Answer:
Explanation:
total weight acting downwards
= 3g + 10g
13 g
volume of lead = 10 / 11.3 = .885 cm³
Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g
Buoyant force on lead = .885 x 1 x g
total buoyant force = vg + .885 g
For floating
vg + .885 g = 13 g
v = 12.115 cm³
total volume of bobber
= 4/3 x 3.14 x 2³
= 33.5 cm³
fraction of volume submerged
= 12.115 / 33.5
= .36
= 36 %
The fraction of the bobber's volume submerged as a percent of the total volume is 36.2 %.
The given parameters;
diameter of the bobber, d = 4 cmmass of the bobber, m = 3 gmass of the lead, m = 10 gdensity of the lead, ρ = 11.3 g/cm³The volume of the bobber is calculated as follows;
[tex]V = \frac{4}{3} \pi \times r^3\\\\V = \frac{4}{3} \pi \times (2)^3\\\\V = 33.52 \ cm^3[/tex]
The buoyant force experienced by the bobber due to the volume submerged is calculated as follows;
[tex]F _b= \rho Vg\\\\F_b = 1 \times V \times g\\\\F_b = Vg[/tex]
The volume of the lead is calculated as follows;
[tex]V = \frac{mass}{density} \\\\V = \frac{10}{11.3} \\\\V = 0.885 \ cm^3[/tex]
The buoyant force experienced by the lead due to the volume submerged is calculated as follows
[tex]F_b = \rho Vg\\\\F_b = 0.885 g[/tex]
The total buoyant force is calculated as;
[tex]Vg + 0.885g = (3+ 10)g\\\\g(V + 0.885) = 13g\\\\V+ 0.885 = 13\\\\V = 13 -0.885\\\\V = 12.12 \ cm^3[/tex]
The fraction of the bobber's volume submerged as a percent of the total volume is calculated as follows;
[tex]= \frac{12.12}{33.52} \times 100\%\\\\= 36.2 \ \%[/tex]
Learn more here:https://brainly.com/question/17009786
Now consider a different electromagnetic wave, also described by: Ex(z,t) = Eocos(kz - ω t + φ) In this equation, k = 2π/λ is the wavenumber and ω = 2π f is the angular frequency. In this case, though, assume φ = +30o and Eo = 1 kV/m. What is the value of Ex(z,t) when z/λ = 0.25 and ft = 0.125?
Answer:
Explanation:
Ex(z,t) = Eocos(kz - ω t + φ)
k = 2π/λ , ω = 2π f
φ = +30° , E₀ = 10³ V .
z/λ = 0.25 , ft = 0.125
Ex(z,t) = Eocos(2πz/λ - 2πf t + φ)
Putting the values given above
Ex(z,t) = 10³ cos ( 2π / 4 - 2π x .125 + 30⁰ )
= 1000cos (90⁰ - 45+30)
= 1000 cos 75
=258.8 V .
A gas in a closed container is heated with 12J of energy, causing the lid of the container to rise 3m with 5N of force. What is the total change in energy?
Answer:
27J
Explanation:
From conservation of Thermal energy, the total internal energy is the total sum of energy supplied or taken from the system plus work done for or on the system.
Now the change in internal energy would be the sum of the received energy substended in the gas plus the work done by the system which is workdone that it will sustend in pushing the lid. This is expressed mathematically as;
U = Q + (F×d);
U- change in internal energy
Q is the energy received by the system and is positive when energy is received by the system.
Fxd is the workdone and is positive since the gas pushes up the lid- the system does work.
U=12+(3×5)= 27J
When Marcel finds the distance L from the previous part, it turns out to be greater than Lend, the distance from the pivot to the end of the seesaw. Hence, even with Jacques at the very end of the seesaw, the twins Gilles and Jean exert more torque than Jacques does. Marcel now elects to balance the seesaw by pushing sideways on an ornament (shown in red) that is at height h above the pivot. (Figure 3)With what force in the rightward direction, Fx, should Marcel push? If your expression would give a negative result (using actual values) that just means the force should be toward the left.Express your answer in terms of W, Lend, w, L2, L3, and h.
Answer:
Fx = - (1/h)( wL2 + wL3 - wLend )
Explanation:
Assuming The twins Gilles and Jean has a weight ( w ) each
The torque that would balance the equation would be = wL2 + wL3 -------- 1
THEREFORE the ccw torques are = wLend + Fh ----------- 2
hence equation 2 equals equation 1
= wLend + Fh = wL2 + wL3 --------- 3
equation 3 can as well be represented as
F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4
From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative
therefore equation 4 is represented as
Fx = - (1/h)( wL2 + wL3 - wLend )
Jason takes off from rest across level water on his jet-powered skis. The combined mass of Jason and his skis is 75 kg (the mass of the fuel is negligible). The skis have a thrust of 200 N and a coefficient of kinetic friction on water of 0.10. Unfortunately, the skis run out of fuel after only 75 s. What is Jason's top speed?
Answer:
v = 126 m / s
Explanation:
Let's analyze this exercise a little, they give us the thrust that is the applied force and the time that it lasts, and they ask us for the final speed, so we can use the Impulse ratio and the variation of the amount of movement
I = F t = Dp
F t = pf -p₀
Now let's use Newton's second law to find the net thrust
F = E - fr
the friction force has the formula
fr = μ N
let's write Newton's second law on the y-axis
N-W = 0
N = W
we substitute
fr = μ mg
we look for the net out
F = 200 - μ mg
With the skater starting from rest, the initial speed is zero (vo = 0)
we substitute
(200 - very m g) t = m v
v = (200 µm - very g) t
let's calculate
v = (200/75 - 0.10 9.8) 75
v = 126 m / s
A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No
Answer:
0.19rad/s and Yes
Explanation:
From the principle of conservation of momentum it means momentum before and after collision is the same.
Momentum before collision is 0.700 kg×12 = 8.4Ns
Momentum of the door = mass of door × velocity of door
8.4Ns = mass of door × velocity of door
Velocity of door = 8.4Ns/45 =0.19m/s
But velocity V= w×r ;
w-angular velocity
r- raduis = width
w= 0.19/1m = 0.19rad/s
2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.
Explain why it is necessary to have a high voltage
Answer:
SO THAT
EACH APPLIANCE CAN GET SUFFICIENT POTENTIAL DIFF. TO RUNWhen a fuel is burned in a cylinder fitted with a piston, the volume expands from an initial value of 0.250 L against an external pressure of 2.00 atm. The expansion does 288 J of work on the surroundings. What is the final volume of the cylinder
Answer:
Vf = 0.0017 m³ = 1.7 L
Explanation:
The work done by the system on the surrounding at constant pressure is given by the following formula:
W = PΔV
W = P(Vf - Vi)
where,
W = Work done = 288 J
P = Constant Pressure = (2 atm)(101325 Pa/atm) = 202650 Pa
Vf = Final Volume f Cylinder = ?
Vi = Initial Volume of Cylinder = (0.25 L)(0.001 m³/ 1 L) = 0.00025 m³
Therefore,
288 J = (202650 Pa)(Vf - 0.00025 m³)
Vf = 288 J/202650 Pa + 0.00025 m³
Vf = 0.0017 m³ = 1.7 L
Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI
Answer:First option
Explanation:
hope it helped
Two vectors having magnitudes of 5.00 and 9.00 respectively. If the value of their dot product is 12.0, find the angle between the two vectors.
Answer:
C = 74.53°
Explanation:
Let the magnitudes of 5.00 and 9.00 be vectors A and B respectively, hence the dot product of this vector is defined as
A.B = |A||B|cosC; let C be the angle between the vectors
12 = 5×9 cos C
Hence cos C = 12/45
C = cos^-1(12/45)
C = 74.53°
Astronaut Flo wishes to travel to a star 20 light years away and return. Her husband Malcolm, who was the same age as Flo when she departs, stays home (baking cookies). If Flo travels at a constand speed of 80% of the speed of light (except for a short time to turn around), how much younger than Malcolm will Flo be when she returns? How long does Malcolm sit around baking cookies? How far is the distance to Flo?
Answer:
a. about 20 years younger
b. Malcolm sits around for 49.94 years
c. 2.268x[tex]10^{17}[/tex] m
Explanation:
light travels 3x[tex]10^{8}[/tex] m in one seconds
in 20 years that will be 3x[tex]10^{8}[/tex] x 20 x 60 x 60 x 24 x 365 = 1.89x[tex]10^{17}[/tex] m
for the to and fro journey, total distance covered will be 2 x 1.89x[tex]10^{17}[/tex] = 3.78x[tex]10^{17}[/tex] m
Flo's speed = 80% of speed of light = 0.8 x 3x[tex]10^{8}[/tex] = 2.4x[tex]10^{8}[/tex] m/s
time that will pass for Malcolm will be distance/speed = 3.78x[tex]10^{17}[/tex] /2.4x[tex]10^{8}[/tex]
= 1575000000 s = 49.94 years
the relativistic time t' will be
t' = t x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
t' = 49.94 x [tex]\sqrt{1 - 0.8^{2} }[/tex]
t' = 49.94 x 0.6 = 29.96 years this is the time that has passed for Flo
this means that Flo will be about 20 years younger than Malcolm when she returns
relativistic distance is
d' = d x [tex]\sqrt{1 - \frac{v^{2} }{c^{2} } }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x [tex]\sqrt{1 - 0.8^{2} }[/tex]
d' = 3.78x[tex]10^{17}[/tex] x 0.6
d' = 2.268x[tex]10^{17}[/tex] m this is how far it is to Flo
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB
Complete question is:
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 m/s. Neglect friction.
Answer:
(V_A) = 31.32 m/s
Explanation:
We are given;
car's mass, m = 1200 kg
h_A = 100 m
h_B = 150 m
v_B = 0 m/s
From law of conservation of energy,
the distance from point A to B is;
h = 150m - 100 m = 50 m
From Newton's equations of motion;
v² = u² + 2gh
Thus;
(V_B)² = (V_A)² + (-2gh)
(negative next to g because it's going against gravity)
Thus;
(V_B)² = (V_A)² - (2gh)
Plugging in the relevant values;
0² = (V_A)² - 2(9.81 × 50)
(V_A) = √981
(V_A) = 31.32 m/s
If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina
Answer:
The distance from the cornea vertex to the retina is 2.36 cm
Explanation:
The question is incomplete.
The complete question is as follows;
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
Solution.
We use image-object reaction to calculate the distance from the cornea vertex to the retina.
Mathematically;
n1/s + n2/s’ = n2-n1/R
From the question, we identify the following;
n1 ; Refractive index of air = 1
n2 ; Refractive index of lens = 1.4
S ; Object Distance = 36 cm
S’ = ?
R ; Radius of curvature of the cornea = 0.65
Substituting these values into the equation above;
1/36 + 1.4/S’ = (1.4-1)/0.65
{S’+ 36(1.4)}/36S’ = 0.4/0.65
{S’ + 50.4}/36S’ = 0.62
S’ + 50.4 = 22.32S’
50.4 = 22.32S’ -S’
21.32S’ = 50.4
S’ = 50.4/21.32
S’ = 2.36 cm
Inside a stereo speaker, you will find two permanent magnets: one on the cone and one near the cone. True of false?
Answer:
false
Explanation:
The equation for distance is d= st. if a car has a speed of 20 m/s how long will it take to go 155m
Answer:
It will take 7.75 seconds for the car to go 155m
Explanation:
From the question, we can understand that the distance covered by the moving car is got by a product of its speed and the time it travels.
i.e distance = speed X time.
However, in this case, we have the distance travelled and the speed of the car, and we are looking for the time of travel
TO solve this, we will simply make the travel time the subject of the formula in the equation above.
i.e time = distance / speed
time = 155/20= 7.75 seconds.
Hence, it will take 7.75 seconds for the car to go 155m
A 330-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,110 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Answer:
t = 402 years
Explanation:
To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:
[tex]I=nqv_dA[/tex] (1)
n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3
q: charge of the electron = 1.6*10^-19 C
vd: drift velocity of electron in the metal = ?
A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2
I: current in the wire = 1110 A
You solve the equation (1) for vd:
[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]
Next, you calculate the time by using the information about the length of the line transmission:
[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]
hence, the electrons will take aproximately 402 years in crossing the line of transmission
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
its mass density is 5.0 g/m2, determine the tension of the wire.
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g[tex]/m^{2}[/tex] = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f = [tex]\frac{1}{2L}[/tex][tex]\sqrt{\frac{T}{M} }[/tex]
⇒ T = 4[tex]L^{2}[/tex][tex]f^{2}[/tex]M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × [tex]2^{2}[/tex] ×[tex]65^{2}[/tex] × 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven-the driving force is transferred to the object, which oscillates instead of the entire building X 50%
Part (a) What effective force constant, in N/m, should the springs have to make them oscillate with a period of 1.2 s? k = 9.5 * 106 9500000 X Attempts Remain 50%
Part (b) What energy, in joules, is stored in the springs for a 1.6 m displacement from equilibrium?
Answer:
The force constant is [tex]k =1.316 *10^{7} \ N/m[/tex]
The energy stored in the spring is [tex]E = 1.68 *10^{7} \ J[/tex]
Explanation:
From the question we are told that
The mass of the object is [tex]M = 4.8*10^{5} \ kg[/tex]
The period is [tex]T = 1.2 \ s[/tex]
The period of the spring oscillation is mathematically represented as
[tex]T =2 \pi \sqrt{ \frac{M}{k}}[/tex]
where k is the force constant
So making k the subject
[tex]k = \frac{4 \pi ^2 M }{T^2}[/tex]
substituting values
[tex]k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}[/tex]
[tex]k =1.316 *10^{7} \ N/m[/tex]
The energy stored in the spring is mathematically represented as
[tex]E = \frac{1}{2} k x^2[/tex]
Where x is the spring displacement which is given as
[tex]x = 1.6 \ m[/tex]
substituting values
[tex]E = \frac{1}{2} (1.316 *10^{7}) (1.6)^2[/tex]
[tex]E = 1.68 *10^{7} \ J[/tex]
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction pressure is developed on the wing at that point? (b) What is the pressure at the leading edge (a stagnation point) of the wing?
Answer:
P1 = 0 gage
P2 = 87.9 lb/ft³
Explanation:
Given data
Airplane flying = 200 mph = 293.33 ft/s
altitude height = 5000-ft
air velocity relative to the airplane = 273 mph = 400.4 ft/s
Solution
we know density at height 5000-ft is 2.04 × [tex]10^{-3}[/tex] slug/ft³
so here P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
and here
P1 = 0 gage
because P1 = atmospheric pressure
and so here put here value and we get
P1 + [tex]\frac{\rho v1^2}{2}[/tex] = P2 + [tex]\frac{\rho v2^2}{2}[/tex]
0 + [tex]\frac{2.048 \times 10^{-3} \times 293.33^2}{2}[/tex] [tex]= P2 + \frac{2.048 \times 10^{-3} \times 400.4^2}{2}[/tex]
solve it we get
P2 = 87.9 lb/ft³
Two very large parallel sheets a distance d apart have their centers directly opposite each other. The sheets carry equal but opposite uniform surface charge densities. A point charge that is placed near the middle of the sheets a distance d/2 from each of them feels an electrical force F due to the sheets. If this charge is now moved closer to one of the sheets so that it is a distance d/4 from that sheet, what force will feel
Answer:
the force we will feel is F
Explanation:
According to the Gauss law, electric field due to very large sheet of charge is as follows.
[tex]E = \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
where,
[tex]\sigma[/tex] = charge per unit area
Since, it is given that there are two sheets of equal and opposite charge. Therefore, electric field between the plates will be as follows. [tex]E = \frac{\sigma}{2 \times \epsilon_{o}} + \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
Also, we know that relation between force and electric field is as follows.
F = qE
Hence, force felt by the charge present inside the plates will be as follows.
[tex]F = q \times \frac{\sigma}{2 \times \epsilon_{o}}[/tex]
This depicts that force is not dependent on the distance and the charge is kept from one of the plate. Therefore, force F felt by the charge is same when it is placed at a distance d/2 and at a distance d/4 from one of the plate.
HOW CAN I SOLVE THIS QUESTION? PLEASE HELP The movement of a locomotive piston in the cylinder is limited to 0.76 m. Assume that the piston makes a simple harmonic movement that makes 180 revolutions per minute, and find its maximum speed.
Answer:
7.2 m/s
Explanation:
The maximum speed is the amplitude times the frequency.
v = Aω
v = (0.76 m / 2) (180 rev × 2π rad/rev / 60 s)
v = 7.2 m/s
What is the resistance of a circuit with a voltage of 10 V in a current of 5 A use almond law to create the resistance
Answer:
2Ω
Explanation:
Ohm's law:
V = IR
10 V = (5 A) R
R = 2 Ω
Find the equivalent resistance from the indicated terminal pair of the networks in the attached doc
Answer:
a) R = 2.5 Ω, b) R = 1 Ω, c) R = 2R / 3 Ω
Explanation:
The resistance configuration can be in series or in parallel, for each one the equivalent resistance can be calculated
series, the equivalent resistance is the sum of the resistances
parallel, the inverse of the equivalent resistance is the inverse of the sum of the resistances
let's apply these principles to each case
case a)
equivalent series resistance
R₁ = 1 +4 = 5 ohm
R₂ = 2 +3 = 5 ohn
these two are in parallel
1 / R = 1/5 +1/5
1 / R = 2/5
R = 2.5 Ω
case B
we solve the parallel
1 / R₁ = ½ + ½ = 1
R₁ = 1 Ω
we solve the resistors in series
R₂ = 1 + 1
R₂ = 2 Ω
finally we solve the last parallel
1 / R = ½ +1/2 = 1
R = 1 Ω
case C
we solve house resistance pair in series
R₁ = R + 2R = 3R
we go to the next mesh
R₂ = R + 2R = 3R
R₃ = R + 2R = 3R
last mesh
R₄ = R + R = 2R
now we solve the parallel of this equivalent resistance
1 / R = 1 / R₁ + 1 / R₂ + 1 / R₃ + 1 / R₄
1 / R = 1 / 3R + 1 / 3R + 1 / 3R + 1 / 2R
1 / R = 3 / 3R + 1 / 2R = 1 / R + 1 / 2R
1 / R = 3 / 2R
R = 2R / 3 Ω
A camera takes a picture that is the correct brightness and the correct zoom level, but the depth-of-focus is too small. One way to increase the depth-of-focus is to increase the f-number. Assuming that we will make changes that have the overall effect to:
1. increase the f-number, and
2. keep the brightness and the zoom level the same, which changes should we make to the aperture diameter and to the shutter time? (keep in mind we're talking about the time the shutter is open; we aren't talking about the shutter speed)
a. Increase the aperture diameter, decrease the shutter time
b. Decrease the aperture diameter, increase the shutter time
c. Increase both the aperture diameter as well as the shutter time
d. Decrease both the aperture diameter as well as the shutter time
5.
The solar system coalesced due to rotational forces and
gravity.
heat.
radioactivity.
solar wind.
Answer:
Gravity
Explanation:
The solar system is held together by rotational forces and gravity. This can be seen from billions of years ago when the solar system was a cloud of dust and gas. This cloud of dust and gas is known as the solar nebula. All of these dust and gas were brought together by the rotational movement as well as the action of gravity which brought all the particles together to form a larger one. This alone brought about the sun's formation in the center of the nebula as well the formation of other planetary bodies, etc.
Cheers.
If two twins (54 kg each) were 0.02 m apart, what is the force of gravity between them?
Answer:
Force, [tex]F=4.86\times 10^{-4}\ N[/tex]
Explanation:
We have,
Masses of two twins are 54 kg each
They are placed at a distance of 0.02 m
It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
plugging all the known values:
[tex]F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N[/tex]
So, the force of gravity between them is [tex]4.86\times 10^{-4}\ N[/tex].
An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time? Group of answer choices
Answer:
Dear Kaleb
Answer to your query is provided below
Acceleration of the vehicle is 12m/s^2
Explanation:
Explanation for the same is attached in image
Four long wires are each carrying 6.0 A. The wires are located
at the 4 corners of a square with side length 9.0 cm. All of
these wires are carrying current out of the page. The
magnetic field (in T) at one corner of the square is:
Answer:
[tex]B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
Explanation:
To find the magnitude of the magnetic field, you use the following formula for the calculation of the magnetic field generated by a current in a wire:
[tex]B=\frac{\mu_oI}{2\pi r}[/tex]
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
I: current = 6.0 A
r: distance to the wire in which magnetic field is measured
In this case, you have four wires at corners of a square of length 9.0cm = 0.09m
You calculate the magnetic field in one corner. Then, you have to sum the contribution of all magnetic field generated by the other three wires, in the other corners. Furthermore, you have to take into account the direction of such magnetic fields. The direction of the magnetic field is given by the right-hand side rule.
If you assume that the magnetic field is measured in the up-right corner of the square, the wire to the left generates a magnetic field (in the corner in which you measure B) with direction upward (+ j), the wire down (down-right) generates a magnetic field with direction to the left (- i) and the third wire generates a magnetic field with a direction that is 45° over the horizontal in the left direction (you can notice that in the image attached below). The total magnetic field will be:
[tex]B_T=B_1+B_2+B_3\\\\B_{T}=\frac{\mu_o I_1}{2\pi r_1}\hat{j}-\frac{\mu_o I_2}{2\pi r_2}\hat{i}+\frac{\mu_o I_3}{2\pi r_3}[-cos45\hat{i}+sin45\hat{j}][/tex]
I1 = I2 = I3 = 6.0A
r1 = 0.09m
r2 = 0.09m
[tex]r_3=\sqrt{(0.09)^2+(0.09)^2}m=0.127m[/tex]
Then you have:
[tex]B_T=\frac{\mu_o I}{2\pi}[(-\frac{1}{r_2}-\frac{cos45}{r_3})\hat{i}+(\frac{1}{r_1}+\frac{sin45}{r_3})\hat{j}}]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[(-\frac{1}{0.09m}-\frac{cos45}{0.127m})\hat{i}+(\frac{1}{0.09m}+\frac{sin45}{0.127m})]\\\\B_T=\frac{(4\pi*10^{-7}T/A)(6.0A)}{2\pi}[-16.67\hat{i}+16.67\hat{j}]\\\\B_T=2.0*10^-5[-\hat{i}+\hat{j}]T[/tex]
Penny is adjusting the position of a stand up piano of mass mp = 150 kg in her living room. The piano is lp = 1.35 m in length. The piano is currently at an angle of θp = 36 degrees to the wall. Penny wants to rotate the piano across the carpeted floor so that it is flat up against the wall. To move the piano, Penny pushes on it at the point furthest from the wall. This piano does not have wheels, so you can assume that the friction between the piano and the rug acts at the center of mass of the piano.
Required:
a. Write an expression for the minimum magnitude of the force FS in N Penny needs to exert on the piano to get it moving. Assume the corner of the piano on the wall doesn't slide and the static friction between the rug and the piano is µs.
b. The coefficient of kinetic friction between the carpet and the piano is uk = 0.27. Once the piano starts moving, calculate the torque τ in N·m that Penny needs to apply to keep moving the piano at a constant angular velocity.
c. Calculate the amount of work Wp, in J Penny does on the piano as she rotates it.
Answer:
a) Fs = (μs*mp*g)/2 | b) τ = Fs*lp | c) Wτ,constant = τΘ
Explanation:
a) Fs = (μs*mp*g)/2
b) τ = Fs*lp
c) Wτ,constant = τΘ
A person sits on a freely spinning lab stool that has no friction in its axle. When this person extends her arms, A. her moment of inertia decreases and her angular speed decreases B. her moment of inertia decreases and her angular speed increases C. her moment of inertia increases and her angular speed decreases D. her moment of inertia increases and her angular speed decreases E. her moment of inertia increases and her angular speed remains the same.
Answer:
C. her moment of inertia increases and her angular speed decreases
D. her moment of inertia increases and her angular speed decreases
Explanation:
The moment of inertia of a body is the sum of the products of an increment of mass and the square of its distance from the center of rotation. When a spinning person extends her arms, part of her mass increases its distance from the center of rotation, so increases the moment of inertia.
The kinetic energy of a spinning body is jointly proportional to the moment of inertia and the square of the angular speed. Hence an increase in moment of inertia will result in a decrease in angular speed unless there is a change in the rotational kinetic energy.
This effect is used by figure skaters to increase their spin rate by drawing their arms and legs closer to the axis of rotation. Similarly, they can slow the spin by extending arms and legs.
When the person extends her arms, her moment of inertia increases and her angular speed decreases.
_____
Note to those looking for a letter answer
Both choices C and D have identical (correct) wording the way the problem is presented here. You will need to check carefully the wording in any problem you may think is similar.
A proton moving along the x axis has an initial velocity of 4.0 × 106 m/s and a constant acceleration of 6.0 × 1012 m/s2. What is the velocity of the proton after it has traveled a distance of 80 cm? Group of answer choices
Answer:
5.06*10^6 m/s
Explanation:
Given that
Initial velocity, u = 4*10^6 m/s
Acceleration, a = 6*10^12 m/s²
Distance traveled, s = 80 cm
Final velocity, v = ?
We can find the final velocity by using one of the equations of motion.
v² = u² + 2as
On substituting the values, we have
v² = (4*10^6)² + 2 * 6*10^12 * 0.8
v² = 2.56*10^13
v = √2.56*10^13
v = 5.06*10^6 m/s
Therefore, the final velocity of the proton is adjudged to be 5.06*10^6 m/s
The final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex].
The given parameters;
initial velocity of the proton, u = 4 x 10⁶ m/sacceleration of the proton, a = 6 x 10¹² m/s²distance traveled by the proton, s = 80 cm = 0.8 mThe final velocity of the proton over the given distance is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v^2 = (4\times 10^6)^2 \ + \ 2(6.0 \times 10^{12})(0.8)\\\\v^2 = 2.56 \times 10^{13} \\\\v = \sqrt{2.56 \times 10^{13} } \\\\v = 5.06 \times 10^6 \ m/s[/tex]
Thus, the final velocity of the proton over the given distance is [tex]5.06 \times 10^6 \ m/s[/tex]
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