A parallel-plate capacitor is constructed of two horizontal 16.8-cm-diameter circular plates. A 1.8 g plastic bead, with a charge of -4.4 nC is suspended between the two plates by the force of the electric field between them.
a. Which plate, the upper or the lower, is positively charged?
b. What is the charge on the positive plate?
Answer:
Explanation:
Given that:
diameter of the circular plates = 16.8 cm
mass of the plastic bead = 1.8 g
charge q = -4.4 nC
From above, the area of the circular plates is:
[tex]Area = \pi r^2[/tex]
[tex]Area = \pi (\dfrac{d}{2})^2[/tex]
[tex]Area = \pi (\dfrac{16.8}{2*100} m)^2 \[/tex]
Area = 0.022 m²
Thus, as the plastic beads glide between the two plates of the capacitor, there exists a weight acting downwards while the weight is balanced by the force of the plates acting upwards.
Hence, this can be achieved only when the upper plate is positively charged.
b)
Recall that
Force (F) = qE
where;
F = mg
mg = qE
[tex]E = \dfrac{mg}{q}[/tex]
[tex]E = \dfrac{1.8*10^{-3}*9.8}{4.4*10^{-9}}[/tex]
E = 4.0 × 10⁶ N/C
From the electric field;
[tex]E = \dfrac{\Big(\dfrac{Q}{A}\Big)}{e_o}[/tex]
[tex]4.0*10^{6} = \dfrac{\Big(\dfrac{Q}{0.022}\Big)}{8.85*10^{-12}}[/tex]
[tex]4.0*10^{6}*8.85*10^{-12} = {\Big(\dfrac{Q}{0.022}\Big)}{}[/tex]
[tex]Q= 4.0*10^{6}*8.85*10^{-12}*0.022[/tex]
Q = 7.788 × 10⁻⁷ C
Q = 779 nC
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
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
At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magnetic B=B k and electric E=−E k fields with the initial velocity v=v0 i. The magnitudes of the fields: B=0.18 T, E=278 V/m, and the initial speed v0=2.1 m/s are given. Find at what time t, the particle's speed would become equal to v(t)=3.78·v0:
Answer:
10.78 s
Explanation:
The force on the charge is computed by using the equation:
[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]
F = ma
∴
[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]
[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]
[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]
At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]
The velocity of the charge after the time t(sec) is expressed by using the formula:
[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]
calculate the force on an object with mass of 50kg and gravity of 10
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is 60 cm2, determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle
a) The mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle is 23.6 cm².
a) The mass flow rate through the nozzle can be calculated with the following equation:
[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]
Where:
[tex]v_{i}[/tex]: is the initial velocity = 20 m/s
[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²
[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³
[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]
Hence, the mass flow rate through the nozzle is 0.27 kg/s.
b) The exit area of the nozzle can be found with the Continuity equation:
[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]
[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]
[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]
Therefore, the exit area of the nozzle is 23.6 cm².
You can find another example of mass flow rate here: https://brainly.com/question/13346498?referrer=searchResults
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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.
We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:
[tex]\dot m_{in} = \dot m_{out}[/tex] (1)
Where:
[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.
[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.
Given that air flows at constant rate, we expand (1) by dimensional analysis:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)
Where:
[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.
[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.
[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.
a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:
[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]
[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]
[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]
The mass flow rate through the nozzle is 0.265 kilograms per second.
b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:
[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]
[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]
[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]
[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]
[tex]A_{out} = 23.202\,cm^{2}[/tex]
The exit area of the nozzle is 23.202 square centimeters.
If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?
a. The magnitude of the change is smaller for the proton.
b. The magnitude of the change is larger for the proton.
c. The signs Of the two changes in potential energy are opposite.
d. They are equal in magnitude.
e. The signs of the two changes in potential energy are the same.
Answer: They are equal in magnitude.
- The signs of the two changes in potential energy are opposite
Explanation:
When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.
Also, it should be noted that the signs of the two changes in potential energy are opposite.
Current is a measure of…
Astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c relative to an observer on Earth. When Jill left Earth, the spaceship was equipped with all kinds of scientific instruments, including a meter stick. Now that Jill is underway, how long does she measure the meter stick to be?
A) 0.280 m
B) 1.00 m
C) 0.960 m
D) 1.28 m
E) 1.04 m
(B) 1.00 m
Explanation:
Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.
When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.
What is length contraction?Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.
When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.
Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.
From the given,
speed of the spaceship = 0.280c (c is the speed of the light)
Length contraction, L = L₀ √(1-v²/c²)
The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter. Thus, the ideal solution is option B.
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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
Using the given temperature and pressures, determine: a) the diameter of the water scale model balloon (m), b) the weight of the scale model, c) the specific gravity of the buoyant material such that the model conditions will be similar to the full-scale balloon.
Complete Question
Complete Question is attached below
Answer:
a) [tex]D=0.7[/tex]
b) [tex]W=1787.5N[/tex]
c) [tex]\rho'=998.19kg/m^3[/tex]
Explanation:
From the question we are told that:
Hot air:
Temperature [tex]T_a=360K[/tex]
Pressure [tex]P_a=100kPa[/tex]
Distance [tex]d=12m[/tex]
Weight [tex]W=1400N[/tex]
Water:
Temperature [tex]T_w=300K[/tex]
Pressure [tex]P_w=100kPa[/tex]
Since we have The same Reynolds number
a)
Generally the equation for equal Reynolds number is mathematically given by
Re_{air}=Re_{water}
Therefore
[tex]\frac{\rho V D}{\mu_{air}}=\frac{p_{water}*V*D}{\mu_{water]}}[/tex]
[tex]\frac{100*12}{300*0.28*1.81*10^{-3}}}=\frac{998*D}{0.000890}[/tex]
[tex]D=0.7[/tex]
b)
Generally the equation for Weight of scale is mathematically given by
[tex]W=\rho*V*g[/tex]
[tex]W=998*\frac{4}{3}*\pi*0.35^3*9.81[/tex]
[tex]W=1787.5N[/tex]
c)
Generally the equation for Density of buoyant material is mathematically given by
[tex]\rho'=\frac{w}{g*V}[/tex]
[tex]\rho'=\frac{1781.5}{\frac{4}{3}*\pi*0.35^3*9.81}[/tex]
[tex]\rho'=998.19kg/m^3[/tex]
9. From this lab, we learn that the electric field and electric potential depend on both, the magnitude of the source charge (q), and the distance from the source charge (r). If we were to increase the magnitude of our source charge from 1 nC to 5 nC, then the magnitudes of the electric field and electric potential would be ____.(you can test this on the animation by dragging five 1 nC charges on top of each other and measuring E and V at a distance of 1 m)
Answer:
the electric field and the electric potential increase 5 times
Explanation:
The electric field created by a point charge is
E = k q / r²
in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C
with the electric field is proportional to the charge
E₅ = 5 E₁
the electrical power for a point charge is
V = k q / r
as the electric power is proportional to the charge
V₅ = 5 V₁
consequently both the electric field and the electric potential increase 5 times
An electric field is produced by a charged object:
[tex]\to E = \frac{k q}{r^2}\\\\[/tex]
In this situation, the charge shifts:
[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]
so with electric field remaining proportional to the charge
[tex]\to E_5 = 5 E_1[/tex]
The electrical power consumed for a point charge:
[tex]\to V = \frac{k q}{ r}[/tex]
since the electric power is related to the charge
[tex]\to V_5 = 5 V_1[/tex]
As a result, both the electric field as well as the electric potential increase by 5 times.
Learn more:
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A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?
a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.
Answer:
the correct one is d
Explanation:
The filament of the bulb fulfills the law of ohm
V = i R
indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,
When examining the different statements, the correct one is d
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball
Answer:
the work done on the ball is 0
Explanation:
Given the data in the question;
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.
circle circumference = 2πr = 2.1 m
radius r will be; r = 2.1 m / 2π = 0.33 m
Tension force = 34.4 N
one half revolution means, displacement of the ball is;
d = 2r = 2 × 0.33 = 0.66 m
Now, Work done = force × displacement × cosθ
we know that, the angle between the tension force on string and displacement of object is always 90.
so we substitute
Work done = 34.4 N × 0.66 m × cos(90)
Work done = 34.4 N × 0.66 m × 0
Work done = 0 J
Therefore, the work done on the ball is 0
The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas.
a. True
b. False
Answer:
true
Explanation:
The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance
I hope this helps
A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly
Answer:
EMF = 51.01 Volt
Explanation:
A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ∘ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s
We have,
Number of turn in the coil, N = 50
The diameter of the coil, d = 15 cm
Radius, r = 7.5 cm = 0.075 m
Initial magnetic field, [tex]B_i=0.5\ T[/tex]
The plane of the coil makes an angle of 30° with the magnetic field.
The magnetic field reduced to zero in 0.1 seconds
We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,
[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]
So, the induced emf in the coil is 51.01 V.
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?
Answer:
1.5kg
Explanation:
Given data
mass m1= 2.5kg
mass m2=??
velocity of mass one v1= 1.5m/s
velocity of mass two v2= 4.1m/s
common velocity after impact v= 2.5m/s
Let us apply the formula for the conservation of linear momentum for inelastic collision
The expression is given as
m1v1+ m2v2= v(m1+m2)
substitute
2.5*1.5+ m2*4.1= 2.5(2.5+m2)
3.75+4.1m2= 6.25+2.5m2
collect like terms
3.75-6.25= 2.5m2-4.1m2
-2.5= -1.6m2
divide both sides by -1.6
m2= -2.5/-1.6
m2= 1.5 kg
Hence the second mass is 1.5kg
Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3, is stirred until its temperature is 500 K. Assuming the ideal gas model, for the air, and ignoring kinetic and potential energy, determine
Answer:
The final pressure in bar will be "[tex]\frac{10}{3} \ Bar[/tex]".
Explanation:
As we know,
PV = nRT
[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} =CONST[/tex]
then,
⇒ [tex]\frac{2 \ bar}{300 \ K} = \frac{P_2}{500 \ K}[/tex]
⇒ [tex]P_2=(\frac{2}{300}\times 500 )Bar[/tex]
[tex]=\frac{10}{3} \ Bar[/tex]
Thus the above is the correct answer.
A man standing in an elevator holds a spring scale with a load of 5 kg suspended from it. What would be the reading of the scale, if the elevator is accelerating downward with an acceleration 3.8 m/s?.
Answer:
3.1 kg
Explanation:
Applying,
R = m(g-a)..................... Equation 1
Where R = weight of the scale when the elevator is coming down, a = acceleration of the elevator, g = acceleration due to gravith.
From the question,
Given: m = 5 kg, a = 3.8 m/s²
Constant: g = 9.8 m/s²
Substitute these values into equation 1
R = 5(9.8-3.8)
R = 5(6)
R = 30 N
Hence the spring scale is
m' = R/g
m' = 30/9.8
m' = 3.1 kg
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
CHEGG Over the course of a multi-stage 4820-km bicycle race, the front wheel of an athlete's bicycle makes 2.40x106 revolutions. How many revolutions would the wheel have made during the race if its radius had been 1.4 cm larger?
Answer:
θ' = 14.44 × [tex]10^{6}[/tex]
Explanation:
given data
total distance is d = 4820
radius = 1.4 cm
solution
we get here total angle by which the wheel rotates traveling is express as
⇒ [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex] ................1
and
total angle (θ) and the total distance (d) express as
⇒ d = r × θ ...............2
here r is radius
and here rotated through some other angle θ' so put value in given equation and find revolutions
⇒ d = (r+r)θ' ........3
here r = d/θ
so
⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]
so put value and get θ'
⇒ θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]
⇒ θ' = 14.44 × [tex]10^{6}[/tex] revany four difference between speed and acceleration
Speed:
Measured in distance / timeIt tells the speed of a bodyIt is a scalar quantitySpeed is always positiveIt's unit is meter per secondAcceleration:
Measured in velocity / timeIt tells the change in the velocityImplied on change in a velocityIt is a vector quantityIt's unit is meter per second square
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
A nearsighted person has a near point of 50 cmcm and a far point of 100 cmcm. Part A What power lens is necessary to correct this person's vision to allow her to see distant objects
Answer:
P = -1 D
Explanation:
For this exercise we must use the equation of the constructor
/ f = 1 / p + 1 / q
where f is the focal length, p and q is the distance to the object and the image, respectively
The far view point is at p =∞ and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image
[tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]
f = 1 m
P = 1/f
P = -1 D
a sperical ballon with a diameter of 6 m filled with helium at 20 degree centigrade and 200kpa determine mole number and the mass of helium
Answer:
A. 9280.78 moles.
B. 37123.12 g.
Explanation:
We'll begin by calculating the volume of the spherical balloon. This can be obtained as follow:
Diameter (d) = 6 m
Radius (r) = d/2 = 6/2 = 3 m
Pi (π) =3.14
Volume (V) =?
V = 4/3πr³
V = 4/3 × 3.14 × 3³
V = 4/3 × 3.14 × 27
V = 113.04 m³
Next, we shall convert 20°C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
T(°C) = 20°C
T(K) = 20°C + 273
T(K) = 293 K
Next, we shall convert 200 KPa to Pa. This can be obtained as follow:
1 KPa = 1000 Pa
Therefore,
200 KPa = 200 KPa × 1000 Pa / 1 KPa
200 KPa = 2×10⁵ Pa
A. Determination of the number of mole of He in the spherical balloon.
Volume (V) = 113.04 m³
Temperature (T) = 293 K
Pressure (P) = 2×10⁵ Pa
Gas constant (R) = 8.314 m³Pa/Kmol
Number of mole (n) =?
PV = nRT
2×10⁵ × 113.04 = n × 8.314 × 293
22608000 = n × 2436.002
Divide both side by 2436.002
n = 22608000 / 2436.002
n = 9280.78 moles
B. Determination of the mass of He.
Mole of He (n) = 9280.78 moles
Molar mass of He = 4 g/mol
Mass of He =?
Mass = mole × molar mass
Mass of He = 9280.78 × 4
Mass of He = 37123.12 g
A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels
Answer: [tex]53\ rev/min[/tex]
Explanation:
Given
angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]
Another wheel of 9 times the rotational inertia is coupled with initial wheel
Suppose the initial wheel has moment of inertia as I
Coupled disc has [tex]9I[/tex] as rotational inertia
Conserving angular momentum,
[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
Two harmonic sound waves reach an overseveer simulatenouslt. the obsever hears the sound intensity rise and fall with a time of 0.2 between the maximmu intensity and the successing minimum intensity. What is the difference in frequency of the two sound waves?
Answer:
[tex]dF=2.5Hz[/tex]
Explanation:
From the question we are told that:
Time [tex]T=0.2sec[/tex]
Generally the Period is given as
[tex]T= 2 * 0.2 = 0.4[/tex]
Therefore difference in frequency dF
[tex]dF=\frac{1}{T}[/tex]
[tex]dF=\frac{1}{0.4}[/tex]
[tex]dF=2.5Hz[/tex]
