Answer:
Change in momentum = 2.197 kgm/s
Explanation:
Momentum = MV
Initial momentum = MU
Final momentum = MV
Computation:
⇒ Change in momentum = MV - MU
⇒ Change in momentum = M (V - U)
⇒ Change in momentum = 0.13(-10.3 - 6.6)
⇒ Change in momentum = 0.13(16.9)
⇒ Change in momentum = 2.197 kgm/s
A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with oil and capped at both ends with tight-fitting pistons. The wider arm of the U-tube has a radius of 18.0 cm and the narrower arm has a radius of 5.00 cm. The car rests on the piston on the wider arm of the U-tube. The pistons are initially at the same level. What is the initial force that must be applied to the
Answer:
F₂ = 925.92 N
Explanation:
In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,
σ₁ = σ₂
F₁/A₁ = F₂/A₂
F₂ = F₁ A₂/A₁
where,
F₂ = Initial force that must be applied to narrow arm = ?
F₁ = Load on Wider Arm to be raised = 12000 N
A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²
A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²
Therefore,
F₂ = (12000 N)(25π cm²)/(324π cm²)
F₂ = 925.92 N
The Bohr radius a0 is the most probable distance between the proton and the electron in the Hydrogen atom, when the Hydrogen atom is in the ground state. The value of the Bohr Radius is: 1 a0 = 0.529 angstrom. One angstrom is 10-10 m. What is the magnitude of the electric force between a proton and an electron when they are at a distance of 2.63 Bohr radius away from each other?
Answer:
The electric force is [tex]F = 11.9 *10^{-9} \ N[/tex]
Explanation:
From the question we are told that
The Bohr radius at ground state is [tex]a_o = 0.529 A = 0.529 ^10^{-10} \ m[/tex]
The values of the distance between the proton and an electron [tex]z = 2.63a_o[/tex]
The electric force is mathematically represented as
[tex]F = \frac{k * n * p }{r^2}[/tex]
Where n and p are charges on a single electron and on a single proton which is mathematically represented as
[tex]n = p = 1.60 * 10^{-19} \ C[/tex]
and k is the coulomb's constant with a value
[tex]k =9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.[/tex]
substituting values
[tex]F = \frac{9*10^{9} * [(1.60*10^{-19} ]^2)}{(2.63 * 0.529 * 10^{-10})^2}[/tex]
[tex]F = 11.9 *10^{-9} \ N[/tex]
Suppose a disk drive has the following characteristics:
1. Four surfaces
2. 1,024 tracks per surface
3. 128 sectors per track
4. 512 bytes/sector
5. Track-to-track seek time of 5ms
6. Rotational speed of 5,000rpm 1.
Required:
a. What is the capacity of the drive?
b. What is the access time?
Answer:
a. 256MB
b. 11 ms
Explanation:
a. The capacity of the drive is
As we know that
[tex]= Number\ of\ surface \times tracks\ per\ surface \times sectors\ per\ surface \times bytes\ or\ sector[/tex]
[tex]= 4 \times 1024 \times 128 \times 512\ bytes[/tex]
= 256MB
b. Now the access time is
As we know that
Average access time= seek time + average rotational latency
where,
seek time = 5ms
And, average rotational latency is
[tex]= \frac{rotational latency}{2}[/tex]
[tex]= \frac{60}{RPM}[/tex]
[tex]= \frac{60}{5,000}[/tex]
= 12 ms
So, average rotational latency is 6 ms
So, average access time is
= 5 ms + 6 ms
= 11 ms
We simply applied the above formulas
The planet should move around the elliptical orbit, and two segments of the orbit should become shaded in green. What aspect(s) of the orbit and shaded segments are the same?
Answer: not sure
Explanation:
1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground
Answer:
(a) 7.67 m/s.
(b) 4.9 J
Explanation:
(a) From the law of conservation of energy,
P.E = K.E
mgh = 1/2(mv²)
therefore,
v = √(2gh)....................... Equation 1
Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.
Given: h = 3 m, g = 9.8 m/s²
Substitute into equation 1
v = √(2×9.8×3)
v = √(58.8)
v = 7.67 m/s.
(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J
Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J
Amount of energy converted to heat = 14.7-9.8 = 4.9 J
Which of the following examples describes mechanical efficiency?
A. greasing a surface helps reduce friction
OB. a longer lever helps lift more weight
O C. a fixed pulley changes the direction you
pull
D. pushing the object down a slope is less
work
Answer:
Hello
I think the first one is correct.
Explanation:
Ball B is hanging from an inextensible cord. An identical ball A is released from rest when it is just touching the cord and drops through the vertical distance hA = 8 in. before striking ball B. Assuming e = 0.9 and no friction, determine the resulting maximum vertical displacement hB of ball B.
The image of the ball hanging from the chord is missing, so i have attached it.
Answer:
hb = 0.1016 m
Explanation:
We are given;
Height from which block A is dropped;hA = 8 in = 0.2032 m
Coefficient of restitution;e = 0.9
Now, let us make v_a and v_b the velocities of balls A and B respectively after collision.
If we assume that both balls have the same masses, then from conservation of momentum, v_a = -v_b
Thus;
½m(v_a)² + ½m(v_b)² = m•g•hA
m will cancel out, also, putting -v_b for v_a, we have;
½(-v_b)² + ½(v_b)² = g•hA
½(v_b² + v_b²) = g•hA
½(2v_b²) = g•hA
v_b² = g•hA
v_b = √g•hA
v_b = √(9.81 × 0.2032)
v_b = 1.412 m/s
Now, using conservation of total mechanical energy, we have;
m•g•hb = ½mv_b²
Making hb the subject, we have;
hb = ½v_b²/g
hb = 1.412²/(2 × 9.81)
hb = 0.1016 m
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is the magnitude of the magnetic field at a distance of 19.8 cm from the wire
Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
a ball starts from rest and accelerates at 1.20 m/s2 east. how far will it travel in the first 4.00 seconds
Answer:9.6m
Explanation:
Copper has a specific heat capacity of 3.80 x 102 Jkg-1K-1 What temperature
change would 300 g of copper experience when heated with 560 J of thermal
energy?
Answer:
4.91 K
Explanation:
From the question,
q = cm(Δt)................ Equation 1
Where q = heat, c = specific heat capacity of copper, m = mass of copper, Δt = temperature change.
make Δt the subject of the equation
Δt = q/cm.................. Equation 2
Given: q = 560 J, m = 300 g = 0.3 kg, c = 380 J/kg.K
Substitute these values into equation 2
Δt = 560(0.3×380)
Δt = 560/114
Δt = 4.91 K
A girl throws a ball at an inclined wall from a height of 3 ft, hitting the wall at A with a horizontal velocity v0 of magnitude 45 ft/s. Knowing that the coefficient of restitution between the ball and the wall is 0.9 and neglecting friction, determine the distance d from the foot of the wall to the Point B where the ball will hit the ground after bouncing off the wall.
The diagram that describes the question is attached as a file below
Answer:
[tex]d = 62.9 s[/tex]
Explanation:
[tex]v_0 = 45 ft/s\\e = 0.9[/tex]
Tangent velocity, [tex](v_A)_t = v_0 sin (90 - 60)[/tex]
[tex](v_A)_t = 45 sin(30) = 22.5 ft/s\\[/tex]
[tex]v_0 sin(30 - \alpha) = 22.5[/tex]....................(1)
Along the normal:
[tex]e = \frac{0 - (v_a)_n}{(v_a)_n - 0} \\0.9 = \frac{0 - v_0 cos(30 - \alpha)}{-45 cos 30} \\[/tex]
[tex]v_0 cos(30 - \alpha) = 35.07[/tex].................(2)
Dividing equation (1) by equation (2)
[tex]tan(30 - \alpha) = 22.5/35.07\\30 - \alpha = tan^{-1}(22.5/35.07)\\30 - \alpha = 32.68\\\alpha = - 2.68^0\\tan(\alpha - 30) = 22.5/35.07\\\alpha = 62.68^0[/tex]
[tex]v_0 = \frac{35.07}{ cos32.69} \\v_0 = 41.67 ft/s\\h = \frac{(v_0)_y^2}{2g} \\h = \frac{(41.67sin62.68)^2}{2*32.2}\\h = 21.28 ft\\[/tex]
Time taken from A to E, t₁
[tex]t_1 = \frac{(v_0)_y}{g} \\t_1 = \frac{41.67sin62.68}{32.2}\\t_1 = 1.15 s[/tex]
Time taken from E to B, t₂
[tex]t_2 = \sqrt{\frac{2(21.28+3)}{9.8}} \\t_2 = 2.23 s[/tex]
Total time, t = 1.15 + 2.23
t = 3.38 s
[tex]d = (v_0)_x t - CD\\d = [(41.67 cos 62.68)*3.38] - \frac{3}{tan 60} \\d = 62.9 s[/tex]
The distance ( d ) from the foot of the wall to the Point B where the ball hits the ground after bouncing off the wall : 62.9
Given data :
Height = 3 ft
Horizontal velocity ( V₀ ) = 45 ft/s
Coefficient of restitution = 0.9
Determine the distance from foot of wallTangent velocity ( Vₐ ) t = 45 sin(30) = 22.5 ft/s
∴ V₀sin ( 30 - ∝ ) = 22.5 ------ ( 1 )
along the normal plane
V₀cos ( 30 - ∝ ) = 35.07 ------ ( 2 )
step 1 : Divide equation ( 1 ) by ( 2 )Tan ( 30 - ∝ ) = ( 22.5 / 35.07 )
∴ ∝ = 30 - 32.68
= -2.68° ≈ 62.68°
V₀ is also expressed as : ( 35.07 ) / ( cos 32.69 )
∴ V₀ = 41.67 ft/s
step 2 : Determine the value of hh = [tex]\frac{(Vo)^{2}y }{2*g}[/tex] = ( 41.67 * sin62.68 )² / ( 2 * 32.2 )
= 21.28 ft
step 3 : Determine time taken from one point to anotherTime taken from point A to E
T₁ = ( 41.67 * sin62.68 ) / 32.2
= 1.15 secs
Time from point E to B
T₂ = [tex]\sqrt{\frac{2(21.28+3)}{9.8} }[/tex] = 2.23 secs
Therefore the Total time taken = 2.23 + 1.15 = 3.38 secs
Final step : Determine the distance d
d = ( V₀ )x t - CD
= [ ( 41.67*cos62.68 ) * 3.38 ] - [tex]\frac{3}{tan 60}[/tex]
= 62.9
Hence we can conclude that The distance ( d ) from the foot of the wall to the Point B where the ball hits the ground after bouncing off the wall is 62.9
Learn more about ball bouncing off the wall : https://brainly.com/question/21293362
Attached below is the missing data related to your question
A 0.47-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstretched length triples. What is the mass of the second block
Answer:
The mass of the second block is 0.94 kg
Explanation:
Given;
mass of the first block, m₁ = 0.47 kg
let the mass of the first block and second block = m₂
According to Hook's law;
F = kx
where;
F is the applied load (force)
k is the elastic constant
x is the extension of the elastic material
According to Newton's law of motion;
F = ma = mg
Thus,
mg = kx
[tex]k = \frac{mg}{x} \\\\\frac{m_1g}{x_1} = \frac{m_2g}{x_2} \\\\But, x_2 \ is \ triple \ of \ x_1,\\x_2 = 3x_1\\\\\frac{m_1g}{x_1} = \frac{m_2g}{3x_1}\\\\\frac{m_1}{x_1} = \frac{m_2}{3x_1}\\\\Substitute \ in \ the \ value \ of \ m_1\\\\\frac{0.47}{x_1} = \frac{m_2}{3x_1}\\\\\frac{0.47}{1} = \frac{m_2}{3}\\\\m_2 = 3 *0.47\\\\m_2 = 1.41 \ kg[/tex]
mass of the second block alone = 1.41 - 0.47 = 0.94 kg
Therefore, the mass of the second block is 0.94 kg
A block slides down a ramp with friction. The friction experienced by the block is 21 N. The mass of the block is 8 kg. The ramp is 6 meters long (meaning, the block slides across 6 meters of ramp with friction). The block is originally 2 meters vertically above the ground (the bottom of the ramp). What is the change in energy of the block due to friction, expressed in Joules
Complete Question
The complete question is shown on the first uploaded image
Answer:
the change in energy of the block due to friction is [tex]E = -126 \ J[/tex]
Explanation:
From the question we are told that
The frictional force is [tex]F_f = 21 \ N[/tex]
The mass of the block is [tex]m_b = 8 \ kg[/tex]
The length of the ramp is [tex]l = 6 \ m[/tex]
The height of the block is [tex]h = 2 \ m[/tex]
The change in energy of the block due to friction is mathematically represented as
[tex]\Delta E = - F_s * l[/tex]
The negative sign is to show that the frictional force is acting against the direction of the block movement
Now substituting values
[tex]\Delta E = -(21)* 6[/tex]
[tex]\Delta E = -126 \ J[/tex]
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a charge of (12.5 A) nC when connected to a (34.8 B) V potential difference. Determine the necessary separation in mm. Round your answer to three significant figures.
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;
[tex]V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}[/tex]
Where;
d is the separation or distance between the two parallel plates;
[tex]d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm[/tex]
Therefore, the necessary separation between the two parallel plates is 0.104 mm
A 310-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,010 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 = 166 years
Explanation:
In order to calculate the amount of years that electrons take to cross the complete transmission line. You first calculate the drift speed of the electrons by using the following formula:
[tex]v_d=\frac{I}{nqA}[/tex] (1)
I: current on the wire = 1,010A
n: free charge density = 8.50*10^28 electrons/m^3
A: cross-sectional area of the transmission line = π*r^2
r: radius of the cross-sectional area = 2.00cm = 0.02m
You replace the values of the parameters in the equation (1):
[tex]v_d=\frac{1,010A}{(8.50*10^{28}electron/m^3)(1.6*10^{-19}C)(\pi (0.02m)^2)}\\\\v_d=5.9*10^{-5}\frac{m}{s}[/tex]
Next, you use the following formula:
[tex]t=\frac{x}{v_d}[/tex] (2)
x: length of the line transmission = 310km = 310,000m
You replace the values of vd and x in the equation (2):
[tex]t=\frac{310,000m}{5.9*10^{-5}m/s}=5.24*10^9s[/tex]
Finally, you convert the obtained t to seconds
[tex]t=5.24*10^9s*\frac{1\ year}{3.156*10^7s}=166.03\ years[/tex]
The electrons take approximately 166 years to travel trough the complete transmission line
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of this field through a square of surface area A = 2 m2 oriented parallel to the yz-plane? Group of answer choices
Answer:
∅ = 0 V.m
Explanation:
The electric flux through a surface is given by the following formula:
∅ = E.ΔA
∅ = E ΔA Cosθ
where,
∅ = Electric Flux through the surface = ?
E = Electric Field Intensity = 100 N/C
ΔΑ = Surface Area = 2 m²
θ = Angle between the electric field and the the normal area vector
The electric field is oriented in positive y-axis direction. The surface is parallel to yz-plane, so its normal area vector will be in the direction of x-axis. Therefore, the electric field and normal area vector shall be perpendicular to each other. Therefore,
θ = 90°
Using values in the equation:
∅ = (100 N/C)(2 m²)(Cos 90°)
∅ = (200 N.m²/C)(0)
∅ = 0 V.m
A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases her moment of inertia by two times. What is the skater's final angular speed
Answer:
The final angular speed of the skater is 12 radians per second.
Explanation:
Let consider the skater as a rotating system, given the absence of external forces, the Principle of Angular Momentum Conservation is applied:
[tex]I_{o}\cdot \omega_{o} = I_{f}\cdot \omega_{f}[/tex]
Where:
[tex]I_{o}[/tex], [tex]I_{f}[/tex] - Initial and final moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex] - Angular speed, measured in radians per second.
The final angular speed is cleared afterwards:
[tex]\omega_{f} = \frac{I_{o}}{I_{f}} \cdot \omega_{o}[/tex]
Given that [tex]I_{f} = \frac{1}{2}\cdot I_{o}[/tex] and [tex]\omega_{o} = 6\,\frac{rad}{s}[/tex], the final angular speed is:
[tex]\omega_{f} = \frac{I_{o}}{\frac{1}{2}\cdot I_{o} } \cdot \omega_{o}[/tex]
[tex]\omega_{f} = 2 \cdot \omega_{o}[/tex]
[tex]\omega_{f} = 2 \cdot \left(6\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 12\,\frac{rad}{s}[/tex]
The final angular speed of the skater is 12 radians per second.
The skater's final angular speed is equal to 12 rad/s.
Given the following data:
Mass of skater = 60 kgInitial angular speed of 6.0 rad/sTo determine the skater's final angular speed, we would use the law of conservation of angular momentum:
By applying the law of conservation of angular momentum, we have:
[tex]I_i\omega_i = I_f\omega_f[/tex] ...equation 1.
Making the [tex]\omega_f[/tex] subject of formula, we have:
[tex]\omega_f = \frac{I_i \omega_i}{I_f}[/tex] ...equation 2
Note: The skater decreases her moment of inertia by two times.
Final moment of inertia, [tex]I_f[/tex] = [tex]\frac{I_i}{2}[/tex] ...equation 3.
Substituting eqn. 3 into eqn. 2, we have:
[tex]\omega_f = \frac{I_i \omega_i}{\frac{I_i}{2} }\\\\\omega_f = I_i \omega_i \times \frac{2}{I_i }\\\\\omega_f = 2\omega_i \\\\\omega_f = 2 \times 6\\\\\omega_f = 12\;rad/s[/tex]
Read more: https://brainly.com/question/23153766
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 delivere
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
Two point charges are separated by a distance of 40.0 cm. The numerical value of one charge is twice that of the other. Each charge exerts a force of magnitude 65.0 N on the other. 1) Find the magnitude of the charge with a smaller magnitude. (Express your answer to three significant figures.)
Answer:
2.4×10⁻⁵ C
Explanation:
From coulomb's law,
F = kqq'/r²..................... Equation 1
Where F = force on each charge, q = magnitude of the first charge, q' = magnitude of the second charge, r = distance between both charges, k = coulomb's constant.
From the question,
q' = 2q
Therefore,
F = 2kq²/r².............. Equation 2
make q the subject of the equation
q = √(Fr²/2k).................... Equation 3
Given: F = 65 N, r = 40 cm = 0.4 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute this values into equation 3
q = √[(65×0.4²)/(2×9×10⁹)]
q = √(5.78×10⁻¹⁰)
q = 2.4×10⁻⁵ C
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units r and v. Rank the wires in order of decreasing electron current. Rank from most to least electron current. To rank items as equivalent, overlap them.
a) radius=3r, drift speed=1v
b) radius=4r, drift speed=0.5v
c) radius=1r, drift speed=5v
d) radius=2r, drift speed=2.5v
Answer:
d > a > b > c
Explanation:
Given that
a) radius = 3r and drift speed = 1v
b) radius = 4r and drift speed = 0.5v
c) radius = 1r and drift speed = 5v
d) radius = 2r and drift speed = 2.5v
Based on the above information, the ranking of the wires for reducing the electron current is
As we can see that the radius i.e to be less and the drift speed that is highest so it should be rank one
And, According to that, other options are ranked
Therefore, the ranking would be d > a > b > c
A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.27 grams. By cocking the lever attached to the stock of the rifle, the spring (k = 1800 N/m) is compressed 8.5 cm from its equilibrium point.
Required:
a. What is the potential energy energy stored in the compressed spring?
b. What is the kinetic energy of the pellet when the spring is released?
Answer:
a.6.5025 J
b.6.5025 J
Explanation:
We are given that
Mass of pellet,m=0.27 g=[tex]0.27\times 10^{-3} kg[/tex]
1 kg=1000 g
Spring constant,k=1800 N/m
x=8.5 cm=[tex]8.5\times 10^{-2} m[/tex]
1m=100 cm
a.Potential energy stored in the compressed spring is given by
P.E=[tex]\frac{1}{2}kx^2[/tex]
[tex]P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2[/tex]
[tex]P.E=6.5025 J[/tex]
b.By using law of conservation of energy
P.E of spring=K.E of the pellet
K.E of the pellet=6.5025 J
An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then charged to a potential difference of 12 V? (ε0 = 8.85 × 10-12 C2 /N ∙ m2 ) (a) How much charge is stored on each of its plates? (b) If the capacitor is then filled with a glass dielectric (K = 5.0), by how much does the charge if it changes at all? The capacitor is not connected to a battery.
Answer:
The charge stored is [tex]Q = 4.25 *10^{-7 } \ C[/tex]
The energy stored is [tex]E = 2.55*10^{-6} \ J[/tex]
Explanation:
From the question we are told that
The area of the plates is [tex]A = 0.40 \ m^2[/tex]
The separation between the plate is [tex]d = 0.10 \ mm =0.0001 \ m[/tex]
The potential difference is [tex]V = 12 \ V[/tex]
The permitivity of free space is [tex]\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2[/tex]
The dielectric constant of glass is K = 5.0
Generally the capacitance of this capacitor is
[tex]C = \frac{\epsilon_o * A}{d}[/tex]
substituting values
[tex]C = \frac{8.85*10^{-12} * 0.40}{0.0001}[/tex]
[tex]C = 3.34 *10^{-8} \ C[/tex]
The charge stored is mathematically evaluated as
[tex]Q = CV[/tex]
substituting values
[tex]Q = (3.54*10^{-8} * 12)[/tex]
[tex]Q = 4.25 *10^{-7 } \ C[/tex]
The energy stored is
[tex]E = 0.5 * CV^2[/tex]
substituting values
[tex]E = 0.5 * (4.25 *10^{-7} * 12^2)[/tex]
[tex]E = 2.55*10^{-6} \ J[/tex]
The Thomson model of a hydrogen atom is a sphere of positive charge with an electron (a point charge) at its center. The total positive charge equals the electronic charge e. Prove that when the electron is at a distance r from the center of the sphere of positive charge, it is attracted with a force F=\frac{e^2r}{4\pi\varepsilon_oR^3} where R is the radius of the sphere.
Answer:
E = (1 / 4π ε₀ ) q r / R³
Explanation:
Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law
F = K q₁ q₂ / r²
we used law Gauss
Ф = ∫ E .dA = q_{int} /ε₀
E 4π r² = q_{int} /ε₀
E = q_{int} / 4π ε₀ r²
we replace the charge inside
E = (1 / 4π ε₀ r²) ρ 4/3 π r³
E = ρ r / 3 ε₀
the density for the entire atom is
ρ = Q / V
V = 4/3 π R³
we substitute
E = (r / 3ε₀ ) Q 3/4π R³
E = (1 / 4π ε₀ ) q r / R³
Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light by total internal reflection at the critical angle for the interface between the core (ncore =1.497) and the cladding (ncladding =1.421)
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]\theta_{max} =18.38^o[/tex]
b
New [tex]n_{cladding} =1.491[/tex]
Explanation:
From the question we are told that
The refractive index of the core is [tex]n_{core} = 1.497[/tex]
The refractive index of the cladding is [tex]n_{cladding} = 1.421[/tex]
Generally according to Snell's law
[tex]n_{core} * sin(90- \theta) = n_{cladding} * sin (90)[/tex]
Where [tex]\theta_{max}[/tex] is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection
[tex]\theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ][/tex]
[tex]\theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ][/tex]
[tex]\theta_{max} =18.38^o[/tex]
Given from the question the the largest angle is 5°
Generally the refraction index of the cladding is mathematically represented as
[tex]n_{cladding} = n_{core} * sin (90 - 5)[/tex]
[tex]n_{cladding} =1.491[/tex]
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of the same mass on the same spring be on the Moon, where the acceleration due to gravity is one sixth that of Earth
Answer:
The displacement of the same mass on the same spring on the Moon is 0.05 m.
Explanation:
Given;
mass suspended from one end of the spring, m = 0.500 kg
displacement on the spring on Earth, x = 0.3 m
Apply Newton's second law of motion;
F = ma = mg
where;
m is mass on the spring
g is acceleration due to gravity
Also, apply Hook's law;
F = Kx
where;
K is force constant
x is extension or diplacement of the spring
Combine the two equations from the two laws;
mg = kx
when the spring in on Earth;
0.5 x 9.8 = 0.3k
4.9 = 0.3k
k = 4.9 / 0.3
k = 16.333 N/m
when the spring is on moon;
mg = kx
mass is the same = 0.5 kg
acceleration due to gravity on moon = ¹/₆ that of Earth = ¹/₆ x 9.8 m/s²
0.5 (¹/₆ x 9.8) = 16.333 x
0.8167 = 16.333 x
x = 0.8167 / 16.333
x = 0.05 m
Therefore, the displacement of the same mass on the same spring on the Moon is 0.05 m.
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.
you can find the answer in this book
physics principles with Applications, Global Edition Problem 67P: Chapter: CH 13 Problem:67p
Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground. When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 1.8 s. What is the water speed as it leaves the nozzle?
Answer:
The speed of the water as it leaves the nozzle is [tex]u =9.7 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the nozzle above the ground is [tex]h = 1.5 \ m[/tex]
The time taken for the flow to stop is t = 1.8 s
According the second law of motion
[tex]h = ut - \frac{1}{2} gt^2[/tex]
making the initial velocity the subject
[tex]u = \frac{h + 0.5gt^2}{t}[/tex]
here [tex]g=9.8 \ m/s^2[/tex]
substituting value
[tex]u = \frac{1.5 + 0.5 * 9.8 * (1.8)^2}{1.8}[/tex]
[tex]u =9.7 \ m/s[/tex]
A lens collects light and focuses it into a small spot. This increases the ________ of the light wave. A lens collects light and focuses it into a small spot. This increases the ________ of the light wave. wavelength energy intensity frequency
Answer:
This increases the intensity of the light wave.
Two blocks are placed at the ends of a horizontal massless board, as in the drawing. The board is kept from rotating and rests on a support that serves as an axis of rotation. The block on the right has a mass of 4.4 kg. Determine the magnitude of the angular acceleration when the system is allowed to rotate.
Answer:
The angular acceleration is [tex]\alpha = 0.788 \ rad/s^2[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image (gotten from study website)
From the question we are told that
The mass of the block on the right is [tex]m_r = 4.4 \ kg[/tex]
The mass of the block on the left is [tex]m_l = 12 \ kg[/tex]
The distance of the left mass to the center is [tex]r_l = 0.6 \ m[/tex]
The distance of the right mass to the center is [tex]r_l = 1.4 \ m[/tex]
At the point the system is allowed to rotate the upward torque is mathematically evaluated as
[tex]\tau = m_l * g * r_l - m_r * g * r_r[/tex]
substituting values
[tex]\tau = 12 * 9.8 * 0.6 - 4.4 * 9.8 * 1.4[/tex]
[tex]\tau = 10.2 \ N m[/tex]
The moment of inertia of the system is mathematically represented as
[tex]I = m_l r_l ^2 +m_r r_r^2[/tex]
substituting values
[tex]I = 12 * (0.6)^2 +(4.4)* (1.4)^2[/tex]
[tex]I = 12.94 \ kg \cdot m^2[/tex]
The angular acceleration is mathematically represented as
[tex]\alpha = \frac{\tau}{I}[/tex]
substituting values
[tex]\alpha = \frac{10.2}{12.94}[/tex]
[tex]\alpha = 0.788 \ rad/s^2[/tex]
