Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
a certain volume of dry air at NTP is allowed to expand five times of it original volume under adiabatic condition.calculate the final pressure.(air=1.4)
Answer:
Final pressure 0.105atm
Explanation:
Let V1 represent the initial volume of dry air at NTP.
under adiabatic condition: no heat is lost or gained by the system. This does not implies that the constant temperature throughout the system , but rather that no heat gained or loss by the system.
Adiabatic expansion:
[tex]\frac{T_1}{T_2} =(\frac{V_1}{V_2} )^{\gamma -1}[/tex]
273/T2=(5V1/V1)^(1.4−1)
273/T2=5^0.4
Final temperature T2=143.41 K
Also
P1/P2=(V2/V1)^γ
1/P2=(5V1/V1)^1.4
Final pressure P2=0.105atm
Two people play tug of war. The 100-kg person on the left pulls with 1,000 N, and the 70-kg person on the right pulls with 830 N. Assume that neither person releases their grip on the rope with either hand at any time, assume that the rope is always taut, and assume that the rope does not stretch. What is the magnitude of the tension in the rope in Newtons
Answer:
The tension on the rope is T = 900 N
Explanation:
From the question we are told that
The mass of the person on the left is [tex]m_l = 100 \ kg[/tex]
The force of the person on the left is [tex]F_l = 1000 \ N[/tex]
The mass of the person on the right is [tex]m_r = 70 \ kg[/tex]
The force of the person on the right is [tex]F_r = 830 \ N[/tex]
Generally the net force is mathematically represented as
[tex]F_{Net} = F_l - F_r[/tex]
substituting values
[tex]F_{Net} = 1000-830[/tex]
[tex]F_{Net} = 170 \ N[/tex]
Now the acceleration net acceleration of the rope is mathematically evaluated as
[tex]a = \frac{F_{net}}{m_I + m_r }[/tex]
substituting values
[tex]a = \frac{170}{100 + 70 }[/tex]
[tex]a = 1 \ m/s ^2[/tex]
The force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by a is mathematically represented as
[tex]m_l * a = F_r -T[/tex]
Where T is the tension on the rope
substituting values
[tex]100 * 1 = 1000 - T[/tex]
=> T = 900 N
Two 2.0-cm-diameter insulating spheres have a 6.70 cm space between them. One sphere is charged to +70.0 nC, the other to -40.0 nC. What is the electric field strength at the midpoint between the two spheres?
Answer:
Explanation:
The distance of middle point from centres of spheres will be as follows
From each of 2 cm diameter sphere
R = 1 + 6.7 / 2 = 4.35 cm = 4.35 x 10⁻² m
Expression for electric field = Q / 4πε R²
Electric field due to positive charge
E₁ = 70 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 33.3 x 10⁴ N/C
Electric field due to negative charge
E₂ = 40 x 10⁻⁹ x 9 x 10⁹ / 4.35² x 10⁻⁴
= 19.02 x 10⁴ N/C
E₁ and E₂ act in the same direction so
Total field = (33.3 + 19.02 ) x 10⁴
= 52.32 x 10⁴ N/C .
A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?
Explanation:
KE = q
½ mv² = mCΔT
ΔT = v² / (2C)
ΔT = (200 m/s)² / (2 × 236 J/kg/°C)
ΔT = 84.7°C
This question involves the concepts of the law of conservation of energy.
The temperature change of the bullet is "84.38°C".
What is the Law of Conservation of Energy?According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.
[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]
where,
ΔT = change in temperature of the bullet = ?C = specific heat capacity of silver = 237 J/kg°Cv = speed of bullet = 200 m/sTherefore,
[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]
ΔT = 84.38°C
Learn more about the law of conservation of energy here:
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hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac slow twitch intermediate fast twitch
Answer:
Dead lifting uses tho muscle fundamentals
Explanation:
Answer:
Fast twitch
Explanation:
Edmentum
When a nucleus at rest spontaneously splits into fragments of mass m1 and m2, the ratio of the momentum of m1 to the momentum of m2 is
Answer:
p₁ = - p₂
the moment value of the two particles is the same, but its direction is opposite
Explanation:
When a nucleus divides spontaneously, the moment of the nucleic must be conserved, for this we form a system formed by the initial nucleus and the two fragments of the fission, in this case the forces during the division are internal and the moment is conserved
initial instant. Before fission
p₀ = 0
since they indicate that the nucleus is at rest
final moment. After fission
[tex]p_{f}[/tex] = m₁ v₁ + m₂ v₂
p₀ = p_{f}
0 = m₁ v₁ + m₂v₂
m₁ v₁ = -m₂ v₂
p₁ = - p₂
this indicates that the moment value of the two particles is the same, but its direction is opposite
A wet shirt is put on a clothesline to dry on a sunny day. Do water molecules lose heat and condense, gain heat and condense or gain heat and evaporate
For a wet shirt is put on a clothesline to dry on a sunny day, water molecules gain heat and evaporate.
When a clothe is placed on a line to dry, the idea is to ensure that the water molecules should evaporate.
For the water molecules to evaporate, they must gain more energy that will enable them to transit from liquid to gaseous state.
Recall that he change from liquid to vapor requires energy, this is why water molecules gain energy when they evaporate.
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1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.
a. Compute the acceleration of the crate?
Answer:
The acceleration of the crate is [tex]0.3356\,\frac{m}{s^2}[/tex]
Explanation:
Recall the formula that relates force,mass and acceleration from newton's second law;
[tex]F=m\,a[/tex]
Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:
[tex]F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}[/tex]
An inquisitive physics student and mountain climber climbs a 47.0-m-high cliff that overhangs a calm pool of water. He throws two stones vertically downward, 1.00 s apart, and observes that they cause a single splash. The first stone has an initial speed of 2.12 m/s.
(a) How long after release of the first stone do the two stones hit the water?
(b) What initial velocity must the second stone have if the two stones are to hit the water simultaneously?
magnitude =
(c) What is the speed of each stone at the instant the two stones hit the water?
first stone =
second stone =
Answer:
a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
Explanation:
a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:
[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]
Where:
[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.
[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.
[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:
[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]
Roots of the polynomial are, respectively:
[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]
Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.
b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:
[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]
[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.
[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.
[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:
[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]
[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]
The initial velocity of the second stone is -16.038 meters per second.
c) The final speed of each stone is determined by the following expressions:
First stone
[tex]v_{1} = v_{1,o} + g \cdot t[/tex]
Second stone
[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]
Where:
[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.
[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.
If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:
First stone
[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]
[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]
Second stone
[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]
[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]
The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.
A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
Yes the monkey will get hit and it will not avoid the dart.
Explanation:
Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.
No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.
What would be the Roche limit (in units of Earth radii) if the Earth had the same mass, but its radius was increased to 1.5 Earth radii?
First calculate the density of this new, larger, Earth. Now use this new density and the new radius in the calculator above to determine the Roche limit for this new larger 'Earth.
Answer:
Roche limit = 1.89 of earth radius
Explanation:
We know that,
Mass of earth = 5.972 × 10²⁷ g
New radius = 1.5(old radius) = 1.5(6.371 × 10⁸) = 9.5565 × 10⁸
Density of earth = 5.5132 g/cm³
New density of earth = Mass of earth / (4/3)πr³
New density of earth = 5.972 × 10²⁷ kg / (4/3)(22/7)( 9.5565 × 10⁸)³
New density of earth = 1.634 g/cm³
Roche limit = [2(Density of earth)/(New density of earth)]¹/³r
Roche limit = 1.89 of earth radius
Given a double slit apparatus with slit distance 1 mm, what is the theoretical maximum number of bright spots that I would see when I shine light with a wavelength 400 nm on the slits
Answer:
The maximum number of bright spot is [tex]n_{max} =5001[/tex]
Explanation:
From the question we are told that
The slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]
The wavelength is [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]
Generally the condition for interference is
[tex]n * \lambda = d * sin \theta[/tex]
Where n is the number of fringe(bright spots) for the number of bright spots to be maximum [tex]\theta = 90[/tex]
=> [tex]sin( 90 )= 1[/tex]
So
[tex]n = \frac{d }{\lambda }[/tex]
substituting values
[tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]
[tex]n = 2500[/tex]
given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as
[tex]n_{max} = 2 * n + 1[/tex]
The 1 here represented the central bright spot
So
[tex]n_{max} = 2 * 2500 + 1[/tex]
[tex]n_{max} =5001[/tex]
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached to an eye anchor embedded in a wall. Rope segment BC creates an angle of ϕ = 51.0 ∘ with the floor and rope segment CD creates an angle θ with the horizontal. If both ropes BCA and CD can support a maximum tensile force Tmax = 120 lb , what is the maximum weight Wmax of the crate that the system can support? What is the
Answer:
Wmax = 63.65 ≈ 64 lb
Explanation:
According to the model in which active galactic nuclei are powered by supermassive black holes, the high luminosity of an active galactic nucleus primarily consists of
Answer:
the high luminosity of an active galactic nucleus primarily consists of light emitted by hot gas in an accretion disk that swirls around the black hole
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.
The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF.A) Estimate the charge that passed through Musschenbroek's body.
B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?
Answer:
a) q = 4.47 10⁻⁵ C
b) ΔV = 4.47 10⁴ V
Explanation:
A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor
U = Q² / 2C
Q = √ (2UC)
let's reduce the magnitudes to the SI system
c = 1 nF = 1 10⁻⁹ F
let's calculate
q = √ (2 1 10⁻⁹-9)
q = 0.447 10⁻⁴ C
q = 4.47 10⁻⁵ C
b) for the potential difference we use
C = Q / ΔV
ΔV = Q / C
ΔV = 4.47 10⁻⁵ / 1 10⁻⁹
ΔV = 4.47 10⁴ V
A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Answer:
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Explanation:
- To find the direction of the conventional current in the wire you use the following formula:
[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex] (1)
i: current in the wire = ?
F: magnitude of the magnetic force on the wire = 15.1N
B: magnitude of the magnetic field = 6.1T
l: length of the wire that is affected by the magnetic field = 0.45m
The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).
The direction of the current must be in the +y direction (+^j). In fact, you have:
^j X ^k = ^i
The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):
[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
How much work will it take to lift a 2-kg pair of hiking boots 2 meters off the
ground and onto a shelf in your closet?
O A. 2.45 J
OB. 4J
C. 39.2 J
D. 20 J
Answer:
Option C - 39.2 J
Explanation:
We are given that;
Mass; m = 2 kg.
Distance moved off the floor;d = 10 m.
Acceleration due to gravity;g = 9.8 m/s².
We want to find the work done.
Now, the Formula for work done is given by;
Work = Force × displacement.
In this case, it's force of gravity to lift up the boots, thus;
Formula for this force is;
Force = mass x acceleration due to gravity
Force = 2 × 9.8 = 19.2 N
∴ Work done = 19.6 × 2
Work done = 39.2 J.
Hence, the Work done to life the boot of 2 kg to a height of 2 m is 39.2 J.
Answer:39.2J
Explanation: I just answered this question and this was the correct answer. 4J is the wrong answer.
Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 38.0° with the normal to the surface, while in the glass it makes an angle of 26.0° with the normal. What is the index of refraction of the glass?
Answer:
n_glass = 1.404
Explanation:
In order to calculate the index of refraction of the light you use the Snell's law, which is given by the following formula:
[tex]n_1sin\theta_1=n_2sin\theta_2[/tex] (1)
n1: index of refraction of vacuum = 1.00
θ1: angle of the incident light respect to normal of the surface = 38.0°
n2: index of refraction of glass = ?
θ2: angle of the refracted light in the glass respect to normal = 26.0°
You solve the equation (1) for n2 and replace the values of all parameters:
[tex]n_2=n_1\frac{sin\theta_1}{sin\theta_2}=(1.00)\frac{sin(38.0\°)}{sin(26.0\°)}\\\\n_2=1.404[/tex]
The index of refraction of the glass is 1.404
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Answer:
D). Uranus.
Explanation:
Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.
As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, option D is the correct answer.
Which of the following is not a benefit of improved cardiorespiratory fitness
Answer:
C - Arteries grow smaller
Explanation:
The option choices are:
A. Faster post-exercise recovery time
B. Lungs expand more easily
C. Arteries grow smaller
D. Diaphragm grows stronger
Explanation:
There are many advantages of cardiorespiratory fitness. It can decrease the risk of heart disease, lung cancer, type 2 diabetes, stroke, and other diseases. Cardiorespiratory health helps develop lung and heart conditions and enhances feelings of well-being.
A particle with charge q is to be brought from far away to a point near an electric dipole. Net nonzero work is done if the final position of the particle is on:__________
A) any point on the line through the charges of the dipole, excluding the midpoint between the two charges.
B) any point on a line that is a perpendicular bisector to the line that separates the two charges.
C) a line that makes an angle of 30 ∘ with the dipole moment.
D) a line that makes an angle of 45 ∘with the dipole moment.
Answer:
Net nonzero work is done if the final position of the particle is on options A, C and D
Explanation:
non zero work is done if following will be the final position of the charges :
A) Any point on the line through the charges of the dipole , excluding the midpoint between the two charges.
C) A line that makes an angle 30° with the dipole moment.
D) A line that makes an angle 45° with the dipole moment.
When a certain capacitor carries charge of magnitude Q on each of its plates, it stores energy Ep. In order to store twice as much energy, how much charge should it have on its plates
2Q
Explanation:
When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;
E = [tex]\frac{1}{2}qV[/tex] ------------(i)
Where;
q = charge between the plates of the capacitor
V = potential difference between the plates of the capacitor
From the question;
q = Q
E = Ep
Therefore, equation (i) becomes;
Ep = [tex]\frac{1}{2} QV[/tex] ----------------(ii)
Make V subject of the formula in equation (ii)
V = [tex]\frac{2E_{p}}{Q}[/tex]
Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;
2Ep = [tex]\frac{1}{2}qV[/tex]
Substitute the value of V into the equation above;
2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])
Solve for q;
[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]
[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]
[tex]q = 2Q[/tex]
Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
An experiment is set up to test the angular resolution of an optical device when red light (wavelength ????r ) shines on an aperture of diameter D . Which aperture diameter gives the best resolution? D=(1/2)????r D=????r D=2????r
Explanation:
As per Rayleigh criterion, the angular resolution is given as follows:
[tex]\theta=\frac{1.22 \lambda}{D}[/tex]
From this expression larger the size of aperture, smaller will be the value of angular resolution and hence, better will be the device i.e. precision for distinguishing two points at very high angular difference is higher.
If a key is pressed on a piano, the frequency of the resulting sound will determine the ________, and the amplitude will determine the ________ of the perceived musical note.
Answer:
If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.
Explanation:
The frequency of a vibrating string is primarily based on three factors:
The sounding length (longer is lower, shorter is higher)
The tension on the string (more tension is higher, less is lower)
The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)
To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.
Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.
And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.
Many things can influence the amplitude.
What is producing the sound?
How far are you from the source of the sound? The farther away the smaller the amplitude.
Intervening material. Sound does not travel through walls as well as air.
Depends on what is detecting the wave sound. Ear vs. microphone.
Answer:
The frequency will determine the pitch
the amplitude will determine the loudness
Explanation:
The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.
The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.
Four point charges have the same magnitude of 2.4×10^−12C and are fixed to the corners of a square that is 4.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.
Answer:
7.2N/C
Explanation:
Pls see attached file
If 2 balls had the same volume but ball a has twice as much mass as babil which one will have the greater density
Two cannonballs are dropped from a second-floor physics lab at height h above the ground. Ball B has four times the mass of ball A. When the balls pass the bottom of a first-floor window at height above the ground, the relation between their kinetic energies, KA and KB, is
Answer:
1:4
Explanation:
The formula for calculating kinetic energy is:
[tex]KE=\dfrac{1}{2}mv^2[/tex]
If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!
The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].
Kinetic EnergyThe kinetic energy of an object depends on the mass and velocity with which it moves.
While under free-fall, the mass of an object does not affect the velocity with which it falls.
So, the velocities of both the balls are the same.
Let the mass of ball A is 'm'
So, the mass of ball B is '4m'
The kinetic energy of ball A is given by;
[tex]KE_{A}=\frac{1}{2} mv^2[/tex]
The kinetic energy of ball B is given by;
[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]
Therefore, the ratio of kinetic energies of A and B is,
[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]
Learn more about kinetic energy here:
https://brainly.com/question/11580018