
Increasing the mass attached to a spring will increase the angular frequency of its vibrations. True or False
Answer: False
Explanation:
Using the angular frequency formula derived from Hooke's law we have :
[tex]\omega =\sqrt{\frac{ k }{ m }} [/tex]
[tex]{\omega}^{ 2 }=\frac {k}{m}[/tex]
where k- is Hooke's constant
m- is mass attached to a spring
According to the formula the mass is inversely proportional to the square of the angular frequency angular which means that increasing the mass attached to the spring will decrease the angular frequency.
The statement is false. Increasing the mass attached to a spring will decrease, not increase, the angular frequency of its vibrations due to the inverse proportional relationship between mass and angular frequency in a mass-spring system.
Explanation:The statement, 'Increasing the mass attached to a spring will increase the angular frequency of its vibrations', is false. According to the principles of simple harmonic motion, the angular frequency of a mass-spring system fundamentally depends on the mass and the force constant (spring constant) of the system. However, the relationship is inversely proportional; meaning that increasing the mass will actually decrease the angular frequency. This is because the angular frequency (ω) is calculated as the square root of the ratio of the force constant (k) to the mass (m: ω = sqrt(k/m). Hence, increasing the mass causes a decrease in the angular frequency, which consequently increases the period of the motion - the time it takes for one complete cycle of vibration to pass a given point.
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A thin rod of length 0.75 m and mass 0.42 kg is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with an angular speed of 4.0 rad/s. neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position, (b) and how far above that position the center of mass rises.
Final answer:
The rod's kinetic energy at its lowest position is 1.26 J, calculated using the formula for rotational kinetic energy. The center of mass of the rod rises to a height of approximately 30.6 cm above the lowest point, using the principle of conservation of mechanical energy to equate potential and kinetic energies.
Explanation:
A thin rod acting as a pendulum is an example of physical principles related to rotational motion and conservation of energy. To find the rod's kinetic energy at its lowest position, we use the formula for rotational kinetic energy, K = (1/2)Iω2, where 'I' is the moment of inertia and 'ω' is the angular speed. For a rod rotating about one end, the moment of inertia, 'I', is given by (1/3)ml2, with 'm' being the mass of the rod and 'l' its length.
We can find the kinetic energy of the rod at its lowest position, where the angular speed is 4.0 rad/s: K = (1/2)Iω2 = (1/2)(1/3)(0.42 kg)(0.75 m)2(4.0 rad/s)2. After calculation, this gives K = 1.26 J (Joules).
To determine how far above the lowest position the center of mass rises, we have to consider the conservation of mechanical energy. At the highest point, all kinetic energy is converted into potential energy (assuming no energy losses), so we have mgh = K, where 'g' is the acceleration due to gravity, 'h' is the height gained, and 'm' is the mass of the rod. Therefore, h = K / (mg). Plugging in the values, we get h = 1.26 J / (0.42kg * 9.81 m/s2) = 0.306 m, or approximately 30.6 cm above the lowest position.
In the chapter on "Assembling the Rooms" in The Place of Houses the houses of Andrea Palladio are discussed in terms of the way their rooms are arranged. (The Villa Rotunda which we discussed in an earlier lecture fits their descripLon perfectly.) Which of the following best describes the observaLons made there?
a.Palladio's villa plans are frequently organized around a Great Room onto which other rooms open. This room provides the basis for arranging everything else.
b.Palladio's villas generally consist of rooms loosely bunched informally along
c.Palladio's villas are always long thin buildings which enfront the street on their long side. All rooms look out on the street.
Answer:
a. Palladio's villa plans are frequently organized around a Great Room onto which other rooms open. This room provides the basis for arranging everything else.
Two 1.0 kg blocks are connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s2 by force F u .
a. What is F?
b. What is the tension at the top end of rope 1?
c. What is the tension at the bottom end of rope 1?
d. What is the tension at the top end of rope 2?
Answer:
a) [tex] F=32.0 N [/tex]
b) [tex] T_{1}=19.2 N [/tex]
c) [tex] T_{2}= 16.0 N [/tex]
d) [tex] T_{3}=3.2 N [/tex]
Explanation:
Let's do the free body diagram per each body. In this particular case the rope has a mass so we have:
First block
[tex] F-T_{1}-m_{1}g=m_{1}a [/tex] (1)
Frist rope
[tex] T_{1}-T_{2}-m_{r}g=m_{r}a [/tex] (2)
Second block
[tex] T_{2}-T_{3}-m_{2}g=m_{2}a [/tex] (3)
Second rope
[tex] T_{3}-m_{r}g=m_{r}a [/tex] (4)
When,
T
₁ is the tension at the top of the rope 1
T
₂ is the tension at the bottom of the rope 1
T
₃ is the tension at the top of the rope 2
Now, if we add all equations from (1) to (4), we will get the value of F,
[tex] F-(m_{1}+m_{2}+2m_{r})g=(m_{1}+m_{2}+2m_{r})a[/tex]
a) Solving this equation for F:
[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]
[tex] F=(m_{1}+m_{2}+2m_{r})(a+g)[/tex]
[tex] F=32.0 N[/tex]
b) Using the equation (1), we can find T₁
[tex] T_{1}=F-m_{1}(g+a)[/tex]
[tex] T_{1}=19.2 N[/tex]
c) Let's use the equation (2) to find T₂
[tex] T_{2}=T_{1}-m_{r}(g+a) [/tex]
[tex] T_{2}= 16.0 N [/tex]
d) Using the equation (4) we can find T₃:
[tex] T_{3}=m_{r}(a+g) [/tex]
[tex] T_{3}=3.2 N [/tex]
Have a nice day!
A very loud train whistle has an acoustic power output of 100 W. If the sound energy spreads out spherically, what is the intensity level in dB at a distance of 100 meters from the train?
The intensity level in dB at a distance of 100 meters from the train is 98.46 dB.
Explanation:To calculate the intensity level in dB at a distance of 100 meters from the train, we need to first calculate the intensity of sound at that distance.
Since the sound energy spreads out spherically, the intensity decreases with the square of the distance from the source.
Using the formula for intensity, I = P/4πr², where P is the acoustic power output and r is the distance from the source, we can calculate the intensity at 100 meters:
I = 100W / (4π * (100m)²) = 0.0796 W/m².
Now, we can use the formula for the intensity level in dB, β = 10log(I/I₀), where I₀ is the reference intensity (10⁻¹² W/m²).
Substituting the values, β = 10log(0.0796 / 10⁻¹²) = 98.46 dB.
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Which fossil fuel is produced as a by-product that occurs when bacteria decompose organic material under anaerobic conditions?
Bio-gas is the naturally produced fossil fuel, a by-product when bacteria decompose organic material under anaerobic conditions.
Explanation:Organic matter particularly waste material is broken down by bacteria through fermentation in an environmental condition without any presence of oxygen. This process of decomposition leads to formation of bio-gas with "carbon dioxide and methane" in a 2:3 ratio.
The above biological process is termed as bio-digestion or anaerobic digestion. Methane is flammable and thus bio-gas can be used as "energy source", a waste-to-energy transformation. The remaining decomposed matter is ideal as manure for plants due to its rich nutrient level.