Answer:
Mass needed to fire is [tex]3 \times 10^4 kg[/tex]
Explanation:
Mass of the spaceship [tex]m_{0} = 5.30 \times 10^4 kg[/tex]
Initial velocity of the spaceship[tex]v_{0} = 5.75 \times 10^3 m/s[/tex]
Final velocity of the spaceship [tex]v_{f} = 8.39 \times 10^3 m/s[/tex]
Take exhaust velocity [tex]u = 4.6\times 10^3 m/s[/tex]
The velocity of the spaceship in the space is
[tex]v_{f}=v_{0}+u\times ln(\frac{m_{0}}{m} )\\8.39 \times 10^3= 5.75 \times 10^3+ 4.6 \times 10^3 \times ln(\frac{5.30 \times 10^4}{m} )\\8.39 \times 10^3-5.75 \times 10^3= 4.6 \times 10^3 \times ln(\frac{5.30 \times 10^4}{m} )\\2.64 \times 10^3=4.6 \times 10^3 \times ln(\frac{5.30 \times 10^4}{m} )\\\frac{2.64 \times 10^3}{4.6 \times 10^3} = ln(\frac{5.30 \times 10^4}{m} )\\0.57=ln(\frac{5.30 \times 10^4}{m} )\\e^{0.57}=\frac{5.30 \times 10^4}{m}\\m=\frac{5.30 \times 10^4}{1.768} \\m=3 \times 10^4 kg[/tex]
Mass needed to fire is [tex]3 \times 10^4 kg[/tex]
1. Impulse is the product of force and distance. True or false?
2. The force necessary to lift an object is g. True or false?
3. A joule is a newton times a second. True or false?
4. Work is done to lift an object in the classroom. True or false?
5. Kinetic energy is the energy of position. True or false?
6. Stopping distance is doubled, if speed is quadrupled.True or false?
Answer:
Explanation:
1. Impulse, I = F.t
The statement impulse is the product of Force and distance is false.
2. F = m g
Force necessary to lift the object depends on the mass of the object.
statement 2 is false.
3. Joule is equal to Newton times meter.
Statement 3 is false.
4. Work done to lift an object is correct statement.
Statement 4 is true.
5. Kinetic energy of an object is due to motion.
Statement 5 is false.
6. Stopping distance is directly proportional to the square of velocity.
If velocity is doubled, stopping distance is quadrupled.
Statement 6 is false.
The correct answer is 1.False; 2.False; 3.False; 4.True; 5.False; 6.False.
1. Impulse is the product of force and time, not force and distance. Therefore, the statement is false.
2. The force necessary to lift an object is not simply g, but weight, which is calculated as mass times the gravitational acceleration g. So, this statement is false.
3. A joule is the unit of work or energy, equivalent to one newton meter (N·m), not a newton times a second. Thus, the statement is false.
4. Work is indeed done to lift an object in the classroom since work is the force applied over a distance in the direction of the force. Therefore, this statement is true.
5. Kinetic energy is the energy of motion, not the energy of position. Thus, this statement is false.
6. If the stopping distance is doubled when speed is quadrupled, it overlooks the fact that stopping distance increases with the square of the speed, meaning a quadrupling of speed typically increases the stopping distance by a factor of sixteen, not two. So, this statement is false.
For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 8.0 cm wide. What is the width of the slit?
Answer:
width of slit(a)≅ 0.1mm
Explanation:
Wave length of laser pointer =λ = 685 nm
Distance between screen and slit = L = 5.5 m
Width of bright band = W=8.0cm=0.08m
width of slit=a
recall the formula;
W=(2λL)/a
a=2λL/W
a=(2 *685*10⁻⁹*5.5m)/0.08m
a=7535*10⁻⁹/0.08
a=94187.5 *10⁻⁹
a=0.0000941875m
a=0.0941875mm
a≅0.1mm
Answer:
The wide of silt is a=93.5×10⁻⁶m
Explanation:
Given data
Wavelength λ=680 nm
Length L=5.5 m
Width w=8.0 cm
To find
Width of slit a
Solution
A single slit of width a has a bright central maximum of width
ω=2λL/a
a=2λL/ω
Substitute the given values
[tex]a=\frac{2*(680*10^{-9}m)5.5m}{0.08m} \\a=93.5*10^{-6}m[/tex]
The wide of silt is a=93.5×10⁻⁶m
A 2.00-nF parallel-plate capacitor is charged to an initial potential
difference AV}: 100 V and is then isolated. The dielectric material
between the plates is mica, with a dielectric constant of 5.00. (a) How
much work is required to withdraw the mica sheet? (b) What is the
potential difference across the capacitor after the mica is withdrawn?
Answer:
The answer to the question is;
(a) The amount of work required to withdraw the mica sheet is 4.0×10⁻⁵ J
(b) The potential difference across the capacitor after the mica is withdrawn is 500 V.
Explanation:
To solve the question, we note that
The Capacitance of the capacitor is proportional to the dielectric constant, that is removal of the dielectric material will reduce the capacitance of the capacitor by a factor of the dielectric constant as follows
Energy stored in a capacitor is given by
[tex]E = \frac{1}{2} CV^{2}[/tex]
Where:
C = Capacitance of the capacitor = 2.00 nF
V = Voltage = 100 V
Therefore before the dielectric material is removed, we have
E = [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 1.0×10⁻⁵ J
Asfter the dielectric material is removed we have
E = 5× [tex]\frac{1}{2}[/tex]×2.00 × 10⁻⁹×100² = 5.0×10⁻⁵ J
Since energy is neither created nor destroyed, we have
Change in energy of system = 5.0×10⁻⁵ J - 1.0×10⁻⁵ J = 4.0×10⁻⁵ J
Work done on the system to remove the mica sheet = energy change of the system = 4.0×10⁻⁵ J.
(b) From the equation [tex]C = \frac{Q}{V}[/tex]
Where
Q = Charge of the capacitor
V= Voltage
C = Capacitance of the capacitor, we have
Q, the charge of the capacitor is constant
and the final capacitance = [tex]\frac{Initial .capacitance}{5} = \frac{C_1}{5}[/tex]
Therefore we have
V₁ = [tex]\frac{Q}{C_1}[/tex]
V₂ = [tex]\frac{Q}{\frac{C_1}{5} }[/tex] = [tex]{5} *\frac{Q}{C_1}[/tex] = 5·V₁
The velocity increases by a factor of 5 and
V₂ = 5×V₁ = 5 × 100 V = 500 V.
The formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with radius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? FThe formula v = √ 2.3 r models the maximum safe speed, v , in miles per hour, at which a car can travel on a curved road with radius of curvature r , in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 370 feet. What is the maximum safe speed? F
The maximum safe speed on a curved road can be calculated using the formula v = √(2.3r), where v represents speed in miles per hour and r is the radius of curvature in feet. By substituting the given radius of curvature into the formula, we can find the maximum safe speed. In this case, the maximum safe speed is approximately 29.18 miles per hour.
Explanation:The formula v = √(2.3r) models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r, in feet. To find the maximum safe speed at a given radius of curvature, we can substitute the value of r into the formula and solve for v. In this case, the radius of curvature is given as 370 feet, so we substitute r = 370 into the formula:
v = √(2.3 * 370)
Simplifying the expression, we get:
v = √(851)
Taking the square root, we find that the maximum safe speed is approximately v ≈ 29.18 miles per hour.
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How is the size of the mri signal related to the applied b0 field?
Answer:
Intensity Of MRI signal is related as square of [tex]B_{0}[/tex]
or
Intensity is related as [tex]B_{0}^{2}[/tex].
Explanation:
MRI simply means Magnetic Resonance Imaging. In this process by the resonance of magnetic field the frequency is produced which is used to image the body parts (Mainly the brain).
The signal Strength is the rate of change of Magnetization (M) with respect to the time. For the calculation of Intensity we use Bloch equation which is given by
[tex]\frac{dH}{dt} = M \times \gamma B[/tex]
On further solving
[tex]\frac{dH}{dt} =\mu B \times \gamma B \\\frac{dH}{dt} = \mu \times \gamma B^{2}[/tex]
So, Intensity is related as [tex]B_{0}^{2}[/tex].
If the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules _______, the average force of an individual collision _________, and the average number of collisions with the wall, per unit area, per second_______.
When the volume of gas is reduced at constant temperature, the average velocity of the molecules, the average force of an individual collision, and the average number of collisions with the wall all increase.
When the volume of a sample of gas is reduced at constant temperature, the average velocity of the molecules increases, the average force of an individual collision between molecules and the container walls increases, and the average number of collisions with the wall, per unit area, per second increases.
When the volume of a gas is reduced at constant temperature, the gas molecules’ average velocity remains unchanged, the force of each collision increases, and collisions with the wall per unit area per second become more frequent.
If the volume of a sample of gas is reduced at constant temperature, according to Boyle's law, the average velocity of the molecules remains the same, the average force of an individual collision increases, and the average number of collisions with the wall, per unit area, per second increases.
At a constant temperature, the kinetic energy of the gas molecules, as well as their average velocity, does not change. This is in accordance with the kinetic theory of gases. However, as the volume decreases, the gas molecules have less space to move around, which means they will encounter the container walls more frequently. Hence, the number of collisions with the wall per unit area per second increases. Since the same amount of molecules exert force over a smaller area of the container wall (because the volume is reduced), the average force per collision also increases.
It's important to clarify that while the average velocity does not change, the frequency of collisions leads to the increased pressure we observe in a confined volume of gas. Consequently, the gas exerts a greater pressure on the walls of the container.
An object with mass 0.900kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00m to the right (the x - direction) to stretch the spring, and released. What is the speed of the object when it is 0.50m to the right of the x
Answer:
7.85 m/s
Explanation:
We are given that
Mass of object=m=0.900 kg
[tex]F(x)=\alpha x-\beta x^2[/tex]
[tex]\alpha=60 N/m[/tex]
[tex]\beta=18N/m^2[/tex]
[tex]F(x)=-60x-18x^2[/tex]
U=0 when x=0
Potential energy=[tex]-\int F(x)dx[/tex]
Substitute the values
[tex]U(x)=-\int (-60x-18x^2)dx[/tex]
[tex]U(x)=60(\frac{x^2}{2})+18(\frac{x^3}{3})+C[/tex]
Using the formula
[tex]\int x^n dx=\frac{x^{n+1}}{n+1}+C[/tex]
Substitute x=0
[tex]U(0)=C\implies C=0[/tex]
[tex]U(x)=30x^2+6x^3[/tex]
[tex]x_1=0.5,x_2=1[/tex]
[tex]v_2=0[/tex]
Using law of conservation energy
[tex]\frac{1}{2}mv^2_1+U(x_1)=\frac{1}{2}mv^2_2+U(x_2)[/tex]
Substitute the values
[tex]\frac{1}{2}(0.9)v^2_1+30(0.5)^2+6(0.5)^3=0+30(1)^2+6(1)^3[/tex]
[tex]\frac{1}{2}(0.9)v^2_1+8.25=36[/tex]
[tex]\frac{1}{2}(0.9)v^2_1=36-8.25=27.75[/tex]
[tex]v^2_1=\frac{27.75\times 2}{0.9}[/tex]
[tex]v_1=\sqrt{\frac{27.75\times 2}{0.9}}[/tex]
[tex]v_1=7.85 m/s[/tex]
You are on an airplane that is landing. The plane in front of your plane blows a tire. The pilot of your plane is advised to abort the landing, so he pulls up, moving in a semicircular upward-bending path. The path has a radius of 450m with a radial acceleration of 17m/s2. What is the plane’s speed?
Answer:
[tex]v=87.46m/s[/tex]
Explanation:
Objects moving in circular path would be have either centripetal or centrifugal force.The force is either to center or away from center. When the object is moving along the circular path the centripetal force is
[tex]F=\frac{mv^{2}}{r}[/tex]
Here m is mass, v is velocity and r is radius of circular path
The acceleration is given by:
[tex]a_{r}=\frac{v^{2}}{r}[/tex]
The point of interest is lowest point on circle.The acceleration of plane at this position point up.The speed of plane from radial acceleration equation is:
[tex]v=\sqrt{a.r}\\[/tex]
Substitute 17 m/s² for a and 450m for r
So
[tex]v=\sqrt{17m/s^{2}*450m }\\ v=87.46m/s[/tex]
The materials that made up the solar nebula can be categorized into the four general types as follows. Rank these materials from left to right based on their abundance in the solar nebula, from highest to lowest.
Answer:
Hydrogen and HeliumRockMetalsHydrogen compoundsRanked from left to right based on their abundance in the solar nebula:
Hydrogen and Helium-Hydrogen compounds-Rocks-Metals
I hope you find this information useful and interesting! Good luck!
The materials in the solar nebula can be ranked from highest to lowest abundance as hydrogen and helium gas, icy and rocky planetesimals, dust and vapor, and meteorites, comets, and asteroids.
Explanation:The materials that made up the solar nebula can be categorized into four general types based on their abundance. These types, ranked from highest to lowest abundance, are:
Hydrogen and Helium Gas: H and He are the most abundant elements in the universe, including the solar nebula. They account for the majority of the mass in the nebula.Icy and Rocky Planetesimals: These are solid bodies made up of a mixture of ice and rock. They are the building blocks of planets and moons.Dust and Vapor: Tiny solid particles and gases make up this category. Dust grains can condense and stick together to form larger particles.Meteorites, Comets, and Asteroids: These are remnants of the original solar nebula. They have survived from the early stages of the solar system's formation.Learn more about materials in the solar nebula here:https://brainly.com/question/31734041
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The circuit to the right consists of a battery ( V 0 = 64.5 V) (V0=64.5 V) and five resistors ( R 1 = 711 (R1=711 Ω, R 2 = 182 R2=182 Ω, R 3 = 663 R3=663 Ω, R 4 = 534 R4=534 Ω, and R 5 = 265 R5=265 Ω). Find the current passing through each of the specified points. -g
To find the Current in a circuit passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.
To find the current passing through each specified point in the circuit, we can use Ohm's law and the known resistances and voltages.
For example, to find the current passing through resistor R₂, we first need to find the voltage applied to it. This can be done by subtracting the voltage drop across resistor R₁ from the battery voltage. Once we have the voltage, we can use Ohm's law to find the current through R₂.
We can repeat this process for the other specified points in the circuit using the appropriate resistances and voltages.
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Which molecule of the gpcr-adenylyl cyclase signal transduction system phosphorylates proteins to create the cellular response?
Answer:
Protein kinase A
Explanation:
Protein kinase A is also called Cyclic AMP- dependent protein kinase or A kinase. It is an enzyme that enhance protein covalently using the phosphate group. The function of this enzyme is that it helps to end the effect of different hormones working via the Cyclic AMP signalling pathway. Protein kinase A can be found in the cytoplasm which phosphorylate proteins.
Protein kinase A helps the cell in regulating sugar, glycogen and lipids metabolism level.
Answer:
The answer is Protein Kinase A.
Refer below for the explanation.
Explanation:
In cell science, protein kinase A is a group of compounds whose movement is reliant on cell levels of cyclic AMP. PKA is otherwise called cAMP-subordinate protein kinase. Protein kinase A has a few capacities in the cell, including guideline of glycogen, sugar, and lipid digestion.
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m. A positive point charge q3 is located at the origin. a) What must the value of q3 be for the net force on thispoint charge to have a magnitude 4.40 μN ?
Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:
[tex]F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\[/tex]
Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:
[tex]F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC[/tex]
In words, the value of q₃ must be 5.3nC.
At one instant of time, a car and a truck are traveling side by side in adjacent lanes of a highway. The car has a greater velocity than the truck has. Does the car necessarily have the greater acceleration?Explain
Answer:
NO
Explanation:
Acceleration is change in velocityΔv in respect to timeΔt
so if the velocity of the car is greater than the truck it does not mean that the car acceleration is greater than the truck.
Sometimes with constant velocity it means no accelaration ,but the truck may have accelaration
so, higher velocity of the car does not mean higher acceleration
0.140-kg baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove
Answer:
-7.8×[tex]10^{2}[/tex] N
Explanation:
see the attachment plz
Two identical loudspeakers separated by distance d emit 200Hz sound waves along the x-axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on.What are the three lowest possible values for d? Assume a sound speed of 340m/s.
Answer:
So three lowest possible values of d ar 0.85m, 2.55m and 4.25m
Explanation:
The wavelength of wave:
λ=v/f
λ=340/200=1.7m
The destructive interference condition is:
Δd=(m+1/2)λ where m=0,1,2,3........
For minimum destructive interference ,the value of m is equal to zero
Δd=(0+1/2)×1.7
Δd=0.85m
For m=1
Δd=(1+1/2)×1.7
Δd=2.55m
For m=2
Δd=(2+1/2)×1.7
Δd=4.25m
So three lowest possible values of d ar 0.85m, 2.55m and 4.25m
The interference of sound waves is causing silence at your location. The three lowest possible values for the distance between two speakers creating this effect, given the given frequency of 200Hz and sound speed of 340m/s, are 0.85m, 2.55m and 4.25m.
Explanation:This is a problem related to sound interference; that is when two or more sound waves overlap, the resultant sound you hear is different than when you hear the sound waves individually.
You are not hearing anything because the sounds from the two speakers are arriving out of phase at your location, causing destructive interference.
At destructive interference, the path difference is (n + 1/2)λ. For the first three lowest values for d, n = 0, 1, 2. The speed of sound v = λf, so λ = v / f. Substituting these values in we get d = (n + 1/2) * v/f.
Now, let's plug in the given values: f=200Hz, v=340m/s.
d for n=0: d = (0 + 1/2) * 340 / 200 = 0.85 md for n=1: d = (1 + 1/2) * 340 / 200 = 2.55 md for n=2: d = (2 + 1/2) * 340 / 200 = 4.25 mLearn more about Sound Interference here:
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Technician A says that a heavy engine load results in high intake manifold vacuum and a high MAP sensor signal voltage. Technician B says that a manifold absolute pressure (MAP) sensor uses a perfect vacuum (zero absolute pressure) in the sensor to determine the pressure. Who is right?
Answer:
Technician B
Explanation:
When a mass of 24 g is attached to a certain spring, it makes 21 complete vibrations in 3.3 s. What is the spring constant of the spring? Answer in units of N/m.
Answer: 3.889N/m
Explanation:
f=(1/2)*√k/m
f=n/t
f= 21/3.3=6.4Hz
6.4*2=√k/m
12.73=√k/m
12.73^2=k/m
162.0529=k/m
Since m=24g
Convert g to kg
m=24/1000
m=0.024kg
K=162.0529*0.024
K=3.889N/m
Final answer:
The spring constant of the spring is approximately 24 N/m.
Explanation:
When a 24-gram mass is attached to a spring, causing it to make 21 complete vibrations in 3.3 seconds, we are looking at a problem that involves simple harmonic motion and can use the properties of a mass-spring system to determine the spring constant (k).
The formula to calculate the frequency (f) is:
f = number of vibrations / time
f = 21 / 3.3 s
f = 6.36363636 Hz
The frequency is related to the spring constant (k) and mass (m) as follows:
f = (1 / (2π)) * (√(k / m))
Given the mass (m) and frequency (f), we can solve for k:
k = (2πf)² * m
k = (2 * π * 6.36363636 Hz)² *0.024 kg
k ≈ 24 N/m
Therefore, the spring constant of the spring is approximately 24 N/m.
Suppose we placed a positive charge Q on the Moon and an equal positive charge Q on the Earth. What value of Q would be needed to neutralize the gravitational attraction of the Moon and the Earth?
Answer: q=5.70 x 10^13 C
Explanation:
gravitational attraction = electrostatic repulsion GMm/d^2 = kQ^2/d^2 as you can see the d^2 cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10^-11 m^3 / kgs^2 M is earth's mass = 5.972 × 10^24 kg m is moon's mass = 7.342×10^22 kg Q is charge on earth and moon. k is coulomb's constant = 9 x10^9 N m^2 /C^2 On solving equation for Q. Q = sqrt (GMm/k) = sqrt ( 6.67 x 10^-11 x 5.972 x 10^24 * 7.342×10^22 / 9 x10^9) = 5.70 x 10^13 C
To neutralize the gravitational attraction between the Moon and the Earth using equal charges, calculate the total gravitational force and equate it to the electrostatic force of the charges. q=5.70 x 10¹³ C
gravitational attraction = electrostatic repulsion
GMm/d² = kQ²/d² as you can see the d² cancel out. that is why lunar distance is irrelevant. G is the universal gravitational constant = 6.67 x 10⁻¹¹ m³ / kgs² M is earth's mass = 5.972 × 10²⁴ kg m is moon's mass = 7.342×10²² kg Q is charge on earth and moon. k is coulomb's constant = 9 x10⁹ N m² /C² On solving equation for Q. Q = √ (GMm/k) = √ ( 6.67 x 10⁻¹¹x 5.972 x 10²⁴ × 7.342×10²² / 9 x10⁹) = 5.70 x 10¹³ C
When the compressor of an air conditioner starts up, it draws a current of 50 A. If the start-up time is 0.60 s, then the amount of charge passing through a cross-sectional area of the circuit during this time is:
Answer:
30 C
Explanation:
Given:
Current flowing in the circuit (I) = 50 A
Start-up time (t) = 0.60 s
Now, we know that, charge drawn in through a cross sectional area of the circuit is given as:
[tex]q=It[/tex]
Where, 'q' is the amount of charge drawn, 'I' is the current and 't' is the start-up time.
Now, plug in 50 A for 'I', 0.60 s for 't' and solve for 'q'. This gives,
[tex]q=50\ A\times 0.60\ s\\\\q=30\ C[/tex]
Therefore, the amount of charge drawn in the circuit at the start-up of the compressor of an air conditioner is 30 C.
Tyrel is learning about a certain kind of metal used to make satellites. He learns that infrared light is absorbed by the metal, X-ray light is transmitted through the metal, and visible light is reflected off the metal. Tyrel wonders of the metal will get warm if he shines the lights on it. Can light cause the metal to get warm?
Answer: yes.
Explanation: The light that will be incidented on that metal is visible light.
It depends on 3 factors:
1. The temperature
2. The specific heat capacity of the metal
3. The thermal conductivity of the metal.
The metal getting warmer also depend on the reflection and the absorption of light energy in which it will surely absorb some energy and not reflect all.
When visible light is absorbed by an object, the object converts the short wavelength light into long wavelength heat. This causes the object to get warmer.
The photoelectric effect causes the metals to get warm when light is shined on them.
Photoelectric effect:
It states that the metals emit electrons when by absorbing electromagnetic radiation.
When light sine on the metal, electrons present on the metals absorb some energy of the light or radiations and leave their atoms. This causes the metal to heat up when light sine on it.
The formula is,
[tex]h \nu= W + E[/tex]
Where,
[tex]h[/tex]- Plank's constant
[tex]\nu[/tex] - Threshold frequency
[tex]W[/tex] - Energy needed to remove the electron
[tex]E[/tex]- Kinetic energy of the emitted electron
Therefore, the metal gets warm due to the Photoelectric effect.
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A diver makes 2.5 revolutions on the way from a 9.3-m-high platform to the water. Assuming zero initial vertical velocity, find the diver's average angular velocity during a dive.
Answer:
11.4 rad/s
Explanation:
The motion of the diver is a free-fall motion, so its center of mass falls down with constant acceleration of
[tex]g=9.8 m/s^2[/tex] towards the water
Therefore, we can use the following suvat equation:
[tex]s=ut-\frac{1}{2}gt^2[/tex]
where:
s = 9.3 m is the vertical displacement of the diver
u = 0 is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex]
And t is the total time of flight. Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(9.3)}{9.8}}=1.38 s[/tex]
So, the diver takes 1.38 s to reach the water.
During this time, the diver makes 2.5 revolutions; since 1 revolution is equal to an angle of [tex]2\pi[/tex] radians, then the total angular displacement is
[tex]\theta=2.5\cdot 2\pi =15.7 rad[/tex]
Therefore, the average angular velocity of the diver is the ratio between the total angular displacement and the time taken:
[tex]\omega=\frac{\theta}{t}=\frac{15.7}{1.38}=11.4 rad/s[/tex]
A 75-hp compressor in a facility that operates at full load for 2500 h a year is powered by an electric motor that has an efficiency of 93 percent. If the unit cost of electricity is $0.11/kWh, the annual electricity cost of this compressor is
The annual electricity cost of the compressor is $15,386.25.
Explanation:To calculate the annual electricity cost of the compressor, we first need to find the energy consumption of the motor. We can use the formula:
Energy Consumption = Power x Time
Given that the compressor operates at full load for 2500 hours a year and has a power of 75 hp, we need to convert the power to watts:
1 hp = 746 watt
So, the power of the motor is 75 x 746 = 55,950 watts.
Now, we can calculate the energy consumption:
Energy Consumption = 55,950 watts x 2500 hours
Next, we need to convert the energy consumption to kilowatt-hours (kWh):
1 kWh = 1000 watt-hours
So, the energy consumption in kWh is 55,950 watts x 2500 hours / 1000 = 139,875 kWh.
Finally, we can calculate the annual electricity cost by multiplying the energy consumption by the unit cost of electricity:
Annual Electricity Cost = 139,875 kWh x $0.11/kWh
Therefore, the annual electricity cost of this compressor is $15,386.25.
Learn more about electricity cost here:https://brainly.com/question/32439156
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The ______ clause in an if statement allows a program to make one choice if the condition is true, and anothere choice if the condition is false.
Answer: The IF clause.
Explanation: This is the IF clause; you can use it as:
IF (something = true) then "something happens"
else "other thing happens"
Some example of it can be, suppose that your program reads a number X that the user inputs, then you can do:
If ( X > 5) then
print: "the number X is bigger than five"
Else
print: "the number X is smaller than five"
Where, of course, the statements depend on the language used, but the "if" clause works almost the same in every language.
A spotlight on a boat is y = 2.2 m above the water, and the light strikes the water at a point that is x = 8.5 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4.0 m. Determine the distance d, which locates the point where the light strikes the bottom.
Answer:
The answer to the question is
The distance d, which locates the point where the light strikes the bottom is 29.345 m from the spotlight.
Explanation:
To solve the question we note that Snell's law states that
The product of the incident index and the sine of the angle of incident is equal to the product of the refractive index and the sine of the angle of refraction
n₁sinθ₁ = n₂sinθ₂
y = 2.2 m and strikes at x = 8.5 m, therefore tanθ₁ = 2.2/8.5 = 0.259 and
θ₁ = 14.511 °
n₁ = 1.0003 = refractive index of air
n₂ = 1.33 = refractive index of water
Therefore sinθ₂ = [tex]\frac{n_1sin\theta_1}{n_2}[/tex] = [tex]\frac{1.003*0.251}{1.33}[/tex] = 0.1885 and θ₂ = 10.86 °
Since the water depth is 4.0 m we have tanθ₂ = [tex]\frac{4}{x_2}[/tex] or x₂ = [tex]\frac{4}{tan\theta_2 }[/tex] =[tex]\frac{4}{tan(10.86)}[/tex] = 20.845 m
d = x₂ + 8.5 = 20.845 m + 8.5 m = 29.345 m.
TheAurora Borealisis a night display in the Northern latitudes caused by ionizing radiation interacting with the Earth's magnetic field and the upper atmosphere. The distinctive green color is caused by the interaction of the radiation with oxygen and has a frequency of 5.38 x 1014units. What is the wavelength of this light?
Answer:
The answer is
The wavelength of the light is 557.2 x 10⁻⁻⁹nm
Explanation:
The wavelength of a wave is the distance between adjacent troughs and crests while the frequency of a wave is the number of completed cycles that pass a given per unit time
Electromagnetic radiation, such as light is usually described n terms of its frequency and wavelength
The equation relating the three quantities of speed of light, frequency and wavelength is as follows
Speed of light, c = Frequency of the light wave, ν × Wavelength of the light, λ
That is c = ν × λ
Where c = 299792458 m/s, v = 5.38 x 10¹⁴ Hz
Therefore the wavelength = [tex]\frac{c}{v}[/tex] = 5.572 x 10⁻⁷ m 557.2 x 10⁻⁻⁹nm
Part C: Quantitative Problems when vf is not 0
8. A 7 kg. bowling ball traveling down a lane at 8 m/s hits a wall, and after 0.05 seconds
rebounds at the same speed.
a. Find (delta)v and m(delta)v.
b. With what impact force did the bowling ball hit the wall?
Answer:
(a)
[tex]\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s[/tex]
(b)
1120 N
Explanation:
Change in velocity, [tex]\triangle v[/tex] is given by subtracting the initial velocity from the final velocity and expressed as [tex]\triangle v= v_f -v_i[/tex]
Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by
[tex]\triangle v=0-8=-8\ m/s[/tex]
To find [tex]m\triangle v[/tex] then we substitute 7 kg for m and -8 m/s for [tex]\triangle v[/tex] therefore [tex]\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s[/tex]
(b)
The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie
[tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}[/tex]
Substituting t with 0.05 s then [tex]a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}[/tex]
Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N
Therefore, the impact force is equivalent to 1120 N
Two blocks with masses M1 and M2 hang one under the other. For this problem, take the positive direction to be upward, and use g for the magnitude of the free-fall acceleration. A. Find T2, the tension in the lower rope. B. Find T1, the tension in the upper rope.C. Find T2, the tension in the lower rope. D Find T1, the tension in the upper rope.
Answer:
A. [tex]T_2=M_2g[/tex]
B. [tex]T_2=(M_1+M_2)g[/tex]
Explanation:
Since the only forces acting on the blocks are the tensions and the weights (both in the vertical direction), and the system has acceleration zero, we can write the equilibrium equations for M₁ and M₂ as:
[tex]T_1-T_2-M_1g=0\\ \\T_2-M_2g=0[/tex]
From the second equation, we get:
[tex]T_2=M_2g[/tex],
Which is the answer to the part A.
Next, we substitute this result in the first equation and obtain:
[tex]T_1-M_2g-M_1g=0\\ \\T_1=(M_1+M_2)g[/tex],
Which is the answer to the part B.
Final answer:
To find the tensions in ropes supporting hanging blocks, one must draw free-body diagrams for each mass, and apply Newton's Second Law. The tension in the lower rope equals the weight of the lower mass or this weight adjusted for acceleration. The tension in the upper rope is the sum of the weight of the upper mass and the tension in the lower rope.
Explanation:
The student's question revolves around the concept of tension in ropes in a physics context, where two blocks are hanging one under the other. The physics involved here includes Newton's second law and the interplay of gravitational and tensional forces in a system. To find the tensions T₁ and T₂, we need to draw free-body diagrams for each block and apply Newton's Second Law, F = ma, where F is the net force, m is the mass and a is the acceleration of the block.
For block M₂, which hangs at the bottom, the tension T₂ is the force counteracting the gravitational pull downward. Hence, assuming the system is at equilibrium or moving at constant speed (acceleration = 0), T₂ would equal the weight of M₂, given by the formula T₂ = M₂ × g.
For block M₁, which is above M₂, the tension T₁ is slightly more complicated to calculate as it must support its own weight as well as the tension from the lower rope T₂. Therefore, the tension T₁ in the upper rope would be T₁= (M₁ + M₂) × g again assuming no net acceleration.
If the entire system is accelerating, we must incorporate the net force necessary to accelerate both masses. This changes the calculations respectively:
For the lower block (M₂), T₂ = M₂ × (g + a) if accelerating upward, or T₂ = M₂ × (g - a) if accelerating downward.
For the upper block (M₁), T₁ = (M₁ × g) + T₂ because T₁ has to support both the weight of M₁ and the tension from the lower rope.
Match the name of each gas law to the properties it compares. Part AA) Temperature and volumeB) Pressure and temperatureC) Pressure and volumePart B1. Boyle's law2. Charles's law 3. Gay-Lussac's law
Answer:
A. CHARLES' LAW
B. GAY-LUSSAC'S LAW
C. BOYLE'S LAW
Explanation:
THE QUESTION SEEKS ANSWER BY MATCHING THE PROPERTIES COMPARED BY A GAS LAW TO THE GAS LAW IN QUESTION.
A. CHARLES' LAW COMPARES TEMPERATURE AND VOLUME.
IT ESTABLISHED THE FACT THAT VOLUME AND TEMPERATURE ARE DIRECTLY PROPORTIONAL AT A FIXED PRESSURE
B. GAY-LUSSAC'S LAW COMPARE PRESSURE AND TEMPERATURE.
IT ESTABLISHED THE FACT THAT PRESSURE AND TEMPERATURE ARE DIRECTLY PROPORTIONAL FOR A GIVEN VOLUME OF GAS
C. BOYLE'S LAW COMPARES PRESSURE AND VOLUME
IT ESTABLISHED THAT BOTH ARE INVERSELY PROPORTIONAL AT A GIVEN TEMPERATURE
Answer:
1) D
2) B
3) E
Explanation: Edge 2020
What is the frequency of highly energetic ultraviolet radiation that has a wavelength of 129 nm? The speed of light is 3 × 108 m/s. Answer in units of Hz.
Answer:
[tex]2.33*10^{15}Hz[/tex]
Explanation:
The relationship between velocity v, frequency f and wavelength for electromagnetic waves is given by;
[tex]v=\lambda f..............(1)[/tex]
Given;
[tex]v=3*10^8m/s\\\lambda=129nm=129*10^{-9}m\\f=?[/tex]
We make substitutions into equation (1) as follows;
[tex]f=\frac{v}{\lambda}\\f=\frac{3*10^8}{129*10^{-9}}[/tex]
Explanation:
Below is an attachment containing the solution.
Calculate the acceleration of the object from 3 seconds to 7 seconds (1pt). Show your work (1pt) and make sure to include the correct units (1pt)! Speed at 3 seconds - 0 m/s Speed at 7 seconds - 8 m/s
Final answer:
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
Explanation:
To calculate the acceleration of the object, you can use the formula:
acceleration = (final velocity - initial velocity) / time
Using the given information, the final velocity is 8 m/s, the initial velocity is 0 m/s, and the time is 7 seconds - 3 seconds = 4 seconds.
Substituting the values into the formula:
acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²
Therefore, the acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².
To calculate the acceleration of the object from 3 seconds to 7 seconds, we use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time Interval
Given:
Initial velocity ([tex]V_{i}[/tex]) at 3 seconds = 0 m/sFinal velocity ([tex]V_{f}[/tex]) at 7 seconds = 8 m/sTime interval (Δt) = 7 seconds - 3 seconds = 4 secondsSo, the calculation is:
Acceleration = (8 m/s - 0 m/s) / 4 s = 2 m/s²
The acceleration of the object from 3 seconds to 7 seconds is 2 m/s².