9514 1404 393
Answer:
1.114 kg/m³
Explanation:
The total mass of the air in the balloon and the balloon + cargo will be the mass of the displaced air. If d is the density of the air in the balloon, then we have ...
2910d +308 = 2910×1.22
Solving for d, we find ...
2910d = 2919(1.22) -308
d = 1.22 -308/2910
d ≈ 1.114 . . . kg/m³
The density of the hot air is about 1.114 kg/m³.
the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp
A
Explanation:
1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.
If a bus travels 50 km in 10 hours, how fast was the
bus travelling?
Answer:
5 kilometers per hour
Explanation:
Speed = distance / time
Distance: 50km
Time: 10 hours
Speed = 50/10 = 5kph
Answer:
5kmph
Explanation:
if the bus traveled 50 km in 10 hours, we have to divide 50 by 10 to see how fast it traveled per hour.
50/10 = 5
therefore, the bus was traveling 5 km per hour
hope this helps :)
Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.
Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.10 V/mV/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
B = 1.03 10⁻⁸ T
Explanation:
For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish
This relational is expressed by the relation
E /B = c
B = E / c
let's calculate
B = 3.10 / 3 10⁸
B = 1.03 10⁻⁸ T
Imagine a spaceship traveling at a constant speed through outer space. The length of the ship, as measured by a passenger aboard the ship, is 28.2 m. An observer on Earth, however, sees the ship as contracted by 16.1 cm along the direction of motion. What is the speed of the spaceship with respect to the Earth
[tex]3.20×10^7\:\text{m/s}[/tex]
Explanation:
Let
[tex]L = 28.2\:\text{m}[/tex]
[tex]L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}[/tex]
The Lorentz length contraction formula is given by
[tex]L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}[/tex]
where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as
[tex]\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}[/tex]
Taking the square of the equation, we get
[tex]1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2[/tex]
or
[tex]1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2[/tex]
Solving for v, we get
[tex]v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}[/tex]
[tex]\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}[/tex]
[tex]\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c[/tex]
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
M
A boy of mass 60 kg and a girl of mass 40 kg are
together and at rest on a frozen pond and push
each other apart. The girl moves in a negative
direction with a speed of 3 m/s. What must be the
total final momentum of the boy AND girl
combined?
A. -120 kgm/s
B. 0 kgm/s
C. -100 kgm/s
D. 120 kgm/s
Answer:
option D thinking so
Explanation:
okay na your whish
Two identical loudspeakers 2.30 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 4.50 m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound.
Required:
What is the lowest possible frequency of sound for which this is possible?
Answer:
Abby is standing (4.5^2 + 2.3^2)^1/2 from the far speaker
D2 = 5.05 m from the far speaker
The difference in distances from the speakers is
5.05 - 4.5 = .55 m (Let y be wavelength, lambda)
n y = 4.5
(n + 1) y = 5.05 for the speakers to be in phase at smallest wavelength
y = .55 m subtracting equations
f = v / y = 340 / .55 = 618 / sec should be the smallest frequency
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
A very long, straight solenoid with a diameter of 3.00 cm is wound with 40 turns of wire per centimeter, and the windings carry a current of 0.235 A. A second coil having N turns and a larger diameter is slipped over the solenoid so that the two are coaxial. The current in the solenoid is ramped down to zero over a period of 0.40 s.
Required:
a. What average emf is induced in the second coil if it has a diameter of 3.5 cm and N = 7?
b. What is the induced emf if the diameter is 7.0 cm and N = 10?
Answer:
a) ε = 14.7 μv
b) ε = 21 μv
Explanation:
Given the data in the question;
Diameter of solenoid; d = 3 cm
radius will be half of diameter, so, r = 3 cm / 2 = 1.5 cm = 1.5 × 10⁻² m
Number of turns; N = 40 turns per cm = 4000 per turns per meter
Current; [tex]I[/tex] = 0.235 A
change in time Δt = 0.40 sec
Now,
We determine the magnetic field inside the solenoid;
B = μ₀ × N × [tex]I[/tex]
we substitute
B = ( 4π × 10⁻⁷ Tm/A ) × 4000 × 0.235
B = 1.1881 × 10⁻³ T
Now, Initial flux through the coil is;
∅₁ = NBA = NBπr²
and the final flux
∅₂ = 0
so, the εmf induced ε = -Δ∅/Δt = -( ∅₂ - ∅₁ ) / Δt
= -( 0 - NBπr² ) / Δt
= NBπr² / Δt
a)
for N = 7
ε = [ 7 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40
ε = 14.7 × 10⁻⁶ v
ε = 14.7 μv
b)
for N = 10
ε = [ 10 × ( 1.1881 × 10⁻³ ) × π( 1.5 × 10⁻² )² ] / 0.40
ε = 21 × 10⁻⁶ v
ε = 21 μv
An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm
Answer:
960 nm
Explanation:
Given that:
wavelength = 640 nm
For the second (2nd) dark spot; the order of interference m = 1
Thus, the path length difference is expressed by the formula:
[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]
[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]
[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]
dsinθ = 960 nm
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density
Answer:
velocity is the answer of this question.
Answer:
Velocity is the right answer ok
What is the largest known star?
Answer:
UY Scuti is slightly larger than VY Canis Majoris
Explanation:
These stars are millions of miles away and cannot be seen by the naked eye.
Beetlejuice is another large star that can be seen by the eye.
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
đổi đơn vị
42 ft2/hr to cm2/s
Answer:
X = 10.8387 cm²/s
Explanation:
In this exercise, you're required to convert a value from one unit to another.
Converting 42 ft²/hr to cm²/s;
Conversion:
1 ft² = 929.03 cm²
42 ft² = X cm²
Cross-multiplying, we have;
X = 42 * 929.03
X = 39019.26 cm²
Next, we would divide by time in seconds.
1 hour = 3600 seconds
X = 39019.26/3600
X = 10.8387 cm²/s
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
which energy does a car travelling 30 m/ph as it slows have:
a). chemical energy
b). thermal energy
c). kinetic energy
please helpp
Answer:
c) kinetic energy
Explanation:
Answer: C) kinetic energy
Explanation:
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
A metal blade of length L = 300 cm spins at a constant rate of 17 rad/s about an axis that is perpendicular to the blade and through its center. A uniform magnetic field B = 4.0 mT is perpendicular to the plane of rotation. What is the magnitude of the potential difference (in V) between the center of the blade and either of its ends?
We are being given that:
The length of a metal blade = 300 cmThe angular velocity at which the metal blade is rotating about its axis is ω = 17 rad/sThe magnetic field (B) = 4.0 mTA pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.
From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).
The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:
[tex]\mathsf{E = B\times l\times v}[/tex]
where;
B = magnetic fieldl = lengthv = relative speedFrom the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.
Then, the velocity (v) = ωx
The potential difference across it can now be expressed as:
[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]
For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.
∴
[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]
[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]
[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]
Recall that,
magnetic field (B) = 4 mT = 4 × 10⁻³ TLength L = 300 cm = 3mangular velocity (ω) = 17 rad/s[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]
[tex]\mathsf{E = 19.13 mV}[/tex]
Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.
Learn more about Faraday's law of induction here:
https://brainly.com/question/13369951?referrer=searchResults
An astronaut is traveling in a space vehicle that has a speed of 0.480c relative to Earth. The astronaut measures his pulse rate at 78.5 per minute. Signals generated by the astronaut's pulse are radioed to Earth when the vehicle is moving perpendicularly to a line that connects the vehicle with an Earth observer. (Due to vehicle's path there will be no Doppler shift in the signal.)
(a) What pulse rate does the Earth-based observer measure? beats/min
(b) What would be the pulse rate if the speed of the space vehicle were increased to 0.940c?
beats/min
Explanation:
The heart rate of the astronaut is 78.5 beats per minute, which means that the time between heart beats is 0.0127 min. This will be the time t measured by the moving observer. The time t' measured by the stationary Earth-based observer is given by
[tex]t' = \dfrac{t}{\sqrt{1 - \left(\dfrac{v^2}{c^2}\right)}}[/tex]
a) If the astronaut is moving at 0.480c, the time t' is
[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.2304c^2}{c^2}\right)}}[/tex]
[tex]\:\:\:\:=0.0145\:\text{min}[/tex]
This means that time between his heart beats as measured by Earth-based observer is 0.0145 min, which is equivalent to 69.1 beats per minute.
b) At v = 0.940c, the time t' is
[tex]t' = \dfrac{0.0127\:\text{min}}{\sqrt{1 - \left(\dfrac{0.8836c^2}{c^2}\right)}}[/tex]
[tex]\:\:\:\:=0.0372\:\text{min}[/tex]
So at this speed, the astronaut's heart rate is 1/(0.0372 min) or 26.9 beats per minute.
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
Define simple harmonic motion. Write down the expressions for the velocity and aceraletion of such motion st different position along its path
Answer:
Simple harmonic motion is a special type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacemen
In the Bohr model of the hydrogen atom, an electron in the 3rd excited state moves at a speed of 2.43 105 m/s in a circular path of radius 4.76 10-10 m. What is the effective current associated with this orbiting electron
Answer:
Current =,charge / time
Charge = e = 1.6E-19 coulombs
t = T time for 1 revolution (period)
v = S / T = distance traveled in 1 revolution / time for 1 revolution
T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14
I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
To learn more about projectile calculations please visit: https://brainly.com/question/14083704
Serena wants to play a trick on her friend Marion. She adds salt, sugar, and vinegar into her glass of water when Marion is out of the room. Why does she know that Marion will drink the water?
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
Someone help me with these questions please!
Answer:
a 25 and b 25
2. 26
60n
