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
Light is a wave made from vibrating electric and magnetic energy. True or False
Answer: oh i actually dont know
Explanation:
But i wish i can help SORRY!
A 1 meter wide door is initially open at an angle of 30o as shown (top view). You push with 20 N force in the middle of the door as shown and the door rotates around the hinge on the left. The door has a rotational inertia =3.0 kg m2. The angular acceleration of the door will be:
Answer:
angular acceleration = 1.67 rad/s²
Explanation:
given data
door wide = 1 m
initially ope angle = 30°
push force = 20 N
rotational inertia = 3.0 kg m²
solution
we apply force at middle so length will be here r1 = [tex]\frac{1}{2}[/tex] = 0.5 m
and
now we get here torque that is express as
torque τ = Force × r1 × sin30 ......................1
put her value and we get
torque τ = 20 × 0.5 × sin30
torque τ = 5 Nm
and we know
torque = rotational inertia × angular acceleration .......................2
put her value and we get angular acceleration
angular acceleration = [tex]\frac{5}{3}[/tex]
angular acceleration = 1.67 rad/s²
what statement is true of AC current?
Answer:
It can easily be transformed from high voltages to low voltages
How many meters are there in 6.50 ly ? Express your answer using three significant figures.
Answer:
6.5 ly is equal to 6.15 x 10¹⁶ m
Explanation:
Light year is the unit of length. It can easily be calculated by multiplying the speed of light in vacuum withe no. of seconds in an year. Let us calculate the value of 1 light year in meters.
1 Light Year = (Speed of Light)(Seconds in 1 Year)
1 Light Year = (3 x 10⁸ m/s)(1 year)(365 days/year)(24 h/day)(3600 s/h)
1 Light Year = 9.46 x 10¹⁵ m
Hence, to get the value of 6.5 Light years, in meters, we will multiply both sides by 6.5.
6.5 Light Years = (6.5)(9.46 x 10¹⁵ m)
6.5 Light Years = 6.15 x 10¹⁶ m
OR
6.5 ly = 6.15 x 10¹⁶ m
Therefore, there are 6.15 x 10¹⁶ m in 6.5 ly
Ancient cultures built some impressive structures that incorporated astronomical functions and information (Stonehenge, Chichen Itza, the Great Pyramid). A friend or acquaintance of yours tries to argue that some of these structures and artifacts are evidence of "ancient astronauts" or visits by intelligent aliens. How would you rebut or argue against this idea?
n experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have equal radii, = 0.10 . The spheres are released from rest with their centers a distance 41.0 apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres.
A) When their centers are a distance 26.0 apart, find the speed of the 27.0 sphere.
B) Find the speed of the sphere with mass 107.0 .
C) Find the magnitude of the relative velocity with which one sphere is approaching to the other.
D) How far from the initial position of the center of the 27.0 sphere do the surfaces of the two spheres collide?
Answer:
Explanation:
Apply the law of conservation of energy
[tex]KE_i+PE_i=KE_f+PE_f[/tex]
[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]
from the law of conservation of the linear momentum
[tex]m_1v_1=m_2v_2[/tex]
Therefore,
[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]
[tex]=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ][/tex]
[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]
Substitute the values in the above result
[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]
[tex]=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s[/tex]
B) the speed of the sphere with mass 107.0 kg is
[tex]v_2=\frac{m_1v_1}{m_2}[/tex]
[tex]=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s[/tex]
C) the magnitude of the relative velocity with which one sphere is
[tex]v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s[/tex]
D) the distance of the centre is proportional to the acceleration
[tex]\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962[/tex]
Thus,
[tex]x_1=3.962x_2[/tex]
and
[tex]x_2=0.252x_1[/tex]
When the sphere make contact with eachother
Therefore,
[tex]x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r[/tex]
And
[tex]x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r[/tex]
The point of contact of the sphere is
[tex]32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m[/tex]
(A) The speed of the sphere of 27 kg is [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].
(B) The speed of sphere of mass 107 kg is [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].
(C) The magnitude of the relative velocity with which one sphere is approaching to the other is [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].
(D) The distance from the initial position of the center of the 27.0 sphere is 20.506 m.
Given data:
The mass of sphere 1 is, [tex]m_{1} = 27.0 \;\rm kg[/tex].
The mass of sphere 2 is, [tex]m_{2} = 107.0 \;\rm kg[/tex].
The radius of each spheres are, r = 0.10 m.
The distance between the centers of each sphere is, d = 41.0 m.
(A)
In this part, we can apply the conservation of energy to find the speed at given distance of 26.0 m. So,
Total energy at initial = Total energy at final
[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2}v^{2}_{2})[/tex]
Now, as per the conservation of momentum,
[tex]m_{1}v_{1}=m_{2}v_{2}\\\\v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}[/tex]
Therefore,
[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2} \times [m_{1}v_{1}/m_{2}]^{2})\\\\\\\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{m_{1}v^{2}_{1}}{2} \times (\dfrac{m_{1}+m_{2}}{m_{2}})[/tex]
Modifying as,
[tex]v^{2}_{1}=[\dfrac{2Gm^{2}_{2}}{m_{1}+m_{2}}] \times [\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}}][/tex]
Substitute the values in the above result
[tex]v^{2}_{1}=[\dfrac{2 \times 6.67 \times 10^{-11} \times 107^{2}}{27+107}] \times [\dfrac{1}{26}-\dfrac{1}{41}]\\\\v_{1}=\sqrt{1.6038 \times 10^{-5}}\\\\v_{1}=1.2664 \times 10^{-5} \;\rm m/s[/tex]
Thus, the speed of the sphere of 27 kg is [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].
(B)
The sphere with mass 107 kg is calculated as,
[tex]v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}\\\\v_{2}=\dfrac{27 \times 1.2664 \times 10^{-5} \;\rm m/s }{107}\\\\v_{2}=3.195 \times 10^{-6} \;\rm m/s[/tex]
Thus, the speed of sphere of mass 107 kg is [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].
(C)
The magnitude of the relative velocity with which one sphere is
[tex]v_{r} = v_{1} +v_{2}\\\\v_{r} =( 1.2664 \times 10^{-5})+ (3.195 \times 10^{-6})\\\\v_{r}=15.85 \times 10^{-6} \;\rm m/s[/tex]
Thus, the magnitude of the relative velocity with which one sphere is approaching to the other is [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].
(D)
The distance of the centre is proportional to the acceleration,
[tex]\dfrac{x_{1}}{x_{2}}=\dfrac{m_{2}}{m_{1}}\\\\\\\dfrac{x_{1}}{x_{2}}=\dfrac{107}{27}\\\\\\\dfrac{x_{1}}{x_{2}}=3.962\\\\\\x_{1}=3.962 \times x_{2}[/tex]
As per the given problem,
[tex]x_{1}+x_{2}+2R = d\\\\3.962x_{2}+x_{2}+(2R) = 41 \\\\x_{2} = 8.262-0.403R[/tex]
And,
[tex]x_{1}=0.252x_{2}\\\\x_{1}=0.252 \times (8.262-0.403R)\\\\x_{1}=32.747-1.597R[/tex]
Then for point of contact of the sphere:
[tex]x_{1} = x_{2}\\\\32.747-1.597R = 8.262-0.403R\\\\R =20.506 \;\rm m[/tex]
Thus, the distance from the initial position of the center of the 27.0 sphere is 20.506 m.
Learn more about the conservation of linear momentum here:
from the initial position of the center of the 27.0 sphere
The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a burette the inner diameter of whose opening is 1.8 mm
Answer:
0.06563 N/m
The percentage difference between the calculated and tabulated value of water surface tension is 9.78%
Explanation:
Solution:-
- Surface tension is the ability of any fluid to resist any external force due to cohesive nature of the fluid molecules.
- Surface tension ( γ ) is defined as the force imparted per unit length on the fluid.
γ = F / L
Where,
F: Force
L: The length over which force is felt
- The mass ( M = 3.78g ) of ( n = 100 ) water droplets are carefully burette. We need to determine the mass of a single droplet as follows:
m = M / n
m = 3.78 / 100
m = 0.0378 g
- Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity ( Weight of the droplet ):
F = m*g
F = 0.0378*9.81*10^-3
F = 0.000370818 N
- The length over which the force is felt can be modeled into a circular patch with diameter ( d ) = equal to the diameter ( d ) of the single water drop. The length ( L ) is defined as the circumference of the patch:
L = π*d
L = π*( 0.0018 )
L = 0.00565 m
- The surface tension would be:
γ = F / L
γ = 0.000370818 / 0.00565
γ = 0.06563 N / m
- The given value of water's surface tension is given as follows:
γa = 0.07275 N/m
- To compare the two values we will determine the percentage difference between the value evaluated and tabulated value as follows:
% [tex]diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\diff = 9.78 \\\\[/tex]
- The percentage difference ( 9.78% ) between the two values is within practical limits of 10%. Hence, the calculated can be accepted.
Emma and Lily jog in the same direction along a straight track. For 0≤t≤15, Emma’s velocity at time t is given by E(t)=7510t2−7t+80.22 and Lily’s velocity at time t is given by L(t)=12t3e−0.5t. Both E(t) and L(t) are positive for 0≤t≤15 and are measured in meters per minute, and t is measured in minutes. Emma is 10 meters ahead of Lily at time t=0, and Emma remains ahead of Lily for 0
Answer:
a) 103.176 m / min
b) 1751.28 meters
Explanation:
Given:-
- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:
[tex]E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t[/tex]
- Where, E ( t ) and L ( t ) are given in m / min
- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:
( 0 ≤ t ≤ 15 ) mins
- It is known that Emma is 10 meters ahead of Lily at time t = 0.
Find:-
a) Find the value of [tex]\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt[/tex] using correct units, interpret the meaning of
b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?
Solution:-
- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:
[tex]f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx[/tex]
- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:
[tex]E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\[/tex]
- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:
[tex]E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\[/tex]
- Complete the square in the denominator:
[tex]E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\[/tex]
- Use the following substitution:
[tex]u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du[/tex]
- Substitute the relations for (u) and (dt) in the above E_avg expression.
[tex]E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du[/tex]
- Use the following standard integral:
[tex]arctan(u) = \int {\frac{1}{u^2 + 1} } \, du[/tex]
- Evaluate:
[tex]E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |[/tex]
- Apply back substitution for ( u ):
[tex]E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\[/tex]
- Plug in the limits and find Emma's average velocity:
[tex]E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}[/tex]
Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.
- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of 0≤t≤15 is given by the relation:
[tex]S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\[/tex]
- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of 0 ≤ t ≤ 15 is given by the relation:
[tex]S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\[/tex]
Apply integration by parts:
[tex]S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\[/tex]
Re-apply integration by parts 2 more times:
[tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\[/tex] [tex]S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\[/tex]
[tex]S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\[/tex]
- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L ) from S ( E ):
[tex]S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\[/tex]
- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.
- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:
[tex]\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671[/tex]
- We will plug in each value of t and evaluate the displacement function S(t) for each critical value:
Help me out on this? Grades due in a couple days and I need to get everything done on time-
Answer:
75mL
...............
The question is full of inconsistencies and conceptual errors. Students should not waste their time trying to answer it ... it will only mislead and confuse them.
The question is poor. It was written by someone unclear on the concepts. It should be ignored.
Which of the following are vectors?
Check all that apply.
O A. energy
B. displacement
O C. force
D. mass
E. momentum
F. speed
G. time
O H. velocity
Answer:
Displacement, force, momentum, and velocity are all vectors
Explanation:
Vectors are quantities that show direction; which all of the above do. The rest of the terms are scalar quantities
A 0.15 kg baseball has a kinetic energy of 18 J. What is its speed? (Round you answer to one decimal place)
Answer: The speed is 15.5 m/s
Explanation:
The kinetic energy can be written as:
K = (1/2)*m*v^2
where m is the mass and v is the speed.
Then we have that:
18 j = (1/2)*0.15kg*v^2
Now we solve this for v.
√(18*2/0.15) = v
15.5 m/s = v
During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ± 0.001) kg with a speed of (3.2 ± 0.01) ms-1. The second trolley is moving away with a distance of (2.5 ± 0.01) ms-1.
What is the absolute uncertainty of the ratio of momentum of the two trolleys X/Y?
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ[tex]P_{y}[/tex] = m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ / [tex]P_{y}[/tex]
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
jodi wants to sell colorful ribbons specifically to dancers. she signs up for a vendor booth at a local dance convention. what marketing concept should she use ?
Answer:
Target Marketing
Explanation:
The Marketing Concept is one of five concepts adopted by organizations when marketing their products to customers. The other four are the Production, Selling, Products, and Social Marketing concepts. The Marketing concept aims at satisfying the needs of the buyers through their products. This approach employs targeted marketing to gain competitive advantage over other firms.
Target Marketing would help Jodi sell colorful ribbons specifically to dancers. Jodi has a target market. These are the Dancers. So, after setting up her booth, Jodi would do well to concentrate her efforts on the dancers whom she hopes to sell the ribbons to. This would enable her establish competitive advantage over other sellers.
A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) carrying 65.0 uC of charge is placed such that its center is 40.0 cm away from the center of the first sphere (A). a. If the two conducting spheres are now connected by a thin conducting wire, what are the new charges on the spheres? b. Now, the conducting wire is cut and the spheres are released from rest. What are their speeds when they are far apart (infinite distance away) from each other? Take the mass of each conductor to be 80.0 grams. Ignore gravity. Assume that the potential is zero at infinity.
Answer:
a) 50μC
b) 37.45 m/s
Explanation:
a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.
Thus, you have:
[tex]Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C[/tex]
Hence, each sphere has a charge of 50μC.
b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:
[tex]\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J[/tex]
where you have used the Coulomb constant = 8.98*10^9 Nm^2/C^2
Next, you equal the total work to the change in K:
[tex]\frac{1}{2}mv^2=56.125J\\\\v=\sqrt{\frac{2(56.125J)}{m}}=\sqrt{\frac{2(56.125J)}{0.08kg}}=37.45\frac{m}{s}[/tex]
hence, the speed of the spheres is 37.45 m/s
Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3585 Hz. To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 V (the same voltage as the car battery). To produce a sufficiently loud sound, the capacitor must store 0.0163 J of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?
Answer:
Explanation:
Energy stored in a capacitor = 1/2 C V² where , C is capacitance and V is potential of capacitor.
Putting the values
.5 x C x 12² = .0163
C = 226.4 x 10⁻⁶ F .
Frequency of L-C circuit = [tex]\frac{1}{2\pi}\sqrt{\frac{1}{LC} }[/tex]
where L is inductance of inductor and C is capacitance of capacitor.
putting the values
3585 = [tex]\frac{1}{2\pi}\sqrt{\frac{1}{L\times226.4\times10^{-6}} }[/tex]
506.87 x 10⁶ = [tex]\frac{1}{L\times226.4\times10^{-6}}[/tex]
[tex]L = \frac{1}{114755.37}[/tex]
= 8.7 x 10⁻⁶ H.
List of priceless your bodies from largest to smallest in terms of their distance from earth
Answer:
hi
Explanation:
hi
Infer the effect of deforestation on the carrying capacity of the Amazon rainforest.
Answer:
More than 20% of the amazon has been destroyed, affects biodiversity.
Explanation:
The biggest issues that the amazon is facing is the deforestation and it involves the clearing of land areas for logging, farming, and other land-use changes. Effects that are seen are droughts, floods, destruction of habitats of flora and fauna along with the valuable services of the ecosystem. This leads to a decrease in the carrying capacity of the species and the decline of the essential resources that are necessary to survive.HELP 25 POINTS MULTIPLE CHOICE
The energy due to the motion of an object is....
A)non mechanical B) potential C) thermal D) kinetic
Answer:
D) Kinetic.
Explanation:
Kinetic energy is energy in motion.
Answer: D
Explanation: kinetic is also called energy of motion , it is defined at the work needed to accelerate a body of a given mass from rest to its stated velocity.
The electric potential difference of a 125 F capacitor is measured across the terminals of the capacitor and found to be 9.0 V. The potential energy of the capacitor, rounded to two significant figures, is
Answer:
E=1/2CV^2
where E is potential energy, C is capacitance and V is the voltage.
E=1/2×125×9^2
E=1/2×125×81
E=5062.5
E=5100J to 2 significant figures.
Answer:
5.1
Explanation:
ed2020
As a prank, your friends have kidnapped you in your sleep, and transported you out onto the ice covering a local pond. Since you're an engineer, the first thing you do when you wake up is drill a small hole in the ice and estimate the ice to be 6.7cm thick and the distance to the closest shore to be 30.5 m. The ice is so slippery (i.e. frictionless) that you cannot seem to get yourself moving. You realize that you can use Newton's third law to your advantage, and choose to throw the heaviest thing you have, one boot, in order to get yourself moving. Take your weight to be 588 N. (Lucky for you that, as an engineer, you sleep with your knife in your pocket and your boots on.)
1)(a) What direction should you throw your boot so that you will most quickly reach the shore? away from the closest shore perpendicular to the closest shore straight up in the air at your friend standing on the closest shore
2)(b) If you throw your 1.08-kg boot with an average force of 391 N, and the throw takes 0.576 s (the time interval over which you apply the force), what is the magnitude of the force that the boot exerts on you? (Assume constant acceleration.)
391 N
3)(c) How long does it take you to reach shore, including the short time in which you were throwing the boot?
Just number 3
Answer:
1a) The direction to throw the boot is directly away from the closest shore.
2b) The magnitude of the force that the thrown boot exerts on the engineer = 391 N
3c) Time taken to reach shore = 8.414 s
Explanation:
1a) Newton's third law of motion explains that for every action, there is an equal and opposite reaction.
The force generated by throwing the boot in one direction is exerted back on the engineer as recoil in the opposite direction.
Hence, the best direction to throw the boot is opposite the direction that the engineer intends to move towards.
2b) Just as explained in (1a) above, the force exerted in one direction always has a reaction of the same magnitude in the opposite direction.
Hence, the force exerted by the boot on the engineer is equal to the force exerted by the engineer on the boot = 391 N.
3c) For this part, we analyze the total motion of the engineer.
The force exerted by the boot on the engineer initially accelerates the engineer until the engineer reaches a constant velocity dictated the impulse of the initial force (since impulse is equal to change in momentum), this constant velocity then takes the engineer all the way to shore, since the ice surface is frictionless.
The weight of the engineer = W = 588 N
W = mg
Mass of the engineer = (W/g) = (588/9.8) = 60 kg
Force exerted on the engineer by the thrown boot = F = 391 N
F = ma
Initial acceleration of the engineer = (F/m) = (391/60) = 6.52 m/s²
We can then calculate the distance covered during this acceleration
X₁ = ut + ½at₁²
u = initial velocity of the engineer = 0 m/s (the engineer was initially at rest)
t₁ = time during which the force acts = 0.576 s
a = acceleration during this period = 6.52 m/s²
X₁ = 0 + 0.5×6.52×0.576² = 1.08 m
For the second part of the engineer's motion, the velocity becomes constant.
So, we first calculate this constant velocity
Impulse = Change in momentum
F×t = mv - mu
F = Force causing motion = 391 N
t = time during which the force acts = 0.576 s
m = mass of the engineer = 60 kg
v = final constant velocity of the engineer = ?
u = initial velocity of the engineer = 0 m/s
391 × 0.576 = 60v
v = (391×0.576/60) = 3.7536 m/s.
The distance from the engineer's initial position to shore is given as 30.5 m
The engineer covers 1.08 m during the time the force causing motion was acting.
The remaining distance = X₂ = 30.5 - 1.08 = 29.42 m
We can then calculate the time taken to cover the remaining distance, 29.42 m at constant velocity of 3.7536 m/s
X₂ = vt₂
t₂ = (X₂/v) = (29.42/3.7536) = 7.838 s
Time taken to reach shore = t₁ + t₂ = 0.576 + 7.838 = 8.414 s
Hope this Helps!!!
The surface tension of water was determined in a laboratory by using the drop weight method. 100 drops were released from a burette the inner diameter of whose opening is 1.8 mm. The mass of the droplets was 3.78 g.
i. Determine the surface tension of the water and, comparing it with the tabulated value,
Answer:
γ = 0.06563 N / m
9.78% difference
Explanation:
Solution:-
- Surface tension is the ability of any fluid to resist any external force which causes a decreases in surface area of the impact area due to inward compressive forces. These compressive forces occur due to cohesive nature of the fluid molecules.
- Mathematically, surface tension ( γ ) is defined as the force felt per unit length by the fluid.
γ = F / L
Where,
F: Force imparted
L: The length over which force is felt
- We are given the mass ( M ) of ( n = 100 ) water droplets to e 3.78 g. The mass of a single droplet ( m ) can be evaluated as follows:
m = M / n
m = 3.78 / 100
m = 0.0378 g
- The force ( F ) imparted by a single drop of water from the burette can be determined from the force balance on a single droplet. Assuming the distance over which the drop falls is negligible and resistive forces are negligible. Then the only force acting on the droplet is due to gravity:
F = m*g
F = 0.0378*9.81*10^-3
F = 0.000370818 N
- The length over which the force is felt can be magnified into a circular area with diameter equal to that of a single droplet ( d ). The circumferential length ( L ) of the droplet would be as follows:
L = π*d
L = π*( 0.0018 )
L = 0.00565 m
- Then the surface tension would be:
γ = F / L
γ = 0.000370818 / 0.00565
γ = 0.06563 N / m
- The tabulated value of water's surface tension is given as follows:
γa = 0.07275 N/m
- We will determine the percentage difference between the value evaluated and tabulated value as follows:
[tex]p.diff = \frac{gamma_a - gamma}{gamma_a} * 100\\\\p.diff = \frac{0.07275- 0.06563}{0.07275} * 100 \\\\p.diff = 9.78 %[/tex]
- The %difference between is within the allowable practical limits of 10%. Hence, the evaluated value ( γ = 0.06563 N / m ) can be accepted with 9.78% error.
The curve (The Load elongation) for rabbit tendon tested to failure in tension is given in the figure above. Please describe the behavior of the tendon in each interval.
Answer:
Explanation:
Elasticity is the tendency of a material to regain its original shape and size after being deformed by an external force. Hooke's law of elasticity state that; provided the elastic limit of a material is not exceeded, the extension is proportional to the force applied.
The given graph shows the elastic property of the rabbit's tendon when a force (load) is applied.
At point 1 on the graph, a given value of load was applied to the tendon. At this point, Hooke's law is obeyed i.e the tendon supports the load.
At point 2, the value of the load was increased and tendon obeys Hooke's law. This implies that as the load is increased, the the tendon was able to support it.
At point 3, a further increase in the value of load causes the elastic limit of the tendon to be exceeded. This means that the tendon would not return to its original shape and size if the load is removed, Hooke's law is no more obeyed.
At point 4, the tendon breaks because it can no longer sustain any value of load.
To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.) You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a. Let the time at which the dragster starts to accelerate be t = 0.
(A) What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
Sone one explain equilibrium to me what is it
Answer:
In econ: When supply and Demand are equal
Explanation:
Equilibrium is when something evens out
A boat crosses a 200 m wide river at 3 ms-1, north relative to water. The river flows at 1 ms-1 as shown.
What is the velocity of the boat as observed by a stationary observer on the river back from which the boat departed?
Answer:
3.16 m·s⁻¹ at an angle of 71.6°
Explanation:
Assume that the diagram is like Fig. 1 below.
The boat is heading straight across the river and the current is directed straight downstream.
We have two vectors at right angles to each other.
1. Calculate the magnitude of the resultant
We can use the Pythagorean theorem (Fig. 2).
R² = (3 m·s⁻¹)² + (1 m·s⁻¹)² = 9 m²·s⁻² + 1 m²·s⁻² = 10 m²·s⁻²
R = √(10 m²·s⁻²) ≈ 3.16 m·s⁻¹
2. Calculate the direction of the resultant
The direction of the resultant is the counterclockwise angle (θ) that it makes with due East .
tanθ = opposite/adjacent = 3/1 = 3
θ = arctan 3 = 71.6°
To an observer at point O, the velocity of the boat is 3.16 m·s⁻¹ at an angle of 71.6°.
define centre of mass
Answer:
a point representing the mean position of the matter in a body or system.
Explanation:
Consider the situation illustrated in the figure below. If θ = 35° and the masses of blocks B and A are, respectively, Mb and Ma= 0.4Mb, (a) what should be the minimum value of the static coefficient of friction μs between the table and block B so that the system remains in balance? (b) In this case, what will be the tension in the rope segment connected to (node) and the vertical wall?
Answer:
(a) μ ≈ 0.57
(b) not enough information
Explanation:
Draw a free body diagram.
There are four forces acting on block B:
Weight force Mb g pulling downNormal force N pushing upFriction force Nμ pulling leftTension force T₁ pulling rightThere are three forces acting on the node:
Tension force T₁ pulling leftTension force T₂ pulling 35° above the horizontalWeight force Ma g pulling down(a)
Sum of forces on block B in the y direction:
∑F = ma
N − Mb g = 0
N = Mb g
Sum of forces on block B in the x direction:
∑F = ma
T₁ − Nμ = 0
T₁ = Nμ
T₁ = Mb g μ
Sum of forces on the node in the y direction:
∑F = ma
T₂ sin 35° − Ma g = 0
T₂ sin 35° = Ma g
Sum of forces on the node in the x direction:
∑F = ma
T₂ cos 35° − T₁ = 0
T₂ cos 35° = T₁
T₂ cos 35° = Mb g μ
Divide the previous equation by this equation, eliminating T₂.
tan 35° = Ma g / (Mb g μ)
μ = Ma / (Mb tan 35°)
μ = 0.4 / tan 35°
μ ≈ 0.57
(b) T₂ sin 35° = Ma g = 0.4 Mb g
Without knowing the value of Mb, we cannot find the value of the tension force T₂.
Many pigeons live near Martyn’s air conditioning vents. Martyn isn’t able to clean the vents properly, and they often contain dried pigeon droppings. Martyn eventually falls sick with a respiratory illness. Which mode of disease transmission most likely made Martyn sick? A. oral B. aerosol C. vector D. fomite E. direct contact
Answer:
B) Aerosol
The particles from the pigeon droppings are likely to be transmitted from the air vents directly into Martyn's respiratory system
Answer:
aerosol
Explanation:
Which force does the shuttle overcome to create an unbalanced net force as it initially lifts from the ground?
Air resistance
Friction
Gravity
Thrust of rocket engine
Hello!
The answer is Gravity. Lets take a look at the options:
A. Nope, this would be if the shuttle was going horizontal.
B. Again, this would be if the shuttle was going horizontal.
C. Yes! This is the only force that is being overcome by using the rocket thrusts to move up.
D. No, the shuttle is using its thrust rocks to push against gravity.
I hope this helps! :)
Consider a disk, having mass and radius , that spins rapidly about an axle at its center, spinning with an angular velocity . The disk's axle is connected by a thin, massless rod of length to a central pivot. The end of the massless rod is fixed in place at that pivot, but it can freely rotate in all directions. 1) If the disk's mass and radius were each doubled, while keeping the other quantities (, , and ) fixed, by what factor would the rate of precession change
Answer:
The precision will change by a factor of 1/4
Explanation:
Check attachment