Answer:
The speed of the two cars after coupling is 0.46 m/s.
Explanation:
It is given that,
Mass of car 1, m₁ = 15,000 kg
Mass of car 2, m₂ = 50,000 kg
Speed of car 1, u₁ = 2 m/s
Initial speed of car 2, u₂ = 0
Let V is the speed of the two cars after coupling. It is the case of inelastic collision. Applying the conservation of momentum as :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V[/tex]
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
[tex]V=\dfrac{15000\ kg\times 2\ m/s+0}{(15000\ kg+50000\ kg)}[/tex]
V = 0.46 m/s
So, the speed of the two cars after coupling is 0.46 m/s. Hence, this is the required solution.
The speed of the two cars after coupling is 0.857 m/s.
Explanation:To find the speed of the two cars after coupling, we can apply the law of conservation of momentum. The momentum of both cars before the collision is equal to the momentum of both cars after the collision. Since the second car is initially at rest, its initial momentum is zero. The momentum of the first car is given by its mass times its speed. By setting up an equation with the initial momentum of the first car and the final momentum of both cars combined, we can solve for the final speed of the two cars after coupling.
Using the equation:
mass1 * v1i = (mass1 + mass2) * vf
Substituting the given values:
(15,000 kg) * (2.0 m/s) = (15,000 kg + 50,000 kg) * vf
Simplifying the equation gives:
vf = 0.857 m/s
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A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? (A) The other end of the cylinder will be repelled by the magnet.(B) The other end of the cylinder will not be affected by the magnet.(C) The other end of the cylinder will be attracted more strongly by the magnet.(D) The other end of the cylinder will be attracted less strongly by the magnet.(E) The other end of the cylinder will be attracted by the magnet.
Answer: Option (A) is the correct answer.
Explanation:
A diamagnetic material does not contain any unpaired electrons and therefore, in a magnetic field direction they will always flow in the direction opposite to the magnetic field.
Hence, a diamagnetic material will always repel the magnetic field whenever it comes in contact with that.
Thus, we can conclude that when the other end of the cylinder is brought near the North pole of the magnet then the other end of the cylinder will be repelled by the magnet.
A 12-g wad of sticky clay is hurled horizontally at a 100-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.5 m before coming to rest. If the coefficient of friction between the block and the surface is 0.65, what was the speed of the clay immediately before the impact?
Answer:
91.5 m/s
Explanation:
m = mass of clay = 12 g = 0.012 kg
M = mass of wooden block = 100 g = 0.1 kg
d = distance traveled by the combination before coming to rest = 7.5 m
μ = Coefficient of friction = 0.65
V = speed of the combination of clay and lock just after collision
V' = final speed of the combination after coming to rest = 0 m/s
acceleration caused due to friction is given as
a = - μ g
a = - (0.65) (9.8)
a = - 6.37 m/s²
Using the kinematics equation
V'² = V² + 2 a d
0² = V² + 2(- 6.37) (7.5)
V = 9.8 m/s²
v = speed of clay just before collision
Using conservation of momentum
m v = (m + M) V
(0.012) v = (0.012 + 0.100) (9.8)
v = 91.5 m/s
To find the speed of the clay immediately before the impact, we can use the principle of conservation of momentum and the equation for friction. By solving the two equations simultaneously, we can determine the speed.
Explanation:To find the speed of the clay immediately before the impact, we can use the principle of conservation of momentum. Since there is no external force acting on the clay-block system in the horizontal direction, the total momentum before the impact is equal to the total momentum after the impact.
Before the impact, the clay has a mass of 12 g and a velocity of v. The wooden block has a mass of 100 g and is initially at rest. After the impact, the clay and block stick together and move with a common velocity.
Using the conservation of momentum:
(mass of clay) × (velocity of clay before impact) + (mass of block) × (velocity of block before impact) = (mass of clay and block) × (velocity of clay and block after impact)
(0.012 kg) × v + (0.100 kg) × 0 = (0.112 kg) × v'
Then substituting the given distance the block slides before coming to rest and the coefficient of friction into the equation:
μ × (force of friction) × (distance) = (0.112 kg) × (final velocity after impact)² / 2
By substituting the given values and solving the two equations simultaneously, we can find the speed of the clay immediately before the impact.
Therefore, the speed of the clay immediately before the impact is approximately 2.89 m/s.
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A point charge is moving with speed 2 × 107 m/s parallel to the x axis along strait line at y=2 m. At t = 0, the charge is at x = 0 m. The magnitude of the magnetic field at x = 4 m is B0 at the origin. The magnitude of the magnetic field at at origin when t = 0.15 μs is?
Final answer:
The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge.
Explanation:
The magnitude of the magnetic field at the origin when t = 0.15 μs can be found using the formula for the magnetic field produced by a moving charge:
B = (μ₀Iv)/(2πr)
Where B is the magnetic field, μ₀ is the permeability of free space, I is the current, v is the velocity of the charge, and r is the distance from the charge to the point where we want to find the magnetic field. In this case, the charge is moving parallel to the x-axis, so the current can be calculated using the formula:
I = (Qv)/L
Where Q is the charge and L is the length of the moving charge. Plugging in the values and solving the equations will give the magnitude of the magnetic field at x = 4 m when t = 0.15 μs.
Ideal meters problem. What internal resistance is ideal for a voltmeter? What internal resistance is ideal for an ammeter?
Answer:
Answer to the question:
Explanation:
The value of the internal resistance of an ideal voltmeter is infinite.
The value of the internal resistance of an ideal ammeter is zero.
Final answer:
The ideal internal resistance for a voltmeter is very high to avoid affecting the circuit during voltage measurements, while for an ammeter it should be very low to avoid affecting current measurements. Incorrect use of meters can lead to inaccurate readings and potential damage.
Explanation:
Ideal Internal Resistance of Meters:
When it comes to the ideal internal resistance of a voltmeter, it should be as high as possible. This is because a voltmeter is used in parallel with a component to measure its voltage. If the voltmeter has a high internal resistance, it will minimize the current flowing through the voltmeter, thereby causing minimal change in the overall circuit operation. For example, a digital voltmeter can have an internal resistance in the megaohms (MΩ), which is ideal for not affecting the circuit.
On the other hand, the ideal internal resistance of an ammeter is very low. An ammeter is connected in series with a circuit to measure the current flowing through it. Thus, to prevent it from significantly altering the voltage drop across the circuit or the current itself, an ammeter should have minimal resistance. Digital ammeters can have much lower internal resistance compared to analog meters, making them less intrusive in circuit measurements.
If a meter with the wrong internal resistance is used, it could cause incorrect readings and potentially damage the meter or the circuit. For instance, using a meter in ammeter mode to measure voltage can short circuit the meter, as an ammeter's low resistance is not meant to handle the high potential differences associated with voltage measurements.
Which of the fundamental forces explains the structure of atoms and molecules? gravitational force
electromagnetic force
weak nuclear force
strong nuclear force
pls help me
Answer:
Strong nuclear force
Explanation:
Strong nuclear force (which is the strongest of the forces of the universe) is responsible for the attractive force between quarks to form nucleons (protons and neutrons). It is the reason why the protons (that are positive in charge) do not fly apart due to electromagnetic repulsion in the nucleus of an atom.
A tennis ball of mass 44.0 g is held just above a basketball of mass 594 g. With their centers vertically aligned, both are released from rest at the same moment, to fall through a distance of 1.08 m, (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. (b) Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?
Answer:
u = 4.6 m/s
h = 8.01 m
Explanation:
Given:
Mass of the tennis ball, m = 44.0 g
Mass of the basket ball, M = 594 g
Height of fall, h = 1.08m
Now,
we have
[tex]u^2-u'^2 = 2as[/tex]
where, s = distance = h
a = acceleration
u = final speed before the collision
u' = initial speed
since it is free fall case
thus,
a = g = acceleration due to gravity
u' = 0
thus we have
[tex]u^2-0^2 = 2\times9.8\tiimes1.08[/tex]
or
[tex]u = \sqrt{21.168}[/tex]
or
u = 4.6 m/s
b) Now after the bounce, the ball moves with the same velocity
thus, v = v₂
thus,
final speed ([tex]v_f[/tex]) = v = 4.6 m/s
Then conservation of energy says
[tex]\frac{1}{2}mu_1^2+\frac{1}{2}Mu_2^2 = \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2[/tex]
also
applying the concept of conservation of momentum
we have
mu₁ + Mu₂ = mv₁ + Mv₂
u₁ =velocity of the tennis ball before collision = -4.6 m/s
u₂ = velocity of the basketball before collision= 4.6 m/s
v₁ = velocity of the tennis ball after collision
v₂ = velocity of the basketball after collision
substituting the values in the equation, we get
Now,
solving both the equations simultaneously we get
[tex]v = (\frac{2M}{m+M})u_1+(\frac{m-M}{m+M})u_2[/tex]
substituting the values in the above equation we get
[tex]v = (\frac{2\times594}{44+594})(-4.6)+(\frac{44-594}{44+594})4.6[/tex]
or
[tex]v = -8.565-3.965[/tex]
or
[tex]v = -12.53m/s[/tex]
here negative sign depicts the motion of the ball in the upward direction
now the kinetic energy of the tennis ball
[tex]K.E = \frac{1}{2}mv^2[/tex]
or
[tex]K.E = \frac{1}{2}44\times 10^{-3}kg\times 12.53^2[/tex]
or
K.E = 3.45 J
also at the height the K.E will be the potential energy of the tennis ball
thus,
3.45 J = mgh
or
3.45 = 44 × 10⁻³ × 9.8 × h
h = 8.01 m
The magnitude of the downward velocity with which the basketball reaches the ground is 4.51 m/s. The height to which the tennis ball rebounds after an elastic collision with the basketball can be found by solving for the velocity of the tennis ball after the collision and using the equation h = (v2')² / (2g).
Explanation:To find the magnitude of the downward velocity with which the basketball reaches the ground, we can use the equation v = gt, where g is the acceleration due to gravity. Since both the basketball and tennis ball are released from rest and fall through the same distance, they will reach the ground at the same time. Therefore, the time it takes for the basketball to reach the ground is the same as the time it takes for the tennis ball to fall and bounce back up:
t = sqrt(2h/g)
where h is the distance the balls fall. We can substitute the given values into the equation to find the time:
t = sqrt(2 * 1.08m / 9.8m/s²) = 0.46 s
Therefore, the magnitude of the downward velocity with which the basketball reaches the ground is given by v = gt:
v = 9.8m/s² * 0.46s = 4.51 m/s
For part (b), to find the height to which the tennis ball rebounds after an elastic collision with the basketball, we can use the conservation of momentum and the conservation of kinetic energy. Since the collision is elastic, the total momentum and the total kinetic energy of the two balls before the collision are equal to the total momentum and total kinetic energy after the collision:
m1v1 + m2v2 = m1v1' + m2v2'
1/2m1v1² + 1/2m2v2² = 1/2m1v1'² + 1/2m2v2'²
We can substitute the given values into the equations and solve for v2', the velocity of the tennis ball after the collision:
0.044kg * 0m/s + 0.594kg * 4.51m/s = 0.044kg * v1' + 0.594kg * v2'
1/2 * 0.044kg * 0m/s² + 1/2 * 0.594kg * (4.51m/s)² = 1/2 * 0.044kg * v1'² + 1/2 * 0.594kg * v2'²
From these equations, we can solve for v2', the velocity of the tennis ball after the collision. The height to which the tennis ball rebounds can be found using the equation:
h = (v2')² / (2g)
We can substitute the calculated value of v2' into the equation to find the height:
h = (v2')² / (2 * 9.8m/s²)
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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end.The wire is 0.330 m long and has a mass of 9.60 g. It is fixed at both ends and oscillates in its fundamental mode.By resonance,it sets the air column in the tube into oscillation at that column’s fundamental frequency. Find (a) that frequency and (b) the tension in the wire.
Answer:
Explanation:
tube:
f=v/4L = 343/(4*1.2)= 71.4583Hz tube's fundamental frequency
wire:
f=v/2L -> v=2Lf
v= 2*0.323*71.4583= 46.162m/s
ρ= 0.0095/0.323= 0.02941kg/m
v=√(T/ρ) -> T=v^2*ρ
T= 46.162^2*0.02941= 62.67[N] Tension of wire.
The IKAROS spacecraft, launched in 2010, was designed to test the feasibility of solar sails for spacecraft propulsion. These large, ultralight sails are pushed on by the force of light from the sun, so the spacecraft doesn't need to carry any fuel. The force on IKAROS's sails was measured to be 1.12 mN. If this were the only force acting on the 290 kg spacecraft, by how much would its speed increase after 3.0 months of flight? Assume there are 30 days in each month.
The IKAROS spacecraft, propelled by the force of light from the sun, would have its speed increased by 30 m/s after 3 months of flight under the assumption that this is the only force acting on the 290 kg spacecraft.
Explanation:The increase in speed of the IKAROS spacecraft can be calculated using the formula force = mass x acceleration. The force exerted by the sunlight is given as 1.12 mN or 0.00112 N. The mass of the spacecraft is given as 290 kg. So, the acceleration can be calculated as force/mass which is (0.00112 N)/(290 kg) = 3.86 x 10⁻⁶ m/s². This acceleration is continuous, so over time it imparts a speed increase to the spacecraft.
The time over which the acceleration is acting is 3 months or 3 x 30 x 24 x 60 x 60 seconds = 7.776 x 10⁶ s. Using the formula speed = acceleration x time, the increase in speed can be calculated as (3.86 x 10⁻⁶ m/s²) x (7.776 x 10⁶ s) = 30 m/s. So, the speed of the spacecraft would increase by 30 m/s after 3 months of flight assuming this force is the only one acting on it.
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The speed of the spacecraft would increase by 100800 m/s after 3.0 months of flight.
Explanation:
To calculate the increase in speed of the spacecraft after 3.0 months of flight, we can use the equation:
F = m * a
Where F is the force, m is the mass, and a is the acceleration. Rearranging the equation, we get:
a = F / m
Plugging in the force value of 1.12 mN and the mass of the spacecraft which is 290 kg, we can calculate the acceleration:
a = 1.12 mN / (290 kg)
Next, we need to convert the time of 3.0 months into seconds. Since there are 30 days in each month, and each day has 24 hours and 60 minutes, we can calculate:
3.0 months = 3.0 * 30 days * 24 hours * 60 minutes * 60 seconds
After converting the time into seconds, we can use the equation:
v = u + at
Where v is the final velocity, u is the initial velocity (assumed to be 0 since the spacecraft is starting from rest), a is the acceleration, and t is the time. Rearranging the equation, we get:
v = at
Plugging in the calculated acceleration and the converted time, we can calculate the final velocity:
v = (1.12 mN / (290 kg)) * (3.0 * 30 days * 24 hours * 60 minutes * 60 seconds)
Simplifying the calculation, we get:
v = 100800 m/s
Therefore, the speed of the spacecraft would increase by 100800 m/s after 3.0 months of flight, assuming the force from the solar sails is the only force acting on the spacecraft.
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How are positive and negative ions formed
Answer: Ions are formed when atoms from group 1/3/2/17 gains or loses a small number of electrons.
Explanation: Let's take Na(Sodium) and Cl(chlorine), their atomic number are consecutively 11 and 17. Sodium's closest noble element is Neon(10) and Cl's is Argon(18). Now all atoms want to gain stability like the noble gases, they want their outer shell to be filled so that they don't have to wander around with other atoms. So to have stability Na will lose one of its electrons and Cl will take in one. Thus they consecutively will become plus charged and minus charged as the balance between their proton and electron numbers are gone
An ion is formed when a neutral atom gains or loses an electron.
A neutral atom has the same number of protons (+) and electrons (-).
If it gains another electron, it has too many negative charges, so it's a negative ion.
If it loses an electron, it has too many positive charges, so it's a positive ion.
=================================
What if it gains or loses a proton ?
That's a great question. I'm glad you asked.
Under normal circumstances, this hardly ever happens. But if it did ...
- If it somehow lost a proton from its nucleus, first of all, it would change the number of protons in the nucleus, so the atom would immediately become an atom of a different element.
- If it didn't lose an electron at the same time, then it would have too many electrons for the number of protons in the nucleus, so it would be a negative ion of the new element.
- If it somehow added a proton to its nucleus, it would change the number of protons in the nucleus, so the atom would become an atom of a different element. If it didn't add another electron at the same time, then it would have too few electrons for the number of protons in the nucleus, so it would be a positive ion of the new element.
What is the relationship between air pressure and wind velocity
Planets that have strong magnetic fields also have rapid rotation and convective fluid interior zones that are also good electrical conductors. The magnetic field of Venus is so weak that it has not yet been detected, which makes it at least 25,000 times weaker than Earth's magnetic field. Examine the Venus data file in your textbook. What information here may indicate why Venus is lacking a strong magnetic field?
Answer: Because its period of rotation is 243.01 days (retrograde)
Explanation:
In comparison, Earth rotates once in about 24 hours with respect to the Sun, but once every 23 hours, 56 minutes, and 4 seconds with respect to other, distant, stars. Earth's rotation is slowing slightly with time, so in the past a day was shorter.
Venus's weak magnetic field can be attributed to its slow retrograde rotation period, lack of a convective liquid metal core, and lack of significant tectonic activity. These conditions are all contrary to those that generally contribute to generating a strong magnetic field.
Explanation:The weak magnetic field of Venus could be linked to its retrograde rotation period and lack of a convective liquid metal core. Unlike planets with strong magnetic fields, Venus has a retrograde rotation period of 243 days, compared to Earth's 24-hour rotation period. The rotation period plays a key role in generating a planet's magnetic field. A faster rotation aids the creation of a dynamo effect, generating a stronger magnetic field.
Furthermore, the composition of Venus' core could also contribute to its weak magnetic field. A planet's magnetic field is typically caused by the movement of liquid metal within its core, which serves as a good conductor of electricity. It seems that Venus lacks this convective liquid metal core, thereby lacking the necessary conditions for a strong magnetic field.
Additionally, Venus may lack tectonic activity that is known to influence the generation of a magnetic field. While the surface of Venus has been modified by tectonics driven by mantle convection, it does not exhibit the same kind of plate tectonics as Earth.
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Two charged ping-pong balls separated by a distance of 1.63 m exert an electric force of 0.0224 N on each other. What will be the force if the objects are brought closer, to a separation of only 24.0 cm? (in N)
Answer: 1.024 N
Explanation:
Step 1:
Using Coulomb's law: F=kQ1Q2/d^2
We can rearrange the terms to get :
kQ1Q2=Fxr^2
Substituting: kQ1Q2=0.0224N x (1.63m)^2
Thus kQ1Q2=0.0595
We know that this value will not change when the balls are brought closer together.
Step 2:
Change from cm to m : 1cm=0.01m
24cm=0.24 m
Substitute into the new distance and the calculated value for kQ1Q2 into the coulomb law formula to calculate the new force
F=kQ1Q2/d^2=0.059/(0.24)^2
F=1.024N
Final answer:
The force between the ping-pong balls will be approximately 1.665 N when they are brought closer to a separation of 24.0 cm.
Explanation:
The force between two charged objects can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this case, the initial force between the ping-pong balls is 0.0224 N when they are separated by 1.63 m. To find the force when they are brought closer to a separation of 24.0 cm, we can use the fact that the force is inversely proportional to the square of the distance.
We can set up a proportion:
(0.0224 N) / (1.63 m)²= (x) / (0.24 m)²
Solving for x, we find that the force will be approximately 1.665 N.
Urban cities like Atlanta have to contend with a serious problem like pollution. Drivers in California are testing out a car that is fueled by hyrodgen; therefore, the emissions produced are water vapor. Read the article from the link below and answer the following questions:
goo.gl/h93L43
1. What does the author mean by the statement:
“If we’re really going to make a significant reduction in carbon emissions, you can only do that with fuel-cell vehicles in the mix"?
2. What are some of the challenges with these type of vehicles? How can these challenges be overcome?
3. Do you think fuel celled cars are a viable answer to decrease pollution why or why not?
1. With this statement, the author is referring to the fact that the vehicles are one of the largest polluters of the air. In order to reduce the pollution, the vehicles that are used will need to be changed, and with it the pollution will decrease significantly. The reduction of the pollution will come because the vehicles on hydrogen will not cause any pollution, so the enormous amounts of carbon dioxide released from the combustion of the engines will be thing of the past.
2. There are several challenges with this type of vehicles in order for them to replace the fossil fuel driven ones. The big price is one of the factors, as the majority of the people can not afford these cars. Another problem is that these vehicles are not as fast as the fossil fuel driven ones, and lot of people enjoy fast driving, despite it not being safe. There are millions of vehicles out there on the roads, and changing all of them with hydrogen vehicles will take a lot of time, as lot of those vehicles are new ones, so the people will not be willing to just throw them away and leave them rot in their garages. In order for the change of the driving park to be accomplished, the prices should go down, the people to be more serious about the environment and its protection, and patience as several decades will probably be needed for a change like this to be competed.
3. The fuel celled cars are a viable answer to decrease the pollution, as they are not causing any pollution, but instead will stop the process of large emissions of carbon dioxide from the fossil fueled cars. While this method is a good one, it should not be the only, as on its own it can not have the desirable effect, but instead all the major polluters should be included in the process. The industry and the production of energy are one of the major polluters as well, so they will need to follow the example, as if they not, the problem will stay, considering that the industry is constantly growing and the demand for energy is constantly growing too.
A harmonic wave is traveling along a rope. The oscillator that generates the wave completes 42.0 vibrations in 20.8 s. A given crest of the wave travels 371 cm along the rope in a time period of 11.6 s. What is the wavelength? Answer in units of m.
Answer:
15.8 cm
Explanation:
In 20.8 seconds, the number of vibrations = 42
In 1 second, the number of vibrations = 42 / 20.8 = 2.019
Frequency is defined as the number of vibrations in one second.
f = 2.019 Hz
Crest travel 371 cm in 11.6 s
v = 3.71 / 11.6 = 0.319 m / s
Use
v = f x λ
λ = v / f = 0.319 / 2.019 = 0.158 m = 15.8 cm
wo tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is 2.0 × 10−6 kg. The gravitational force that each sphere exerts on the other is balanced by the electric force. determine the charge magnitude.
Answer:
Charge, [tex]q=1.72\times 10^{-16}\ C[/tex]
Explanation:
It is given that, two tiny spheres have the same mass and carry charges of the same magnitude. Let charge on both sphere is q.
Also, the gravitational force that each sphere exerts on the other is balanced by the electric force i.e.
[tex]F_g=F_e[/tex]
[tex]G\dfrac{m^2}{r^2}=k\dfrac{q^2}{r^2}[/tex]
[tex]q=\sqrt{\dfrac{Gm^2}{k}}[/tex]
Where
G is the universal gravitational constant
k is the electrostatic constant
[tex]q=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times (2\times 10^{-6}\ kg)^2}{9\times 10^9\ Nm^2/C^2}}[/tex]
[tex]q=1.72\times 10^{-16}\ C[/tex]
So, the charge on both the spheres is [tex]1.72\times 10^{-16}\ C[/tex]. Hence, this is the required solution.
Answer:
[tex] q =5.439\times 10^{-17}C[/tex]
Explanation:
Given:
Mass of the tiny sphere, M = 2.0 × 10⁻⁶ kg
also masses are equal i.e M₁ = M₂
Now,
the gravitational force between the two masses M₁ and M₂ is given as:
[tex]F_G = \frac{GM_1M_2}{r^2}[/tex]
where,
G is the gravitational force constant = 6.67 x 10⁻¹¹ m³/kg.s²
r = center to center distance between the masses
also,
Electric force between the charges is given as
[tex]F_e=\frac{kq_1q_2}{r^2}[/tex]
where,
q₁ and q₂ are the charges and also it is given that q₁=q₂=q
k is the coulomb's law constant = 9.0 x 10⁹ N.m²/C²
since it is mentioned that [tex]F_G = F_e[/tex]
we have
[tex]\frac{kq_1q_2}{r^2} = \frac{GM_1M_2}{r^2}[/tex]
or
[tex]{9\times 10^9\times q^2} ={6.67\times 10^{-11}(2.0\times 10^{-6})^2}[/tex]
or
[tex] q =5.439\times 10^{-17}C[/tex]
A motorboat travels 92 km in 2 hours going upstream. It travels 132 km going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current?
Answer:
The speed on boat in still water is [tex]56 \frac{km}{h}[/tex] and the rate of the current is [tex]10 \frac{km}{h}[/tex]
Explanation:
Since speed , [tex]v= \frac{Distance\, traveled(D)}{Time\, taken(t)}[/tex]
Therefore speed of motor boat while traveling upstream is
[tex]v_{upstream}=\frac{92}{2}\frac{km}{h}=46\frac{km}{h}[/tex]
and speed of motor boat while traveling downstream is
[tex]v_{downstream}=\frac{132}{2}\frac{km}{h}=66\frac{km}{h}[/tex]
Let speed of boat in still water be [tex]v_b[/tex] and rate of current be [tex]v_w[/tex]
Therefore [tex]v_{upstream}=v_b-v_w=46\frac{km}{h}[/tex] ----(A)
and [tex]v_{downstream}=v_b+v_w=66\frac{km}{h}[/tex] ------(B)
Adding equation (A) and (B) we get
[tex]2v_b= (46+66) \frac{km}{h}=112 \frac{km}{h}[/tex]
=>[tex]v_b= 56 \frac{km}{h}[/tex] ------(C)
Substituting the value of [tex]v_b[/tex] in equation (A) we get
[tex]v_w= 10 \frac{km}{h}[/tex]
Thus the speed on boat in still water is [tex]56 \frac{km}{h}[/tex] and the rate of the current is [tex]10 \frac{km}{h}[/tex]
The natural frequency of an object depends on its
Answer: composition, its size and shape.
Explanation:
All objects have a frequency that characterizes them, which is called natural frequency. This means that if a disturbance or vibration is emitted near the object, it will begin to vibrate due to its natural frequency.
It should be noted that there are objects that have more than one natural frequency, and this depends on the composition of the object, its elasticity, its shape and size.
The natural frequency of an object refers to the frequency at which it naturally tends to vibrate or oscillate when disturbed.
The specific factors that determine the natural frequency of an object are:
1. Mass: The mass of an object plays a significant role in determining its natural frequency.
2. Stiffness: The stiffness or rigidity of an object is another important factor in determining its natural frequency.
3. Geometry: The shape and size of an object also influence its natural frequency.
These factors interact with each other to determine the natural frequency of an object. Generally, objects with higher mass and/or higher stiffness will have lower natural frequencies, while objects with lower mass and/or lower stiffness will have higher natural frequencies.
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A mass weighing 4 pounds is attached to a spring whose constant is 2 lb/ft. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of 6 ft/s. Determine the time at which the mass passes through the equilibrium position. (Use g = 32 ft/s2 for the acceleration due to gravity.)
Answer:
t = 025 s
Explanation:
We know
weight, W = 4 pounds
spring constant, k = 2 lb/ft
Positive damping, β = 1
Therefore mass, m = W / g
m = 4 / 32
= 1 / 8 slug
From Newtons 2nd law
[tex]\frac{d^{2}x}{dt^{2}}=-kx-\beta .\frac{dx}{dt}[/tex]
where x(t) is the displacement from the mean or equilibrium position. The equation can be written as
[tex]\frac{d^{2}x}{dt^{2}}+\frac{\beta }{m}.\frac{dx}{dt}+\frac{k}{m}x=0[/tex]
Substituting the values, the DE becomes
[tex]\frac{d^{2}x}{dt^{2}}+8\frac{dx}{dt}+16x=0[/tex]
Now the equation is
[tex]m^{2}+8m+16=0[/tex]
and on solving the roots are
[tex]m_{1}[/tex] = [tex]m_{2}[/tex] = -4
Therefore the general solution is [tex]x(t)=e^{-4t}\left ( c_{1}+c_{2}t \right )[/tex]
Now for initial condition x(0) = -1 ft
x'(0)= 8 ft/s
Now we can find the equation of motion becomes,
[tex]x(t)=e^{-4t}\left ( -1+4t \right )[/tex]
Therefore, the mass passes through the equilibrium when
x(t) = 0
[tex]e^{-4t}\left ( -1+4t \right )[/tex] = 0
-1+4t = 0
t = [tex]\frac{1}{4}[/tex]
= 0.25 s
which optical device can spread light apart through refraction
Answer:
CONCAVE
Explanation:
Convex lenses bend light rays so they come together at focus. Concave lenses spread light rays apart so they do not come together at a focus.
hope this answer help u!!
Answer:
Concave lens
Explanation:
Concave lenses are thicker at ends and thinner at middle. When parallel rays of light pass through the concave lens, they are refracted and spread out so that they appear to come from a common point known as principle focus. Due to this reason, Concave lens also called as diverging lens.
A mass of gas at pressure P and temperature T (degrees Kelvin) and volume V satisfies the equation T = KPV where K is a constant. The initial temperature is 300 degrees Kelvin, the inital volume is 4 cubic centimeters, and the inital pressure is 4 [The units are N per square centimeter. This is force per unit area, which is what pressure is.] If the pressure increases by 7 % and the volume increases by 0.07 cubic centimeters, approximately how much does the temperature change by? [hint: use the initial information to find K]
The equation is T = KPV
Using the initial information you have, solve for K:
300 = K(4)(4)
Simplify:
300 = K(16)
Divide both sides by 16
K = 300/16
K = 18.75
Now if pressure increases by 7%, the new pressure would be 4 x 1.07 = 4.28
Volume increases by 0.07, so the new volume would be 4 + 0.07 = 4.07
Now solve for t:
T = 18.75(4.28)(4.07)
T = 326.6175
Now subtract the new temperature from the original one:
326.6175 - 300 = 26.6175 increase ( Round answer as needed).
A rock thrown with speed 7.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 18.0 m before hitting the ground. From what height was the rock thrown? Use the value g = 9.800 m/s2 for the free-fall acceleration.
Answer:
height from where rock was thrown is 27.916 m
Explanation:
speed = 7.50 m/s
θ = 30°
g= 9.8 m/s²
horizontal distance = 18 m
time require for vertical displacement
[tex]time = \frac{distance}{velocity} \\t = \frac{18}{7.5\ cos30^0}[/tex]
t = 2.8 sec
now for calculation of height
s = ut + 0.5 a t²
-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)
-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)
-h = -27.916 m
h= 27.916 m
height from where rock was thrown is 27.916 m
Final answer:
To find the height from which the rock was thrown, we can use the equations of projectile motion to calculate the time it takes for the rock to hit the ground and then find the height using another equation. The rock was thrown from a height of 5.7 meters.
Explanation:
To find the height from which the rock was thrown, first, we need to determine the time it takes for the rock to hit the ground.
Using the equation d = v_iy * t + (1/2) * g * t^2, where d is the vertical distance, v_iy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity, plug in the values:
d = 0 (since the rock starts and ends at the same height)
v_iy = 7.5 m/s * sin(30°)
g = 9.8 m/s^2
Solving for t, we get t = 1.22 s.
Next, we can find the height by using the equation d = v_iy * t - (1/2) * g * t^2 and plugging in the values:
d = ?, the height we want to find
v_iy = 7.5 m/s * sin(30°)
t = 1.22 s
g = 9.8 m/s^2
Solving for d, we get d = 5.7 m.
Therefore, the rock was thrown from a height of 5.7 meters.
What is the resistance of a nichrome wire at 0.0 ∘c if its resistance is 120.00 ω at 11.5 ∘c?
Answer:
28.5 deg
Explanation:
The resistance of a nichrome wire at 0.0 ∘c if its resistance is 120.00 ω at 11.5 ∘c is 28.5 deg.
R = R0[1+ alpha(T-T0)]
R(28.5)=0.2000[1-0.005(28.5-0)]
The resistance of a nichrome is 28.5 deg.
How to find the resistance of a nichrome?The resistance of a nichrome wire at 0.0 ∘c if its resistance is 120.00 ω at 11.5 ∘c is 28.5 deg.
R = R0[1+ alpha(T-T0)]
R(28.5) = 0.2000[1-0.005(28.5-0)]
What's the value resistance of nichrome wire?Nichrome is continuously silvery-grey in shade, is corrosion-resistant, has an excessive melting thing of approximately 1, four hundred °C (2,550 °F), and has an electrical resistivity of around 112 microOhm-cm, that's around 66 times better resistivity than copper of one.678 microOhm-cm.
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A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15°C. What is the volume of the bubble when it reaches the surface where the pressure is 1.0 atm and the temperature is 27°C?
Answer: 624.98 mL
Explanation:
According to the Ideal Gas Law:
[tex]\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}[/tex] (1)
Where:
[tex]P_{1}=2.4 atm[/tex] is the pressure of the bubble under the surface
[tex]V_{1}=250 mL[/tex] is the volume of the bubble under the surface
[tex]T_{1}=15\°C + 273.15=288.15 K[/tex] is the temperature of the bubble under the surface
[tex]P_{2}=1 atm[/tex] is the pressure of the bubble at the surface
[tex]V_{2}[/tex] is the volume of the bubble at the surface
[tex]T_{2}=27\°C + 273.15=300.15 K[/tex] is the temperature of the bubble at the surface
So, we have to find [tex]V_{2}[/tex]:
[tex]V_{2}=\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}[/tex] (2)
[tex]V_{2}=\frac{(2.4 atm)(250 mL)(300.15 K)}{(288.15 K)(1 atm)}[/tex] (3)
Finally:
[tex]V_{2}=624.98 mL[/tex] This is the volume of the bubble when it reaches the surface
The volume of the bubble when it reaches the surface is about 625 mL
Further explanationThe basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
In this problem , we will use Ideal Gas Law as follows:
Given:
Initial Volume = V₁ = 250 mL
Initial Pressure = P₁ = 2.4 atm
Initial Temperature = T₁ = 15 + 273 = 288 K
Final Pressure = P₂ = 1.0 atm
Final Temperature = T₂ = 27 + 273 = 300 K
Unknown:
Final Volume = V₂ = ?
Solution:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]\frac{2.4(250)}{288} = \frac{1.0V_2}{300}[/tex]
[tex]\frac{600}{288} = \frac{1.0V_2}{300}[/tex]
[tex]\frac{25}{12} = \frac{V_2}{300}[/tex]
[tex]V_2 = \frac{25}{12} \times 300[/tex]
[tex]V_2 = 625 \texttt{ mL}[/tex]
[tex]\texttt{ }[/tex]
Learn moreMinimum Coefficient of Static Friction : https://brainly.com/question/5884009The Pressure In A Sealed Plastic Container : https://brainly.com/question/10209135Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454Answer detailsGrade: High School
Subject: Physics
Chapter: Pressure
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure
A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Do your results explain why such players seem to hang in the air at the top of a jump?
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
Part a)
[tex]T = 0.35 s[/tex]
Part b)
[tex]T' = 0.041 s[/tex]
So for top 15 cm the time interval is sufficiently larger than the time interval of last 15 cm
Explanation:
Part a)
Maximum height reached is
H = 76 cm
now the velocity of the player at starting position is given as
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]0 - v^2 = 2(-9.81)(0.76)[/tex]
[tex]v = 3.86 m/s[/tex]
time taken by it in top 15 cm position is given as
[tex]y = \frac{1}{2}gt^2[/tex]
[tex]0.15 = \frac{1}{2}(9.81)t^2[/tex]
[tex]t = 0.175 s[/tex]
so total time in that position is double because in that position first it will go up and then go down
[tex]T = 0.35 s[/tex]
Part b)
Now for bottom position of 15 cm first we will find the time to reach (76 - 15)cm have
[tex]H = \frac{1}{2}gt^2[/tex]
[tex](0.76 - 0.15) = \frac{1}{2}(9.81)t^2[/tex]
[tex]t_1 = 0.352 s[/tex]
now for total time to drop
[tex]t_2 = \sqrt{\frac{2H}{g}}[/tex]
[tex]t_2 = 0.394 s[/tex]
so time interval of last 15 cm is given as
[tex]t' = 0.394 - 0.352[/tex]
[tex]T' = 0.041 s[/tex]
So for top 15 cm the time interval is sufficiently larger than the time interval of last 15 cm
A 66-kg woman cheats on her diet and eats a 625-Calorie (625 kcal) jelly doughnut for breakfast. (a) How many joules of energy are the equivalent of one jelly doughnut? J (b) How many stairs must the woman climb to perform an amount of mechanical work equivalent to the food energy in one jelly doughnut? Assume the height of a single stair is 15 cm. stairs (c) If the human body is only 26% efficient in converting chemical energy to mechanical energy, how many stairs must the woman climb to work off her breakfast? stairs
To find the energy in joules equivalent to one jelly doughnut, multiply the Calorie value by 4.184. The number of stairs required to perform the same work as the energy in one jelly doughnut is calculated by dividing the work done to move up the stairs by the work done to move one stair height. To account for the body's efficiency in converting energy, divide the number of stairs calculated in part (b) by the efficiency.
Explanation:(a) To find the number of joules of energy equivalent to one jelly doughnut, we need to convert the Calorie value to joules. One Calorie (capital C) is equal to 1000 calories (lowercase c), which is equal to 4.184 joules. Therefore, one jelly doughnut contains 625 calories × 4.184 joules/calorie = 2615 joules.
(b) The work done by climbing stairs is equal to the product of the force applied and the distance moved. The force can be calculated using the weight of the woman, which is equal to her mass multiplied by the acceleration due to gravity (66 kg × 9.8 m/s2). Dividing the work done to move up the stairs by the work done to move one stair height (15 cm) gives the number of stairs that the woman must climb. This is equal to 1764 joules ÷ (66 kg × 9.8 m/s2 × 0.15 m) = 181 stairs.
(c) Since the human body is only 26% efficient in converting chemical energy to mechanical energy, we need to divide the work calculated in part (b) by the efficiency to account for this loss. Therefore, the number of stairs the woman must climb to work off her breakfast is 181 stairs ÷ 0.26 = 696 stairs.
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A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and the temperature is 304 K.
Answer:
[tex]p_2 = 1.76 atm[/tex]
Explanation:
given data:
v_1 = 2.37 L
v_2 = 1.68 L
p_1 =1.15 atm
p_2 = ?
t_1 = 280 K
t_2 = 304 K
from Gas Law Equation
, WE HAVE
[tex]\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}[/tex]
Putting the values
[tex]\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}[/tex]
[tex]9.733*10^{-3} = \frac{p_2 *1.68}{304}[/tex]
[tex]9.733*10^{-3}*304 = p_2*1.68[/tex]
[tex]\frac{9.733*10^{-3}*304}{1.68} =p_2[/tex]
[tex]p_2= 1.76 atm[/tex]
Can someone please help me with this physics problem? I'd really appreciate any help I can get!
Answer:
560 m
Explanation:
"A projectile is fired at time t = 0.0 s from point 0 at the edge of a cliff, with initial velocity components of v₀ₓ = 30 m/s and v₀ᵧ = 100 m/s. The projectile rises, and then falls into the sea at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible. What is the height of the cliff?"
Use constant acceleration equation in the y direction.
y = y₀ + v₀ t + ½ gt²
0 = h + (100 m/s) (25 s) + ½ (-9.8 m/s²) (25 s)²
h = 560 m
The height of the cliff is 560 m.
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?
Answer:
0.6375 m/s
Explanation:
Let x be the distance of the man from the building
from the figure attached
initially the value of x=12
Given:
[tex]\frac{dx}{dt}=-1.7m/s[/tex]
where the negative sign depicts that the distance of the man from the building is decreasing.
Now, Let The length of the shadow be = y
we have to calculate [tex]\frac{dy}{dt}[/tex] when x=4
from the similar triangles
we have,
[tex]\frac{2}{12-x}=\frac{y}{12}[/tex]
or
[tex]y=\frac{24}{12-x}[/tex]
Differentiating with respect to time 't' we get
[tex]\frac{dy}{dt}=-\frac{24}{12-x}^2\frac{-dx}{dt}[/tex]
or
[tex]\frac{dy}{dt}=\frac{24}{12-x}^2\frac{dx}{dt}[/tex]
Now for x = 4, and [tex]\frac{dx}{dt}=-1.7m/s[/tex] we have,
[tex]\frac{dy}{dt}=\frac{24}{12-4}^2\times (-1.7)[/tex]
or
[tex]\frac{dy}{dt}=-0.6375m/s[/tex]
here, the negative sign depicts the decrease in length and in the question it is asked the decreasing rate thus, the answer is 0.6375m/s
In a total-immersion measurement of a woman's density, she is found to have a mass of 63 kg in air and an apparent mass of 0.0975 kg when completely submerged with lungs almost totally empty.
A. What mass, in kg, of water does she displace?
B. What is her volume in m^3?
C. What is her average density in kg/m^3?
D. If her lung capacity is 1.75 L, is she able to float without treading water with her lungs filled with air? Assume the density of air is 1.29 kg/m3.
Answer:
Explanation:
A) mass of displaced water = loss of mass in water = 63 - .0975 = 62.9025 kg
B) her volume = volume of displaced water
= mass of displaced water / density of water = 62.9025 / 1000 m³.
= 62.9025 x 10⁻³ m³
C) her average density = her mass / her volume = 63 / 62.9025 x 10⁻³
= 1.00155 x 10³ kg / m³
D) lung capacity = 1.75 L = 1.75 x 10⁻³ m³
buoyant force created by air in the lungs = 1.75 x 10⁻³ ( 1000- 1.29 ) = 1.74 kg
As it is more than 0.0975 kg , the apparent weight in water so woman will float in water .
A. Mass of Water Displaced: ≈ 62.99 kg
B. Volume of Woman: ≈ 0.06299 m³
C. Average Density: ≈ 1000 kg/m³
D. Yes, she can float with lungs filled with air.
We can solve this problem using the concepts of buoyancy and density. Here's how to find the answers:
A. Mass of Water Displaced:
1. Buoyant Force: The difference between her mass in air and apparent mass underwater represents the buoyant force exerted by the water.
2. Buoyant Force = Weight in Air - Apparent Weight = (63 kg * 9.81 m/s²) - (0.0975 kg * 9.81 m/s²) (We can multiply mass by gravitational acceleration to convert it to weight, but for this problem, it cancels out)
3. Buoyant Force ≈ 618.27 N (Newtons)
Since buoyant force equals the weight of the water displaced, the mass of water displaced is:
Mass of Displaced Water = Buoyant Force / Gravitational Acceleration
Mass of Displaced Water ≈ 618.27 N / 9.81 m/s² ≈ 62.99 kg (rounded to two decimal places)
B. Volume of Woman:
1. Density of Water: We can assume the density of water to be 1000 kg/m³ (a standard value).
2. Volume of Displaced Water = Mass of Displaced Water / Density of Water
3. Volume ≈ 62.99 kg / 1000 kg/m³ ≈ 0.06299 m³ (rounded to five decimal places)
C. Average Density of Woman:
1. Woman's Volume (from part B): 0.06299 m³
2. Woman's Mass: 63 kg
3. Density = Mass / Volume
4. Average Density ≈ 63 kg / 0.06299 m³ ≈ 1000 kg/m³ (rounded to three decimal places)
D. Ability to Float:
1. Density of Air: 1.29 kg/m³
2. Volume of Air in Lungs: 1.75 L = 0.00175 m³ (convert liters to cubic meters)
3. When lungs are filled with air, her effective volume increases by the volume of air in her lungs.
4. Effective Volume with Air: 0.06299 m³ (woman's volume) + 0.00175 m³ (air in lungs) ≈ 0.06474 m³
5. When considering air in her lungs, her effective density becomes:
Effective Density = Mass / Effective Volume
Effective Density ≈ 63 kg / 0.06474 m³ ≈ 973.3 kg/m³ (rounded to one decimal place)
Since her effective density (with air in lungs) is lower than the density of water (1000 kg/m³), she would be able to float with her lungs filled with air.
Answers:
A. Mass of Water Displaced: ≈ 62.99 kg
B. Volume of Woman: ≈ 0.06299 m³
C. Average Density: ≈ 1000 kg/m³
D. Yes, she can float with lungs filled with air.
A certain sound level increases by an additional 20 db. By how much does the intensity increase?
Answer:
The intensity increases by a factor 100.
Explanation:
Given that,
Sound intensity = 20 db
We need to calculate the intensity factor
[tex]F=10^{\dfrac{LI}{10}}[/tex]
Where, LI = sound intensity
Put the value into the formula
[tex]F=10^{\dfrac{20}{10}}[/tex]
[tex]F=10^2[/tex]
[tex]F=100[/tex]
Hence, The intensity increases by a factor 100.
Final answer:
A 20 dB increase in sound level corresponds to a 100 times increase in sound intensity, as each 10 dB increase represents a tenfold increase in intensity.
Explanation:
A sound level increase of 20 dB signifies a substantial increase in sound intensity. Given that each 10 dB increase corresponds to a tenfold increase in intensity, a 20 dB increase would mean the intensity is 10 times 10, or 100 times greater than the original level. This is derived from the logarithmic scale used to measure sound intensity in decibels (dB), where an increase of 10 dB represents a tenfold increase in sound intensity. Therefore, we can say that a 20 dB increase in sound level would result in the new sound being 100 times more intense than the original. To provide an example, if one sound measures 60 dB and another sound measures 80 dB, the latter is 100 times more intense than the former.