Answer:
the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg
the initial acceleration is a = 753.47 m/s²
the an instrument package could they carry a required mass of 4.64 kg
Explanation:
a) What should be the radius of these balloons so they just hover above the surface of Mars?
Given that :
The density of the Martian atmosphere, ρ = 0.0154 kg/m³
The volume of the sphere, V = (4/3)πr³
The area of the sphere, A = 4πr²
The mass of the balloon is m = (4.60 g/m²)A
m = (4.60×10⁻³ kg/m²)(4πr²)
The formula for the buoyant force is expressed as :
F = ρVg
m×g = ρ×V×g
m = ρ×V
Now;
(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³
r = 3(4.60×10⁻³ kg/m²)/ ρ
r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³
r = 0.896 m
Thus; the radius of the balloon r = 0.896 m
The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)
m = (4.60×10⁻³ kg/m²)(4π×0.896²)
m = 4.64 × 10⁻² kg
b) what would be its initial acceleration assuming it was the same size as on Mars?
The density of the air on earth, ρ = 1.20 kg/m³
The volume of the balloon is V = (4/3)(π)(0.896 m)³
V = 3.01156 m³
Considering the net force acting on the balloon ; we have
ΣF = ρVg - mg = ma
However; making the initial acceleration a of the balloon the subject ; we have:
a = (ρVg - mg)/m
a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)
a = 753.47 m/s²
c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
The volume of the total system is V' = (4/3)π(5r)³
V' = (4/3)π(5)³(0.896 m)³
V' =376.446 m³
The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)
m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²
m = 1.159587 kg
We can then say that the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:
F = (m + m')g
ρV'g = (m + m')g
ρV' = (m + m')
Thus; the required mass m' is = ρV' - m
m' = ρV' - m
m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)
m' = 4.64 kg
Thus; the an instrument package could they carry a required mass of 4.64 kg
An electricity-generating station needs to deliver 20 MW of power to a city 1.0 km away. A common voltage for commercial power generators is 22 kV, but a step-up transformer is used to boost the voltage to 230 kV before transmission.
(a) If the resistance of the wires is 2.0 ohms and the electricity costs about 10 cents/kWh, estimate what it costs the utility company to send the power to the city for one day?
Answer:
Explanation:
Power to be transferred = 20 x 10⁶ W .
Voltage at which power is transferred V = 230 x 10³ V .
If current in the carrying wire is I
V x I = Power
230 x 10³ x I = 20 x 10⁶
I = 86.9565 A
Power loss in the transmission line
I² R , I is current and R is resistance
= 86.9565² x 2
= 15122.86 W
= 15.123 KW
In one day power loss
= 15.123 x 24 kWH .
= 363 kWH .
Cost = 363 x 10
= 3630 cent
= 36.30 dollar .
If the frequency of a wave is tripled, what happens to the period of the wave?
Answer:
if the frequency of the wave if tripled then period of wave gets tripled
When the wave frequency is tripled, the period of the wave becomes one-third of its original.
Relationship between frequency and period of a waveThe frequency (f) of a wave is inversely proportional to the period (T) of the wave.
Mathematically,
[tex]$f \propto \frac{1}{T}$[/tex]
Thus, when frequency increases, the period decrease and when frequency decreases, the period of the wave increases.
So when frequency of a wave increases by three time, the period of the wave decreases by three times.
Hence, when the frequency is tripled, the period of the wave becomes one-third of its original value.
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What was the revolution in atomic theory produced by the discovery of the electron?
Explanation:
the atom are comprised of particle. this atomic theory also proves that atoms cannot be destroyed nor created and it composed of many particles. it is also said that atoms of the same element can be identical.
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
(a) Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon? Express your answer in terms of G, m, and r.
(b) Since the gravitational force between two bodies decreases with distance, the acceleration a_near experienced by a unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration a_far experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general expression for the quantity a_near - a_far. Express your answer in terms of G, m, r, and re.
Answer:
Explanation:
radius of earth = re
mass of the noon = m
mass of the earth = E
distance between earth and moon = r
acceleration of earth ae
force on earth = GMm / r²
acceleration of the earth
ae = force / mass
= GMm / (r² x M )
= Gm / r²
b ) The point on the earth nearest to moon will be at a distance of r - re
a_near = Gm / ( r - re)²
The point farthest on the earth to moon will be at a distance of r + re
a_ far = Gm / ( r + re )²
Consider a spinning plate is dropped onto a stationary plate (which is itself at rest on a frictionless surface). Both plates have a radius of 30cm and a mass of 1kg. The spinning plate is initially spinning at a rate of 0.7 revolutions per second. Hint: This is like a totally-inelastic collision.
Required:
a. After a sufficiently long time, what is the angular velocity of the initially-spinning plate? What about the initially-stationary plate?
b. Assume that the period of velocity matching happens over a course of 2 seconds. Further, assume that the torque exerted by each plate on the other is constant over time. In that case, what is the magnitude of the acceleration that each plate feels during those two seconds? Hint: Use the rotational impulse-momentum theorem.
Answer:
The final angular velocity is [tex]w_f = 2.1994 rad/sec[/tex]
The angular acceleration is [tex]\alpha = 1.099 \ rad/sec^2[/tex]
Explanation:
From the question we are told that
The radius of each plate is [tex]r = 30 \ cm = \frac{30}{100} = 0.3 \ m[/tex]
The mass of each plate is [tex]m_p = 1 \ kg[/tex]
The angular speed of the spinning plate is [tex]w = 0.7 \ rev \ per \ sec = 0.7 * 2 \pi = 4.3988 \ rad/sec[/tex]
From the law of conservation of momentum
[tex]L_i = L_f[/tex]
Where [tex]L_i[/tex] is the initial angular momentum of the system (The spinning and stationary plate ) which is mathematically represented as
[tex]L_i = I_1 w + 0[/tex]
here [tex]I_1[/tex] is the moment of inertia of the spinning plate which mathematically represented as
[tex]I_1 = \frac{m_pr^2}{2}[/tex]
and the zero signify that the stationary plate do not have an angular momentum as it is at rest at the initial state
[tex]L_f[/tex] is the final angular momentum of the system (The spinning and stationary plate) , which is mathematically represented as
[tex]L_f = (I_1 + I_2 ) w_f[/tex]
Where
[tex]I_2[/tex] is the moment of inertia of the second plate (This was stationary before but now it spinning due to the first pate ) and is equal to [tex]I_1[/tex]
and [tex]w_f[/tex] is the final angular speed
So we have
[tex]I_1 w = (I_1 + I_2)w_f[/tex]
[tex]\frac{m_p r^2}{2} * w = 2 * \frac{m_p r^2}{2} * w_f[/tex]
[tex]w = 2 * w_f[/tex]
substituting values
[tex]4.3988 = 2 * w_f[/tex]
[tex]w_f = \frac{4.3988 }{2}[/tex]
[tex]w_f = 2.1994 rad/sec[/tex]
The the rotational impulse-momentum theorem can be mathematially represented as
[tex]\tau * \Delta t = 0.09891[/tex]
Where [tex]\tau[/tex] is the torque and [tex]\Delta t[/tex] is the change in time
So at [tex]\Delta t = 2 \ sec[/tex]
[tex]\tau = \frac{0.09891}{2}[/tex]
[tex]\tau = 0.0995 \ Nm[/tex]
now the angular acceleation is mathematically represented as
[tex]\alpha = 2 * \frac{\tau}{m_p * r^2 }[/tex]
substittuting values
[tex]\alpha = 2 * \frac{0.0995}{1 * 0.3^2}[/tex]
[tex]\alpha = 1.099 \ rad/sec^2[/tex]
Una furgoneta circula por una carretera a 55km/h. Diez km atrás , un coche circula en el mismo sentido a 85km/h ¿ En cuanto tiempo alcanzará el coche a la furgoneta? ¿A qué distancia se producirá el encuentro?
Answer:
t = 0.33h = 1200s
x = 18.33 km
Explanation:
If the origin of coordinates is at the second car, you can write the following equations for both cars:
car 1:
[tex]x=x_o+v_1t[/tex] (1)
xo = 10 km
v1 = 55km/h
car 2:
[tex]x'=v_2t[/tex] (2)
v2 = 85km/h
For a specific value of time t the positions of both cars are equal, that is, x=x'. You equal equations (1) and (2) and solve for t:
[tex]x=x'\\\\x_o+v_1t=v_2t\\\\(v_2-v_1)t=x_o\\\\t=\frac{x_o}{v_2-v_1}[/tex]
[tex]t=\frac{10km}{85km/h-55km/h}=0.33h*\frac{3600s}{1h}=1200s[/tex]
The position in which both cars coincides is:
[tex]x=(55km/h)(0.33h)=18.33km[/tex]
A lion and a pig participate in a race over a 2.20 km long course. The lion travels at a speed of 18.0 m/s and the pig can do 2.70 m/s. The lion runs for 1.760 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.
(a) How far (in m) is the pig from the finish line when the lion resumes the race? (b) For how long in time (in s) was the lion stationary?
The lion covers a distance of [tex]\left(18.0\frac{\rm m}{\rm s}\right)t[/tex] after [tex]t[/tex] seconds, so it reaches the 1.760 km mark at time
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=1760\,\mathrm m\implies t\approx97.8\,\mathrm s[/tex]
The pig travels a distance of [tex]\left(2.70\frac{\rm m}{\rm s}\right)t[/tex], so that it has moved
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(97.8\,\mathrm s)=264\,\mathrm m[/tex]
in the time it takes for the lion to move 1.760 km.
(a) The lion has 0.44 km left in the race, which would take it
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=440\,\mathrm m\implies t\approx24.4\,\mathrm s[/tex]
to finish.
In order for the lion and pig to cross the finish line at the same time, the lion needs to resume running once the pig has 24.4 s remaining to the finish line; this happens when it is
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(24.4\,\mathrm s)\approx\boxed{65.9\,\mathrm m}[/tex]
away from the end.
(b) The lion is stationary for as long as it takes the pig to cover the distance between [65.9 m away from the finish line] and [264 m from the starting line], or (2.20 km - 65.9 m) - 264 m = 1.87 km, which takes it
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)t=1870\,\mathrm m\implies t=\boxed{935\,\mathrm s}[/tex]
(a) The distance of the pig from the finish line is 428.3 m
(b) The lion was stationary for 4.32 s
The given parameters;
total distance covered by both animals, d = 2.20 km = 2,200 mspeed of the lion, [tex]v_l[/tex] = 18 m/sspeed of the pig, [tex]v_p[/tex] = 2.7 m/sinitial distance traveled by the lion, = 1.760 km = 1,760 mWhen the pig meets the lion, both have covered a total distance of 1,760 m
The remaining distance to be covered = 2,200 m - 1,760 m = 440 m
Let the time both animals finished the remaining distance = tApply relative velocity concept; as the pig is moving, the lion is closing the gap between them until the finish line.
[tex](V_l - V_p)t = 440\\\\(18- 2.7)t = 440\\15.3 t = 440\\\\t = \frac{440}{15.3} \\\\t = 28.76 \ s[/tex]
If the lion had maintained a constant motion without stopping, the time it would have finished the remaining race is calculated as;
[tex]t_l = \frac{440}{18} = 24.44 \ s[/tex]
(b) This show that the lion was stationary for (28.76 - 24.44 )s = 4.32 s
(a) The distance traveled by the pig during the 4.32 s that the lion was stationary is calculated as;
d = 4.32 x 2.7 = 11.7 m
Thus, the distance of the pig from the finish line is (440 - 11.7)m = 428.3 m.
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Explain the force responsible for the formation of our solar system.
Answer:
gravity
Approximately 4.5 billion years ago, gravity pulled a cloud of dust and gas together to form our solar system.
Explanation:
Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction between the snow and the sled, and there is air resistance. Looking at the entire time that they spend on the hill (from when they start from rest at the top of the hill until they reach the bottom moving fast), indicate whether each of the following quantities is positive, negative, or zero, by writing either, -, or 0 for the questions below. We take Calvin, the tiger, and the sled as our system.
a. The work done by the normal force that the snow exerts on the system?
b. The work is done by the frictional force that the snow exerts on the system?
c.The change in the kinetic energy of the system?
d.The change in the gravitational potential energy of the system?
e. The total work is done on the system?
Answer:
a) W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d) the variation of the potential enrgy is negative,
e) total work is positive
Explanation:
Work in physics is defined by the scalar scalar product of force by displacement
W = F. dx
The bold are vectors; this can be written in the form of the mules of the quantities
W = F dx cos θ
where θ is the angle between force and displacement.
a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is
θ = 90 cos 90 = 0
W=0
In conclusion the work is zero
b) The friction force opposes the displacement whereby the angle is
θ = 180 cos 190 = -1
W = - fr d
Work is negative
c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy
m g sin θ L = ΔK
this magnitude is positive since the angle is zero cos 0 = 1
how the system starts from rest ΔK = Kf -K₀= + Kf -0
work is positive and scientific energy variation is positive
d) change in potential energy
The potential energy is is ΔU = Uf -U₀
we fix the reference system in the bases of the plane so Uf = 0
ΔU = -U₀
the variation of the potential enrgy is negative
e) The total work is formed by the work of the weight component, the work of the friction force
W_Total = W_weight - W_roce
as the body moves down
W_Total> 0
Therefore the total work is positive
How much time is needed to push a 5,000 N car 50 meters if you are using a machine with a power of 4,500 W?
Answer:
55.56
Explanation:
5000N * 50 = 250000/4500= 55.55555555 or 55.56
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.220 T/s.
Required:
What is the magnitude of the electric field induced in the ring?
Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
[tex]\int E_{in} dl=-\frac{\Phi_B}{dt}[/tex]
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
[tex]E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)[/tex]
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
[tex]E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}[/tex]
hence, the induced electric field is 2.75*10^-3 N/C
Which forms of energy are this a blender?
Answer:
Electrical energy.
Explanation:
Electrical energy comes into the blender, which turns into magnetic energy in the electric motor. This magnetism repels permanent magnets. The motor spins as well as the blade. So the magnetism is converted to kinetic energy.
When you turn down the volume on the television, you reduce the _______ carried by the sound waves, so you also reduce their ________
Answer:
when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
Explanation:
When you turn down the volume of the television, you are actually reducing the intensity of the sound wave, which is directly proportional to the amplitude of the sound. Amplitude is height of the sound wave.
Therefore, when you turn down the volume on the television, you reduce the intensity carried by the sound waves, so you also reduce their amplitude.
A car is travelling at 16.7m/s. If the car can slow at a rate of 21.5m/s^2, how much time does the driver need in order to stop the red light?
Answer:
About 0.7767 seconds
Explanation:
[tex]\dfrac{16.7m/s}{21.5m/s^2}\approx 0.7767s[/tex]
Hope this helps!
Let’s discuss some particle physics. In 1897, physicist J.J. Thomson conducted a seminalexperiment with a cathode ray tube. In the experiment, a beam of an unknown particle of chargeqandmassmis shot through the tube onto a fluorescent screen, at some speedvand in thex-direction. Uponimpact the beam leaves a bright spot on the screen. Sincevis very fast, effects of gravity are negligible andthe beam forms essentially a straight line normal to the screen.
To eliminate this deflection, an external uniform magnetic field of magnitudeBis turned on as well. What should be the direction and magnitudeBof this field such that ∆z= 0?
Answer:
The direction of the magnetic field should be in the x and y coordinates alone, and their should be no change towards the z coordinate. Also the magnitude of the magnetic field in the z direction would be zero if there is no magnetic field in that coordinate.
The coordinate should be:
B = bi + bj
Which shows evidence of active transport?
A scientist places four identical cells into four different
liquids, each with different concentrations of magnesiuni.
Celll
w
Description of Liquid
Slightly more magnesium than the
cell
The least amount of magnesium
O
O
O
O
cells W and Z
cell W only
cell Y only
cells X and Y
Result
Took in
magnesium
Took in
magnesium
Took in
magnesium
Took in
magnesium
Slightly less magnesium than the
cell
The most amount of magnesium
Answer: D
X and Y
Explanation:
X The least amount of magnesium Took in magnesium
Y Slightly less magnesium than the cell Took in magnesium
Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.
Answer:
Cells X and Y
Explanation:
Active transport occurs when a substance moves across a membrane against its concentration gradient.
Cells W and Z were placed in a liquid containing more magnesium than the cells. Magnesium therefore, diffuses passively down it's concentration gradient into the cells.
However, cells X and Y were placed in a solution containing less magnesium than the cells, yet these cells took in magnesium against this concentration gradient. This, shows that active transport had taken place.
1. What is the potential energy of a 5.0-kg
object located 2.0 m above the ground?
A. 2.5 J
C. 98 J
B. 10 J
D. 196 J
Answer:
C. 98 J
Explanation:
The appropriate formula is ...
PE = mgh . . . . . m is mass; below, m is meters
PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2
PE = 98 J
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.
What value of g will you report back to headquarters?
Answer:
The value of g is [tex]g =76.2 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the weight is [tex]m = 1.30 kg[/tex]
The spring constant [tex]k = 1.73 g/m = 1.73 *10^{-3} \ kg/m[/tex]
The second harmonic frequency is [tex]f = 100 \ Hz[/tex]
The number of oscillation is [tex]N = 200[/tex]
The time taken is [tex]t = 315 \ s[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
[tex]l = \lambda[/tex]
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is [tex]L[/tex]
So [tex]l = \frac{L}{2}[/tex]
The velocity of the vibrating string is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension on the string which can be mathematically represented as
[tex]T = mg[/tex]
So
[tex]v = \sqrt{\frac{mg}{k} }[/tex]
Then
[tex]f = \frac{v}{\frac{L}{2} }[/tex]
=> [tex]v = \frac{fL }{2}[/tex]
=> [tex]\sqrt{\frac{mg}{k} } = \frac{fL}{2}[/tex]
=> [tex]g = \frac{f^2 L^2 \mu}{4m}[/tex]
substituting values
[tex]g = \frac{(100) * (1.73 *10^{-3} )}{(4 * 1.30)} L^2[/tex]
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
Generally the period of oscillation is mathematically represented as
[tex]T_p = 2 \pi \sqrt{\frac{L}{g} }[/tex]
=> [tex]L = \frac{T^2 g}{4 \pi ^2}[/tex]
The period can be mathematically evaluated as
[tex]T_p = \frac{t}{N}[/tex]
substituting values
[tex]T_p = \frac{315}{200}[/tex]
[tex]T_p = 1.575 \ s[/tex]
Therefore
[tex]L = \frac{1.575^2 * g }{4 \pi ^2}[/tex]
[tex]L = 0.0628 ^2 g[/tex]
so
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
substituting for L
[tex]g = 3.326 ((0.0628) g)^2[/tex]
=> [tex]g = \frac{1}{(3.326)* (0.0628)^2}[/tex]
[tex]g =76.2 m/s^2[/tex]
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) [tex]F_{net} \text {on wire }3=0[/tex]
[tex]\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm[/tex]
position of wire = 50 - 1.2
= 48.8cm
b) [tex]F_{net} \text {on wire }1=0[/tex]
[tex]\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A[/tex]
Direction ⇒ downward
how much external energy is required to bring three identical point charges (20uc) from infinity and place them at the corners of an equilateral triangle with side of 2 meter length:
Answer:
U = 269.4 kJ
Explanation:
The energy required to place the three charges from infinity is given by:
[tex]U=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}[/tex]
In this case, you have that
q1 = q2 = q3 = q = 20uC
r12 = r13 = r23 = r = 2m
k: Coulomb constant = 8.98*10^9 NM^2/C^2
Then, you replace the values of q, r and k in the equation for the energy U:
[tex]U=3k\frac{q^2}{r}\\\\U=3(8.98*10^9Nm^2/C^2)\frac{(20*10^{-6}C)^2}{2m}=269400\ J=269.4\ kJ[/tex]
hence, the required energy is 269.4 kJ
Which factors affect the resistance of a material?
Explanation:
The resistance of a material is given by the formula as follows :
[tex]R=\rho \dfrac{L}{A}[/tex]
Here,
[tex]\rho[/tex] = resistivity of a material
L is the length of the material
A is the area of cross section of the material
So, the factors on which the resistance of a material depends are :
Length and area of cross section
Revlew Millikan's Photoelectric Experiment Robert A. Mlkan (1868 1953). although best known for his "oil-drop experiment," which measured the charge of an electron, also perfomed pioneering research on the photoelectric effect. In experiments on lithium, for example, Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
Part A What is the work function,W, for lithium, as determined from Milikan's results? Express your answer to three significant figures and include appropriate units.
Part B What maximum kinetic energy do you expet illikan found when he used light with a wavelength of 362.4 TIm? Express your answer to three significant figures and include appropriate units Value Units
Answer:
A.) Work function = 2.3 eV
B.) Max. K.E observed = 1.1 eV
Explanation:
A.) Millikan observed a maximum kinetic energy of 0.550 eV when electrons were ejected with 433.9-nm light. When light of 253.5 m was used, he observed a maximum kinetic energy of 2.57 eV.
work function (f) is the minimum energy required to remove an electron from the surface of the material.
hf = Ø + K.E (maximum)
Where
h = Plank constant 6.63 x 10-34 J s
Ø = work function
hc/λ = Ø + K.E (max)
(6.63×10^-34 × 3×10^8)/433.9×10^-9 = Ø + 0.550 × 1.6×10^-19
4.58×10^-19 = Ø + 8.8×10^-20
Ø = 4.58×10^-19 - 8.8×10^-20
Ø = 3.7 × 10^-19 J
Converting Joule to eV
Ø = 3.7 × 10^-19/1.6×10^-19
Ø = 2.3 eV
B.) When light of wavelength 362.4 m is used
The maximum K.E observed = incident light K.E - (the work function).
Incident K.E = hf = hc/λ
Incident K.E =
(6.63×10^-34 × 3×10^8)/362.4
Incident K.E = 5.5 × 10^-28J
Let's convert joule to eV
Incident K.E = 5.5×10^-28/1.6×10^-19
Incident K.E = 3.4 × 10^-9
Max. K.E observed = 3.4 - 2.3
Max. K.E observed = 1.1 eV
What is the relationship Between frequency and sound?
Answer:
The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave.
Explanation:
Hope this helps : )
You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-cm-long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is 14.5 μg . You then hang a 0.400-kg mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse 24.7 ms to travel the length of the wire.
Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
[tex]v = \sqrt{\frac{T}{m} }[/tex] , T is tension in the wire , m is mass per unit length of wire .
40.48 L = [tex]\sqrt{\frac{3.92}{7.25\times10^{-7}} }[/tex]
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s
if the power developed in an electric circuit is doubled the energy used in one second is
Answer:
Energy is doubled.
Explanation:
Power developed in an electric is the rate of change in time of electric energy travelling throughout circuit. The most common units is the amount of energy used in a second. Therefore, if power is double, the energy used in one second is also doubled.
Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index nwater = 1.333). The Joker stands gloating at the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon y1 = 1.49 m above the surface of the water and fires at an angle of θ1 = 27◦ to the horizontal. He hits the Boy Wonder squarely on the letter "R", which is located y2 = 3.77 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (Fear not, Batfans. The "R" is made of laser-reflective material.) Answer in units of m.
Answer:
x_total = 4.29m
Explanation:
To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.
Let's start by looking for the angles that the laser refracts
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333
θ₂ = sin⁻¹ (n₁ sin θ₁/n₂)
θ₂ = sin⁻¹ (1 sin 27 / 1,333)
θ₂ = sin⁻¹ 0.34057
θ₂ = 19.9º
now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water
tan θ₁ = y₁ / x₁
x₁ = y₁ / tan θ₁
x₁ = 1.49 / tan 27
x₁ = 2,924 m
Now let's look for the waterfall in the water as far as Robin
tan θ₂₂ = y₂ / x₂
x₂ = y₂ / tan θ₂
x₂ = 3.77 / tan 19.9
x₂ = 1,364
the distance from the edge of the pool to Robin is
x_total = x₁ + x₂
x_total = 2,924 + 1,364
x_total = 4.29m
A ball thrown downward by initial speed of 3m/s. It hits the ground after 5 second.
How high is throwing point?
You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa , and the depth of the water will be 13.6 m . The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa .
A) Find the net downward force on the tank's flat bottom, of area 2.15 m^2 , exerted by the water and air inside the tank and the air outside the tank.
Answer:
630.93 kN of force.
Explanation:
Pressure inside the tank is 150 kPa
The acceleration due to gravity on Mars g is 3.71 m/s^2.
The depth of water h is 13.6 m.
Pressure due to air outside tank is 93 kPa
The density of water p is 1000 kg/m^3
Pressure of the water on the tank bottom will be equal to pgh
Pressure of water = pgh
= 1000 x 3.71 x 13.6 = 50456 Pa
= 50.456 kPa.
Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.
Pt = (150 + 50.456 + 93) = 293.456 kPa
Force at the bottom of the tank will be pressure times area of tank bottom.
F = Pt x A
F = 293.456 x 2.15 m^2 = 630.93 kN
What does the speed of sound depend on ?
Answer:
As with any wave the speed of sound depends on the medium in which it is propagating.
Explanation:
Which type of power plant uses the movement of air in nature to generate
electricity?
A. Radiant
B. Wind
C. Coal
D. Hydroelectric
Answer:
Wind
Explanation: