Complete question:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.
Answer:
The ratio of the power delivered to A to power delivered to B is 7 : 1
Explanation:
Cross sectional area of a wire is calculated as;
[tex]A = \frac{\pi d^2}{4}[/tex]
Resistance of a wire is calculated as;
[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]
Resistance in wire A;
[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]
Resistance in wire B;
[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]
Power delivered in wire;
[tex]P = \frac{V^2}{R}[/tex]
Power delivered in wire A;
[tex]P = \frac{V^2_A}{R_A}[/tex]
Power delivered in wire B;
[tex]P = \frac{V^2_B}{R_B}[/tex]
Substitute in the value of R in Power delivered in wire A;
[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]
Substitute in the value of R in Power delivered in wire B;
[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]
Take the ratio of power delivered to A to power delivered to B;
[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]
The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]
The wires are connected across the same potential; [tex]V_A = V_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]
wire A has seven times the diameter and seven times the length of wire B;
[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]
Therefore, the ratio of the power delivered to A to power delivered to B is
7 : 1
A heavy, 6 m long uniform plank has a mass of 30 kg. It is positioned so that 4 m is supported on the deck of a ship and 2 m sticks out over the water. It is held in place only by its own weight. You have a mass of 70 kg and walk the plank past the edge of the ship. How far past the edge do you get before the plank starts to tip, in m
Answer:
about 1 meter
Explanation:
The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m
We are given;
Mass of plank; m = 30 kg
Length of plank; L = 6m
Mass of man; M = 70 kg
Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.
Thus;
Moment of weight of plank about the right support;
τ_p = mg((L/2) - 2)
τ_p = 30 × 9.8((6/2) - 2)
τ_p = 294 N.m
Moment of weight of man about the right support;
τ_m = Mgx
where x is the distance past the edge the man will get before the plank starts to tip.
τ_m = 70 × 9.8x
τ_m = 686x
Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;
τ_m = τ_p
686x = 294
x = 294/686
x = 0.4285 m
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Potential difference of a battery is 2.2 V when it is connected
across a resistance of 5 ohm, if suddenly the potential difference
falls to 1.8V, its internal resistance will be
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
When using a mercury barometer , the vapor pressure of mercury is usually assumed to be zero. At room temperature mercury's vapor pressure is about 0.0015 mm-Hg. At sea level, the height hhh of mercury in a barometer is about 760 mm.Required:a. If the vapor pressure of mercury is neglected, is the true atmospheric pressure greater or less than the value read from the barometer? b. What is the percent error? c. What is the percent error if you use a water barometer and ignore water's saturated vapor pressure at STP?
Answer:
Explanation:
(a)
The true atmospheric pressure will has more value than the reading in the barometer. If Parm is the atmospheric
pressure in the tube then the resulting vapour pressure is
Patm - pgh = Prapor
The final reading ion the barometer is
pgh = Palm - Proper
Hence, the true atmospheric pressure is greater.
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physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x
Answer:
If the particle is an electron [tex]E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton, [tex]E_y = 6.08 * 10^6 N/C[/tex]
Explanation:
Initial speed at the origin, [tex]u = 3 * 10^6 m/s[/tex]
[tex]\theta = 38^0[/tex] to +ve x-axis
The particle crosses the x-axis at , x = 1.5 cm = 0.015 m
The particle can either be an electron or a proton:
Mass of an electron, [tex]m_e = 9.1 * 10^{-31} kg[/tex]
Mass of a proton, [tex]m_p = 1.67 * 10^{-27} kg[/tex]
The electric field intensity along the positive y axis [tex]E_y[/tex], can be given by the formula:
[tex]E_y = \frac{2 m u^2 sin \theta cos \theta}{qx} \\[/tex]
If the particle is an electron:
[tex]E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 3311.13 N/C\\E_y = 3.311 * 10^3 N/C[/tex]
If the particle is a proton:
[tex]E_y = \frac{2 m_p u^2 sin \theta cos \theta}{qx} \\[/tex]
[tex]E_y = \frac{2 * 1.67 * 10^{-27} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\[/tex]
[tex]E_y = 6.08 * 10^6 N/C[/tex]
Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
Answer:
Explanation:
Given that:
The mass of the cell is 9.0 x 10^-14 kg
The charges of the cell is -2.5pC and the other -3.30 pC
[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]
Radius is 3.75 × 10-6 m
The final distance is twice the radius
i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]
The formula for the velocity of the cell is
[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]
[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]
[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]
The maximum acceleration of the cells as they move toward each other and just barely touch is
[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]
[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]
[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]
The answers obtained are;
1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].
2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].
The answer is explained as shown below.
We have, the mass of the red blood cell;
[tex]m=9\times 10^{-14}\,kg[/tex]Also, the charges of the cells are;
[tex]q_1=-2.5\times 10^{-12}\,C[/tex] and[tex]q_2=-3.30\times 10^{-12}\,C[/tex]The distance between the charges when they barely touch will be two times the radius of each charge.
[tex]r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m[/tex]Kinetic Energy of moving charges1. As both the cells are negatively charged they will repel each other.
So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.[tex]\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}[/tex]Where, [tex]k=9\times10^9\,Nm^2/C^2[/tex] is the Coulomb's constant.Now, substituting all the known values in the equation, we get;
[tex](9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}[/tex][tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]
[tex]\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s[/tex]Electrostatic force between two charges2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.
[tex]ma=k\frac{q_1 q_2}{r^2}[/tex]Substituting the known values, we get;
[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]
[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]
[tex]a=1.47\times 10^{10}\,m/s^2[/tex]Find out more information about moving charges here:
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Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box of mass M. There is friction between the surfaces of blocks 2M and 3M so the three blocks accelerate together to the right.
Which block has the smallest net force acting on it?
A) M
B) 2M
C) 3M
D) The net force is the same for all three blocks Submit
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M
[tex]\Sigma F = F - F' = M\cdot a[/tex]
Box with mass 2M
[tex]\Sigma F = F' - F'' = 2\cdot M \cdot a[/tex]
Box with mass 3M
[tex]\Sigma F = F'' = 3\cdot M \cdot a[/tex]
On the third equation, acceleration can be modelled in terms of F'':
[tex]a = \frac{F''}{3\cdot M}[/tex]
An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.
[tex]F' = 2\cdot M \cdot a + F''[/tex]
[tex]F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''[/tex]
[tex]F' = \frac{5}{3}\cdot F''[/tex]
Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:
[tex]F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)[/tex]
[tex]F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''[/tex]
[tex]F = 2\cdot F''[/tex]
[tex]F'' = \frac{1}{2}\cdot F[/tex]
Afterwards, F' as function of the external force can be obtained by direct substitution:
[tex]F' = \frac{5}{6}\cdot F[/tex]
The net forces of each block are now calculated:
Box with mass M
[tex]M\cdot a = F - \frac{5}{6}\cdot F[/tex]
[tex]M\cdot a = \frac{1}{6}\cdot F[/tex]
Box with mass 2M
[tex]2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F[/tex]
[tex]2\cdot M \cdot a = \frac{1}{3}\cdot F[/tex]
Box with mass 3M
[tex]3\cdot M \cdot a = \frac{1}{2}\cdot F[/tex]
As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be 70.0 km directly north of his starting point.
a. What was the length of the third leg of his trip?b. What was the direction of the third leg of his trip?
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
A 2.4-kg ball falling vertically hits the floor with a speed of 2.5 m/s and rebounds with a speed of 1.5 m/s. What is the magnitude of the impulse exerted on the ball by the floor
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
The magnitude of impulse will be "9.6 Ns".
According to the question,
Mass,
m = 2.4 kgFinal velocity,
v = 2.5 m/sInitial velocity,
u = -1.5 m/sBy using Newton's 2nd law of motion, we get
→ Impulse, [tex]I = m(v-u)[/tex]
By substituting the values, we get
[tex]= 2.4[2.5-(1.5)][/tex]
[tex]= 2.4(2.5+1.5)[/tex]
[tex]= 2.4\times 4[/tex]
[tex]= 9.6 \ Ns[/tex]
Thus the above answer is right.
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A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force
Answer:
The magnitude of the electric force on the particle in terms of the variables given is, F = qE
The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)
Explanation:
Given;
a charged particle, q
magnitude of electric field, E
magnitude of magnetic field, B
The magnitude of the electric force on the particle in terms of the variables given;
F = qE
The magnitude of the magnetic force on the particle in terms of the variables given;
F = q (v x B)
where;
v is the constant velocity of the charged particle
Answer:
The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|
The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|
Explanation:
The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).
Electric force = qE
The magnitude of the electric force = |qE|
That is, magnitude of the product of the charge and the electric field vector.
The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).
It is given mathematically as (qv × B)
The magnitude of the magnetic force is the magnitude of the vector product obtained.
Magnitude of the magnetic force = |qv × B|
Hope this Helps!!!
A 300 g bird flying along at 6.2 m/s sees a 10 g insect heading straight toward it with a speed of 35 m/s (as measured by an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.
Required:
What is the bird's speed immediately after swallowing?
Answer:
The velocity of the bird is [tex]v_f = 4.87 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the bird is [tex]m_1 = 300 \ g = 0.3 \ kg[/tex]
The initial speed of the bird is [tex]u_1 = 6.2 \ m/s[/tex]
The mass of the insect is [tex]m_2 = 10 \ g = 0.01 \ kg[/tex]
The speed of the insect is [tex]u_ 2 =-35 \ m/s[/tex]
The negative sign is because it is moving in opposite direction to the bird
According to the principle of linear momentum conservation
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v_f[/tex]
substituting values
[tex](0.3 * 6.2 ) + (0.01 * (-35)) = (0.3 + 0.01 )v_f[/tex]
[tex]1.51 = 0.31 v_f[/tex]
[tex]v_f = 4.87 \ m/s[/tex]
The Final velocity of Bird = 4.87 m/s
Mass of the bird = 300 g = 0.3 kg
Velocity of bird = 6.2 m/s
Momentum of Bird = Mass of bird [tex]\times[/tex] Velocity of Bird = 0.3 [tex]\times[/tex] 6.2 = 1.86 kgm/s
Mass of the insect = 10 g = 0.01 kg
Velocity of insect = - 35 m/s
Momentum of the Insect = Mass of Insect [tex]\times[/tex] Velocity of Insect = - 0.35 kgm/s
According to the law of conservation of momentum We can write that
In the absence of external forces on the system , the momentum of system remains conserved in that particular direction.
The bird opens the mouth and enjoys the free lunch hence
Let the final velocity of bird is [tex]v_f[/tex]
Initial momentum of the system = Final momentum of the system
1.86 -0.35 = [tex]v_f[/tex] ( 0.01 + 0.3 )
1.51 = [tex]v_f[/tex] 0.31
[tex]v_f[/tex] = 4.87 m/s
The Final velocity of Bird = 4.87 m/s
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when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.2525 N. What were the initial charges on the spheres
Answer:
q1 = 7.6uC , -2.3 uC
q2 = 7.6uC , -2.3 uC
( q1 , q2 ) = ( 7.6 uC , -2.3 uC ) OR ( -2.3 uC , 7.6 uC )
Explanation:
Solution:-
- We have two stationary identical conducting spheres with initial charges ( q1 and q2 ). Such that the force of attraction between them was F = 0.6286 N.
- To model the electrostatic force ( F ) between two stationary charged objects we can apply the Coulomb's Law, which states:
[tex]F = k\frac{|q_1|.|q_2|}{r^2}[/tex]
Where,
k: The coulomb's constant = 8.99*10^9
- Coulomb's law assume the objects as point charges with separation or ( r ) from center to center.
- We can apply the assumption and approximate the spheres as point charges under the basis that charge is uniformly distributed over and inside the sphere.
- Therefore, the force of attraction between the spheres would be:
[tex]\frac{F}{k}*r^2 =| q_1|.|q_2| \\\\\frac{0.6286}{8.99*10^9}*(0.5)^2 = | q_1|.|q_2| \\\\ | q_1|.|q_2| = 1.74805 * 10^-^1^1[/tex] ... Eq 1
- Once, we connect the two spheres with a conducting wire the charges redistribute themselves until the charges on both sphere are equal ( q' ). This is the point when the re-distribution is complete ( current stops in the wire).
- We will apply the principle of conservation of charges. As charge is neither destroyed nor created. Therefore,
[tex]q' + q' = q_1 + q_2\\\\q' = \frac{q_1 + q_2}{2}[/tex]
- Once the conducting wire is connected. The spheres at the same distance of ( r = 0.5m) repel one another. We will again apply the Coulombs Law as follows for the force of repulsion (F = 0.2525 N ) as follows:
[tex]\frac{F}{k}*r^2 = (\frac{q_1 + q_2}{2})^2\\\\\sqrt{\frac{0.2525}{8.99*10^9}*0.5^2} = \frac{q_1 + q_2}{2}\\\\2.64985*10^-^6 = \frac{q_1 + q_2}{2}\\\\q_1 + q_2 = 5.29969*10^-^6[/tex] .. Eq2
- We have two equations with two unknowns. We can solve them simultaneously to solve for initial charges ( q1 and q2 ) as follows:
[tex]-\frac{1.74805*10^-^1^1}{q_2} + q_2 = 5.29969*10^-^6 \\\\q^2_2 - (5.29969*10^-^6)q_2 - 1.74805*10^-^1^1 = 0\\\\q_2 = 0.0000075998, -0.000002300123[/tex]
[tex]q_1 = -\frac{1.74805*10^-^1^1}{-0.0000075998} = -2.3001uC\\\\q_1 = \frac{1.74805*10^-^1^1}{0.000002300123} = 7.59982uC\\[/tex]
A ball is thrown straight upward and falls back to Earth. Suppose a y-coordinate axis points upward, and the release point is the origin. Instantaneously at the top its flight, which of these quantities are zero
a. Displacment
b. Speed
c. Velocity
d. Accerlation
Explanation:
A ball is thrown straight upward and falls back to Earth. It means that it is coming to the initial position. Displacement is given by the difference of final position and initial position. The displacement of the ball will be 0. As a result velocity will be 0.
Acceleration is equal to the rate of change of velocity. So, its acceleration is also equal to 0.
Hence, displacement, velocity and acceleration are zero.
Which circuits are parallel circuits?
Answer:
The bottom two lines.
Explanation:
They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.
A skydiver stepped out of an airplane at an altitude of 1000m fell freely for 5.00s opened her parachute and slowed to 7.00m/s in a negligible time what was the total elapsed time from leaving the airplane to landing on the ground
Answer:
t = 17.68s
Explanation:
In order to calculate the total elapsed time that skydiver takes to reache the ground, you first calculate the distance traveled by the skydiver in the first 5.00s. You use the following formula:
[tex]y=y_o-v_ot-\frac{1}{2}gt^2[/tex] (1)
y: height for a time t
yo: initial height = 1000m
vo: initial velocity = 0m/s
g: gravitational acceleration = 9.8m/s^2
t: time = 5.00 s
You replace the values of the parameters to get the values of the new height of the skydiver:
[tex]y=1000m-\frac{1}{2}(9.8m/s^2)(5.00s)^2\\\\y=877.5m[/tex]
Next, you take this value of 877.5m as the initial height of the second part of the trajectory of the skydiver. Furthermore, use the value of 7.00m/s as the initial velocity.
You use the same equation (1) with the values of the initial velocity and new height. We are interested in the time for which the skydiver arrives to the ground, then y = 0
[tex]0=877.5-7.00t-4.9t^2[/tex] (2)
The equation (2) is a quadratic equation, you solve it for t with the quadratic formula:
[tex]t_{1,2}=\frac{-(-7.00)\pm \sqrt{(-7.00)^2-4(-4.9)(877.5)}}{2(-4.9)}\\\\t_{1,2}=\frac{7.00\pm 131.33}{-9.8}\\\\t_1=12.68s\\\\t_2=-14.11s[/tex]
You use the positive value of t1 because it has physical meaning.
Finally, you sum the times of both parts of the trajectory:
total time = 5.00s + 12.68s = 17.68s
The total elapsed time taken by the skydiver to arrive to the ground from the airplane is 17.68s
A 50-kg block is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 150 N. Fp is parallel to the displacement of the block. The coefficient of kinetic friction is 0.25.
a) What is the total work done on the block?
b) If the box started from rest, what is the final speed of the block?
Answer:
a) WT = 137.5 J
b) v2 = 2.34 m/s
Explanation:
a) The total work done on the block is given by the following formula:
[tex]W_T=F_pd-F_fd=(F_p-F_f)d[/tex] (1)
Fp: force parallel to the displacement of the block = 150N
Ff: friction force
d: distance = 5.0 m
Then, you first calculate the friction force by using the following relation:
[tex]F_f=\mu_k N=\mu_k Mg[/tex] (2)
μk: coefficient of kinetic friction = 0.25
M: mass of the block = 50kg
g: gravitational constant = 9.8 m/s^2
Next, you replace the equation (2) into the equation (1) and solve for WT:
[tex]W_T=(F_p-\mu_kMg)d=(150N-(0.25)(50kg)(9.8m/s^2))(5.0m)\\\\W_T=137.5J[/tex]
The work done over the block is 137.5 J
b) If the block started from rest, you can use the following equation to calculate the final speed of the block:
[tex]W_T=\Delta K=\frac{1}{2}M(v_2^2-v_1^2)[/tex] (3)
WT: total work = 137.5 J
v2: final speed = ?
v1: initial speed of the block = 0m/s
You solve the equation (3) for v2:
[tex]v_2=\sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(137.5J)}{50kg}}=2.34\frac{m}{s}[/tex]
The final speed of the block is 2.34 m/s
In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
To know more about friction, follow
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The index of refraction for a certain type of glass is 1.645 for blue light and 1.609 for red light. A beam of white light (one that contains all colors) enters a plate of glass from the air, nair≈1, at an incidence angle of 38.55∘. What is the absolute value of ????, the angle in the glass between blue and red parts of the refracted beams?
Answer:
blue θ₂ = 22.26º
red θ₂ = 22.79º
Explanation:
When a light beam passes from one material medium to another, it undergoes a deviation from the path, described by the law of refraction
n₁ sin θ₁ = n₂ sin θ₂
where n₁ and n₂ are the incident and transmitted media refractive indices and θ are the angles in the media
let's apply this equation to each wavelength
λ = blue
in this case n₁ = 1, n₂ = 1,645
sin θ₂ = n₁/ n₂ sin₂ θ₁
let's calculate
sin θ₂ = 1 / 1,645 sint 38.55
sin θ₂ = 0.37884
θ₂ = sin⁻¹ 0.37884
θ₂ = 22.26º
λ = red
n₂ = 1,609
sin θ₂ = 1 / 1,609 sin 38.55
sin θ₂ = 0.3873
θ₂ = sim⁻¹ 0.3873
θ₂ = 22.79º
the refracted rays are between these two angles
The force a spring exerts on a body is a conservative force because:
a. a spring always exerts a force parallel to the displacement of the body.
b. the work a spring does on a body is equal for compressions and extensions of equal magnitude.
c. the net work a spring does on a body is zero when the body returns to its initial position.
d. the work a spring does on a body is equal and opposite for compressions and extensions of equal magnitude.
e. a spring always exerts a force opposite to the displacement of the body.
Answer:
c. the net work a spring does on a body is zero when the body returns to its initial position
Explanation:
A force is conservative when the net work done over any path that returns to the initial position is zero. Choice C matches that definition.
An ideal spring of the kind used in physics problems has the characteristic that it applies the same force at the same distance always. So any work required to extend or compress the spring is reversed when the reverse motion takes place.
In cricket how bowler and batsman use acceleration?
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands
Complete question is;
After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod x = 0.98 m long that pivots freely about the hip.
Answer:
Tangential speed of foot just before it lands is; v = 5.37m/s
Explanation:
Let U (potential energy) be zero on the ground.
So, initially, U = mgh
where, h = 0.98/2 = 0.49m (midpoint of the leg)
Now just before the leg hits the floor it would have kinetic energy as;
K = ½Iω²
where ω = v/r and I = ⅓mr²
So, K = ½(⅓mr²)(v/r)²
K = (1/6) × (mr²)/(v²/r²)
K = (1/6) × mv²
From principle of conservation of energy, we have;
Potential energy = Kinetic energy
Thus;
mgh = (1/6) × mv²
m will cancel out to give;
gh = (1/6)v²
Making v the subject, we have;
v = √6gh
v = √(6 × 9.81 × 0.49)
v = √28.8414
v = 5.37m/s
g A mass of 2 kg is attached to a spring whose constant is 7 N/m. The mass is initially released from a point 4 m above the equilibrium position with a downward velocity of 10 m/s, and the subsequent motion takes place in a medium that imparts a damping force numerically equal to 10 times the instantaneous velocity. What is the differential equation for the mass-spring system.
Answer:
mass 20 times of an amazing and all its motion
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.
Answer:
μ_k = 0.1773
Explanation:
We are given;
Initial velocity;u = 20 m/s
Final velocity;v = 0 m/s (since it comes to rest)
Distance before coming to rest;s = 115 m
Let's find the acceleration using Newton's second law of motion;
v² = u² + 2as
Making a the subject, we have;
a = (v² - u²)/2s
Plugging relevant values;
a = (0² - 20²)/(2 × 115)
a = -400/230
a = -1.739 m/s²
From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:
F_k = −ma - - - (1)
We also know that F_k is defined by;
F_k = μ_k•N
Where;
μ_k is coefficient of kinetic friction
N is normal force which is (mg)
Since gravity acts in the negative direction, the normal force will be positive.
Thus;
F_k = μ_k•mg - - - (2)
where g is acceleration due to gravity.
Thus,equating equation 1 and 2,we have;
−ma = μ_k•mg
m will cancel out to give;
-a = μ_k•g
μ_k = -a/g
g has a constant value of 9.81 m/s², so;
μ_k = - (-1.739/9.81)
μ_k = 0.1773
The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178
Given the following data:
Initial speed = 20 m/sFinal velocity = 0 m/s (since it came to rest)Distance = 115 mScientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To determine the coefficient of kinetic friction between the hockey puck and ice:
First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.
[tex]V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}[/tex]
Acceleration, a = -1.74 [tex]m/s^2[/tex]
Note: The negative signs indicates that the hockey puck is slowing down or decelerating.
From Newton's Second Law of Motion, we have:
[tex]\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}[/tex]
Coefficient of kinetic friction = 0.178
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Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two options for the propulsion system --- chemical and electric --- each with a different specific impulse. Recall that the relationship between specific impulse and exhaust velocity is: Vex=g0Isp Using the Ideal Rocket Equation and setting g0=9.81m/s2 , calculate the propellant fraction required to achieve the necessary ΔV for each of propulsion system. Part 1: Cryogenic Chemical Propulsion First, consider a cryogenic chemical propulsion system with Isp≈450s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%): incorrect Part 2: Electric Propulsion Next, consider an electric propulsion system with Isp≈2000s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%):
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
Where:
[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse
[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.
The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]
To determine the fraction:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²
Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
Part 1: Isp = 450s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]
ln (MR) = 2.17
MR = [tex]e^{2.17}[/tex]
MR = 8.76
Part 2: Isp = 2000s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]
ln (MR) = 0.49
MR = [tex]e^{0.49}[/tex]
MR = 1.63
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}[/tex]
k = 1.4
[tex]T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K[/tex]
Work done is given as;
[tex]W = \frac{1}{2} *m*(v_i^2 - v_e^2)[/tex]
inlet velocity is negligible;
[tex]v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41 \ m/s[/tex]
Therefore, the exit velocity is 629.41 m/s
How have physicists played a role in history?
A. Physics has changed the course of the world.
B. History books are written by physicists.
C. Physicists have controlled most governments.
D. Most decisions about wars are made by physicists.
Answer:
A. Physics has changed the course of the world.
Explanation:
what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.
Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]
The orbital speed for a satellite is given by the formula as follows :
[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]
So, the orbital speed for a satellite is 349.23 m/s.