-- Water cannot be boled.
-- There is no such thing as a coffe aker.
-- There is no such thing as an immension-type heating element.
Be that as it may, and it very likely still is, as it were . . .
-- The latent heat of vaporization of water at 100°C is about 2250 kilo-joules per kg.
-- To evaporate half of the kg of water in the coffee maker requires 1125 kJ of heat energy.
-- To supply that amount of heat energy over a period of 10 minutes (600 seconds), it must be supplied at a rate of
(1,125,000 Joules / 600 seconds) = 1,875 joules/second.
-- That's 1,875 watts, or 1.875 kilowatts.
-- Choice-a is the choice when the solution is rounded.
A small insect is placed 3.75 cm from a +4.00 −cm -focal-length lens. Calculate the angular magnification.
Final answer:
The angular magnification produced by a +4.00 cm focal length lens with an object at 3.75 cm distance is approximately 7.25 times.
Explanation:
The question presented is a problem in optics, more specifically in the subtopic of lens magnification, which is a part of Physics. To calculate the angular magnification for a lens, we can use the thin lens formula and the magnification formula. Given that the focal length of the lens (f) is +4.00 cm and the object distance (do) is 3.75 cm, we first determine if an image is formed by using the lens equation 1/f = 1/do + 1/di. However, since this equation will not yield a real value for di (image distance) because do is smaller than f, we can't directly use it to find the angular magnification.
Instead, the angular magnification (M) of a simple lens while viewing a close object is approximately given by M = 1 + (25 cm / f), where 25 cm is the near point of a normal human eye. Thus, for this lens:
M = 1 + (25 cm / 4.00 cm)
M = 1 + 6.25
M = 7.25
This means the lens would produce an angular magnification of approximately 7.25 times when the object is placed at 3.75 cm from the lens.
Suppose that 2 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. (a) How much work is needed to stretch the spring from 35 cm to 40 cm? (b) How far beyond its natural length will a force of 30 N keep the spring stretched?
Answer:
(a) 1.042 J
b) 0.108 m
Explanation:
(a)
W = work done to stretch the spring = 2 J
k = spring constant
L₀ = initial length = 30 cm = 0.30 m
L = final length = 42 cm = 0.42 m
x = stretch of the spring = L - L₀ = 0.42 - 0.30 = 0.12 m
Work done to stretch the spring is given as
W = (0.5) k x²
2 = (0.5) k (0.12)²
k = 277.78 N/m
x₀ = initial stretch of the spring = 35 - 30 = 5 cm = 0.05 m
x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m
W' = work needed to stretch the spring from 35 cm to 40 cm
Work needed to stretch the spring from 35 cm to 40 cm is given as
W' = (0.5) k (x² - x₀²)
W' = (0.5) (277.78) (0.10² - 0.05²)
W' = 1.042 J
b)
x = Stretch of the spring beyond natural length
F = force = 30 N
Spring force is given as
F = k x
30 = (277.78) x
x = 0.108 m
(a) The work is needed to stretch the spring from 35 cm to 40 cm is 1.042 J.
(b) Beyond the length of 0.108 m force of 30 N keep the spring stretched.
what is spring force?In order to stretch or compress the spring some amount of work has to be done. Which is given as,
[tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]
where k = spring constant of the spring
x = compression of spring
W= Work required or spring work
Initial length is given by 30 cm = 0.30 m
final length is given by 42 cm = 0.42 m
When spring is stretched change in the length occurs denoted by x
x = final length - initial length =0.42 - 0.30 = 0.12 m
[tex]\rm{W=\frac{1}{2} Kx^{2} }[/tex]
[tex]\rm{2=\frac{1}{2} K(0.12)^{2} }[/tex]
K= 277.78 N/m
(a) work is needed to stretch the spring from 35 cm to 40 cm
stretching length from30cm to 35 is 5 cm which is 0.05 m
stretching length from cm to 30 is 40 cm is 10 cm which is 0.1m
x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m
Work needed to stretch the spring from 35 cm to 40 cm is given
W = (0.5)K([0.1]² - [0.5]²)
W = (0.5) (277.78) (0.10² - 0.05²)
W = 1.042 J
x is stretch of the spring beyond natural length
F = force = 30 N
Spring force is given as
F = k x
30 = (277.78) x
x = 0.108 m
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An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equivalent spring constant of the bow? N/m (b) How much work does the archer do on the string in drawing the bow? J
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J
When an electromagnetic wave falls on a white, perfectly reflecting surface, it exerts a force F on that surface. If the surface is now painted a perfectly absorbing black, what will be the force that the same wave will exert on the surface?
Answer and Explanation:
We know by Newton's second law of motion that Force can be given by the rate of change of momentum i.e.,
F = m[tex]\frac{dp}{dt}[/tex]
where, p is momentum
Now, when EM wave falls on perfectly reflecting body , change in momentum is : -(p+p) = -2p
i.e., after reflection momentum is twice of its initial value
In case of absorption of radiation of EM wave as in perfectly black painted body, change in momentum is half of that in reflection i.e., '-p'
Since, the force, F is equal to the change in momentum, the Force erxerted by the wave will also be half i.e., [tex]\frac{F}{2}[/tex] or 0.5F
Final answer:
When the wave is absorbed, the force is half compared to when the wave is reflected, that's means the force will be F/2
Explanation:
When an electromagnetic wave falls on a white, perfectly reflecting surface, it exerts a force F on that surface. According to Maxwell's predictions, electromagnetic waves carry momentum, and thus when these waves are absorbed by an object, they exert a force in the direction of the wave's propagation, known as radiation pressure. In the case of a perfectly reflecting surface,
the force is twice as great as that on a perfectly absorbing surface because the change in momentum when the wave is reflected is twice as much as when it is absorbed, due to the wave's momentum being reversed while conserving the system's total momentum.
When the same electromagnetic wave falls on a surface that has been painted a perfectly absorbing black, the force exerted on the surface will be half compared to the force on a perfectly reflecting surface.
This is because the absorbing surface does not reflect the electromagnetic waves, and thus, the change in momentum is only due to the wave being absorbed rather than being both absorbed and reflected back.
The property of the surface, whether it is absorbing or reflecting, significantly changes the amount of force exerted by the electromagnetic wave on the surface, that's means the force will be F/2 for the absorbing surface.
An air-track cart is attached to a spring and completes one oscillation every 5.67 s in simple harmonic motion. At time t = 0.00 s the cart is released at the position x = +0.250 m. What is the position of the cart when t = 29.6 s?
Answer:
0.2447 m
Explanation:
Amplitude, A = 0.25 m, T = 5.67 s, t = 29.6 s, y = ?
The general equation of SHM is given by
y = A Sin wt
y = 0.25 Sin (2 x 3.14 t /5.67)
Put t = 29.6 s
y = 0.25 Sin (2 x 3.14 x 29.6 / 5.67)
y = 0.2447 m
An object of weight 8N is moving along a horizontal plane. If the coefficient of kinetic friction is 0.25 find the friction force acting on the object.
Answer:
2 N
Explanation:
Wight of object = mg = 8 N
Coefficient of friction, u = 0.25
The normal reaction acting on the object = N = mg = 8 N
According to the laws of limiting friction,
The friction force, f = uN
f = 0.25 x 8 = 2 N
Calculate the force required to double the length of a steel wire of area of cross section 5 x 10-5 m2? (Y=2x 1011N m2) (a) 10-5 N (b) 10-7 N n's D (c) 107 N (d) 105 N
Answer:
Option C is the correct answer.
Explanation:
We equation for elongation
[tex]\Delta L=\frac{PL}{AE}[/tex]
Here we need to find load required,
We need to double the wire, that is ΔL = 2L - L = L
A = 5 x 10⁻⁵ m²
E = 2 x 10¹¹ N/m²
Substituting
[tex]L=\frac{PL}{5\times 10^{-5}\times 2\times 10^{11}}\\\\P=10^7N[/tex]
Option C is the correct answer.
A circular coil of wire of radius 8 cm and 19 turns is placed with its plane perpendicular to a 5.6 T magnetic field, produced by an electromagnet. The current to the electromagnet is then switched off, causing the magnetic field to collapse to zero in a time of 744 ms. What is the emf generated in the coil while the magnetic field is collapsing, in units of volts?
Answer:
2.87 Volt
Explanation:
N = 19
change in magnetic field = 5.6 T
Time, dt = 744 ms = 0.744 s
Induced emf,
e = N × change in flux / time
e = 19 × 3.14 × 0.08 × 0.08 ×5.6/0.744
e = 2.87 Volt
A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cross section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Answer:
The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
(c) is correct option.
Explanation:
Given that,
Length of steel wire = 4.7 m
Cross section[tex]A = 3\times10^{-3}\ m^2[/tex]
Length of copper wire = 3.5 m
Cross section[tex]A = 4\times10^{-5}\ m^2[/tex]
We need to calculate the ratio of young's modulus of steel and copper
Using formula of young's modulus for steel wire
[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]
[tex]Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}[/tex]....(I)
The young's modulus for copper wire
[tex]Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}[/tex]....(II)
From equation (I) and (II)
The ratio of the young's modulus of steel and copper
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}[/tex]
[tex]\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}[/tex]
Hence, The ratio of the young's modulus of steel and copper is [tex]1.79\times10^{-2}[/tex]
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3.
Answer:
[tex]\dot{W} = 339.84 W[/tex]
Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as
[tex]\dot{W} =\dot{m}\frac{V^{2}}{2}[/tex]
here [tex]\dot{m}[/tex]is mass flow rate and given as
[tex]\dot{m} = \rho*Q[/tex]
[tex]\dot{W} =\rho*Q\frac{V^{2}}{2}[/tex]
Putting all value to get minimum power
[tex]\dot{W} =1.18*9*\frac{8^{2}}{2}[/tex]
[tex]\dot{W} = 339.84 W[/tex]
The minimum power required to accelerate air to a velocity of 8 m/s at a rate of 9 m^3/s with a density of 1.18 kg/m^3 is roughly 339.84 Watts.
Explanation:
The subject of this question is physics, specifically the topic of power in mechanical systems. To find the minimum power required, we use the formula Power = 0.5 * density * volume flow rate * velocity^2. Plugging in the given values: Power = 0.5 * 1.18 kg/m^3 * 9 m^3/s * (8 m/s)^2, we get approximately 339.84 W. Therefore, the minimum power that must be supplied to the fan is roughly 339.84 Watts.
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a) How fast must a meter stick be moving if its length is observed to shrink to 0.6 m? b) With what speed will a clock have to be moving in order to run at a rate that is one-half the rate of a clock at rest?
Answer:
A)
0.8 c
B)
0.87 c
Explanation:
A)
L₀ = Original length of the meter stick = 1 m
L = Length observed = 0.6 m
[tex]v[/tex] = speed of the meter stick
Using the equation
[tex]L = L_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.6 = 1 \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.36 = 1 - \left ( \frac{v}{c} \right )^{2}[/tex]
[tex]v[/tex] = 0.8 c
B)
T₀ = Time of the clock at rest = t
T = Time of the clock at motion = (0.5) t
[tex]v[/tex] = speed of the clock
Using the equation
[tex]T = T_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 t = t \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
[tex]0.5 = \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}[/tex]
v = 0.87 c
Three point charges are arranged along the x-axls. Charge q1 = +4.25 μC Is at the origin, and charge q2 =-5.75 pC ls at x = 0.230 m. charge q3 =-9.00 pC. Where ls σ3 located if the net force on q1 ls 6.00 N In the -x-direction?
Hey there!:
q1 = 4.25 * 10^-6 C is at x1 = 0 m
q2 = - 5.75 * 10^-6 C is at x2 = 0.23 m
q3 = - 9 uC = -9 * 10^-6 C
for the net force on q1 to be in -x direction
q3 must be placed to the left of q1
the net force on q1 , F = k * q1 * q3 /( x3^2) - k * q1 * q2 /( x2 - x3)^2
6 = 9 * 10^9 * 10^-6 * 4.25 * 10^-6 * (9 /x3^2 - 5.75 /( 0.25 - x3 )^2 )
solving for x3
x3 = - 0.22 m
the charge q3 is placed at x3 = - 0.22 m
Hope this helps!
An ideal gas in a sealed container has an initial volume of 2.50 L. At constant pressure, it is cooled to 21.00 ∘C, where its final volume is 1.75 L. What was the initial temperature?
Answer: The initial temperature of the system comes out to be 147 °C
Explanation:
To calculate the initial temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]
where,
[tex]V_1\text{ and }T_1[/tex] are the initial volume and temperature of the gas.
[tex]V_2\text{ and }T_2[/tex] are the final volume and temperature of the gas.
We are given:
[tex]V_1=2.50L\\T_1=?K\\V_2=1.75L\\T_2=25^oC=(25+273)K=294K[/tex]
Putting values in above equation, we get:
[tex]\frac{2.50L}{T_1}=\frac{1.75L}{294K}\\\\T_1=420K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]420=T(^oC)+273\\T(^oC)=147^oC[/tex]
Hence, the initial temperature of the system comes out to be 147 °C
You're driving your pickup truck around a curve that has a radius of 22 m.How fast can you drive around this curve before a steel toolbox slides on the steel bed of the truck?
The maximum speed at which the truck can briefly move around the curve without making the toolbox slide can be found by setting the sliding acceleration equal to the centripetal acceleration (v²/r) and solving for v. Factors like tire friction and flat terrain are also taken into account.
Explanation:To determine how fast the truck can go before the toolbox starts to slide, we need to calculate the maximum static friction (the force that prevents sliding).
The formula to identify the maximum speed that didn't result in sliding, we would use centripetal acceleration, which is the product of tangential speed squared divided by the curve's radius.
The acceleration of the crate on the truck bed must equal centripetal acceleration to prevent sliding. Given that the sliding acceleration is 2.06 m/s², setting this value to equal centripetal acceleration (v²/r where r is 22 m) and solving for v (velocity/speed), would provide the maximum speed at which the truck can move without the toolbox sliding.Tire friction also plays a role, as it allows vehicles to move at higher speeds without sliding off the road.
This also assumes that the road is relatively flat and the truck moves uniformly, similar to the situation illustrated in Figure 24.7 with a motorist moving in a straight direction.
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The speed at which a toolbox in a truck will start to slide when the truck is turning around a curve depends on the balance of the centripetal force and static friction force. These depend on the truck's speed, the curve's radius, the toolbox's mass and the friction coefficient. Without specific values, a numerical answer can't be given.
Explanation:The question is asking about the speed at which the toolbox in the truck would start sliding due to the forces acting upon it when the truck turns around a curve. This is related to centripetal force and frictional force.
When the truck turns, the toolbox experiences a centripetal force which pushes it towards the center of the curve. At the same time, friction between the toolbox and the truck bed is resisting this pushing force, keeping the box in place. At some point, if the truck is going too fast, the centripetal force will overcome friction, and the toolbox will start to slide.
We can use the formula for centripetal force, which is F = mv²/r, where m is the mass of the toolbox, v is the velocity of the truck and r is the radius of the curve. And we know that the maximum static friction force (F_max) is f_s * m * g, where f_s is the static friction coefficient and g is the acceleration due to gravity.
The toolbox starts to slide when F = F_max, so we can find the maximum speed (v_max) before sliding happens by making these two equations equal to each other and solving for v. Without the specific values for the mass of the toolbox and the static friction coefficient, we cannot calculate a numerical answer.
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Given the size, shape and material of a conductor, what else would you need in order to find the current through the conductor?
Explanation:
The number of charges flowing per second inside the conductor is called the current through the conductor. Mathematically, it is given by :
[tex]I=\dfrac{Q}{t}[/tex]
Where
Q is the charge
t is time
Since, Q = ne
n is number of electrons
e is the charge on one electron
So, to find the current through the conductor we must know the number of electrons moving per second through it. Hence, this is the required solution.
Four objects are situated along the y axis as follows: a 2.00-kg object is at 13.00 m, a 3.00-kg object is at 12.50 m, a 2.50-kg object is at the origin, and a 4.00-kg object is at 20.500 m. Where is the center of mass of these objects?
The center of mass of these objects is located at approximately 12.652 meters on the y-axis.
We can find the center of mass (center of gravity) of these objects by calculating the weighted average of their positions along the y-axis. Here's how:
Consider Each Object:
We have four objects with masses m1, m2, m3, and m4 at positions y1, y2, y3, and y4 respectively.
Center of Mass Formula:
The center of mass coordinate (y_cm) is calculated as:
y_cm = (Σ(mi * yi)) / Σ(mi)
Σ (sigma) represents summation over all objects (i = 1 to 4 in this case).
mi is the mass of the i-th object.
yi is the y-axis position of the i-th object.
Apply the formula to our case:
y_cm = [(2.00 kg * 13.00 m) + (3.00 kg * 12.50 m) + (2.50 kg * 0.00 m) + (4.00 kg * 20.500 m)] / (2.00 kg + 3.00 kg + 2.50 kg + 4.00 kg)
Calculate:
y_cm = [26.00 kgm + 37.50 kgm + 0.00 kgm + 82.00 kgm] / 11.50 kg
y_cm ≈ 145.50 kg*m / 11.50 kg ≈ 12.652 m (rounded to four decimal places)
Therefore, the center of mass of these objects is located at approximately 12.652 meters on the y-axis.
In jumping to block a shot, a volleyball player with a weight of 600 N generates an average vertical ground reaction force of 900 N for 0.37 seconds. What is the net vertical impulse that causes her velocity to increase in the upward direction?
Answer:
111 N
Explanation:
weight (downwards) = 600 N
reaction force (upwards) = 900 N
t = 0.37 s
Net force
F = reaction force - weight = 900 - 600 = 300 N
Impulse = force x time = 300 x 0.37 = 111 N
Consider two copper wires. One has twice the length and twice the cross-sectional area of the other. How do the resistances of these two wires compare? A) Both wires have the same resistance. B) The shorter wire has twice the resistance of the longer wire. The longer wire has twice the resistance of the shorter wire. D) The longer wire has four times the resistance of the shorter wire. E) The shorter wire has four times the resistance of the longer wire.
Final answer:
The longer wire will have twice the resistance of the shorter wire.
Explanation:
The resistance of a wire depends on both its length and cross-sectional area. In this case, the longer wire has twice the length and twice the cross-sectional area of the shorter wire. The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area. Therefore, the longer wire will have twice the resistance of the shorter wire.
Nitroglycerin flows through a pipe of diameter 3.0 cm at 2.0 m/s. If the diameter narrows to 0.5 cm, what will the velocity be?
Answer:
72 m/s
Explanation:
D1 = 3 cm, v1 = 2 m/s
D2 = 0.5 cm,
Let the velocity at narrow end be v2.
By use of equation of continuity
A1 v1 = A2 v2
3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2
v2 = 72 m/s
A batter hits a baseball into the air. The formula y=-16x^2+64x+5models the baseball’s height above the ground, y, in feet, x seconds after it is hit.When does the baseball reach its maximum height? What is that height?
Explanation:
A batter hits a baseball into the air. The formula that models the baseball’s height above the ground, y, in feet, x seconds after it is hit is given by :
[tex]y=-16x^2+64x+5[/tex].........(1)
(a) We need to find the maximum height reached by the baseball. The maximum height reached is calculated by differentiating equation (1) wrt x as :
[tex]\dfrac{dy}{dx}=0[/tex]
[tex]\dfrac{d(-16x^2+64x+5)}{dx}=0[/tex]
[tex]-32x+64=0[/tex]
x = 2 seconds
(b) The height of the ball is given by equation (1) and putting the value of x in it as :
[tex]y=-16(2)^2+64(2)+5[/tex]
y = 69 feets
So, at 2 seconds the baseball reaches its maximum height and its maximum height is 69 feets.
Answer:
wow
Explanation:
A 1210 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating? (b) What is the resistance of the heating coil? (c) How much thermal energy is produced in 4.07 h?
Answer:
(a) 10.52 A
(b) 11 ohm
(c) 1.77 x 10^7 J
Explanation:
P = 1210 W, V = 115 V,
Let i be the current.
(a) Use the formula of power
P = V i
1210 = 115 x i
i = 10.52 A
(b) Let the resistance of heating coil is R.
Use the formula given below
V = i x R
R = V / i = 115 / 10.52 = 11 ohm
(c) Use the formula for energy
E = V x i x t
E = 115 x 10.52 x 4.07 x 60 x 60 = 1.77 x 10^7 J
A hydrogen atom contains a single electron that moves in a circular orbit about a single proton. Assume the proton is stationary, and the electron has a speed of 6.1 105 m/s. Find the radius between the stationary proton and the electron orbit within the hydrogen atom.
Answer:
Radius between electron and proton[tex] = 6.804\times 10^{-10}m[/tex]
Explanation:
The motion of the electron is carried out in the orbit due to the balancing of the electrostatic force between the proton and the electron and the centripetal force acting on the electron.
The electrostatic force is given as = [tex]\frac{kq_1q_2}{r^2}[/tex]
Where,
k = coulomb's law constant (9×10⁹ N-m²/C²)
q₁ and q₂ = charges = 1.6 × 10⁻¹⁹ C
r = radius between the proton and the electron
Also,
Centripetal force on the moving electron is given as:
=[tex]\frac{m_eV^2}{r}[/tex]
where,
[tex]m_e[/tex] = mass of the electron (9.1 ×10⁻³¹ kg)
V = velocity of the moving electron (given: 6.1 ×10⁵ m/s)
Now equating both the formulas, we have
[tex]\frac{kq_1q_2}{r^2}[/tex] = [tex]\frac{m_eV^2}{r}[/tex]
⇒[tex]r = \frac{kq_1q_2}{m_eV^2}[/tex]
substituting the values in the above equation we get,
[tex]r = \frac{9\times 10^{9}\times (1.6\times 10^{-19})^2}{9.1\times 10^{-31}\times (6.1\times 10^5)^2}[/tex]
⇒[tex]r = 6.804\times 10^{-10}m[/tex]
You throw a tennis ball straight up (neglect air resistance). It takes 7.0 seconds to go up and then return to your hand. How fast did you throw the ball?
Answer:
Velocity of throwing = 34.335 m/s
Explanation:
Time taken by the tennis ball to reach maximum height, t = 0.5 x 7 = 3.5 seconds.
Let the initial velocity be u, we have acceleration due to gravity, a = -9.81 m/s² and final velocity = 0 m/s
Equation of motion result we have v = u + at
Substituting
0 = u - 9.81 x 3.5
u = 34.335 m/s
Velocity of throwing = 34.335 m/s
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electric field at a point between the two surfaces?
Answer:
Electric field, E = 45.19 N/C
Explanation:
It is given that,
Surface charge density of first surface, [tex]\sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2[/tex]
Surface charge density of second surface, [tex]\sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2[/tex]
The electric field at a point between the two surfaces is given by :
[tex]E=\dfrac{\sigma}{2\epsilon_o}[/tex]
[tex]E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}[/tex]
[tex]E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}[/tex]
E = 45.19 N/C
So, the magnitude of the electric field at a point between the two surfaces is 45.19 N/C. Hence, this is the required solution.
Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.3 kg · m2 and an angular velocity of +6.6 rad/s. Disk B is rotating with an angular velocity of -9.3 rad/s. The two disks are then linked together without the aid of any external torques, so that they rotate as a single unit with an angular velocity of -2.1 rad/s. The axis of rotation for this unit is the same as that for the separate disks. What is the moment of inertia of disk B?
The angular momentum of a rotation object is the product of its moment of inertia and its angular velocity:
L = Iω
L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Apply the conservation of angular momentum. The total angular momentum before disks A and B are joined is:
L_{before} = (3.3)(6.6) + B(-9.3)
L_{before} = -9.3B+21.78
where B is the moment of inertia of disk B.
The total angular momentum after the disks are joined is:
L_{after} = (3.3+B)(-2.1)
L_{after} = -2.1B-6.93
L_{before} = L_{after}
-9.3B + 21.78 = -2.1B - 6.93
B = 4.0kg·m²
The moment of inertia of disk B is 4.0kg·m²
The total electric field consists of the vector sum of two parts. One part has a magnitude of E1 = 1300 N/C and points at an angle θ1 = 35o above the +x axis. The other part has a magnitude of E2 = 1700 N/C and points at an angle θ2 = 55o above the +x axis. Find the magnitude and direction of the total field. Specify the directional angle relative to the x axis.
Answer:
2954.6 N/C, 46.36 degree from positive axis
Explanation:
E1 = 1300 N/C, θ1 = 35 degree
E2 = 1700 N/C, θ2 = 55 degree
Now write the electric fields in vector form
E1 = 1300 ( Cos 35 i + Sin 35 j) = 1064.9 i + 745.6 j
E2 = 1700 ( Cos 55 i + Sin 55 j) = 975.08 i + 1392.6 j
Resultant electric field
E = E1 + E2
E = 1064.9 i + 745.6 j + 975.08 i + 1392.6 j
E = 2039.08 i + 2138.2 j
Magnitude of E
E = sqrt (2039.08^2 + 2138.2^2)
E = 2954.6 N/C
Let it makes an angle Φ from X axis
tan Φ = 2138.2 / 2039.08 = 1.049
Φ = 46.36 degree from positive X axis.
This question is about calculating the overall electric field from two individual fields using vector addition. This involves resolving each field into its components, adding the respective components, and then using Pythagoras' theorem and the tangent inverse relation to determine the overall magnitude and direction of the resultant field.
Explanation:The total electric field can be determined by using vector addition to add together the individual electric fields due to each of the two components mentioned in your question. Firstly, we resolve each field into its x and y components. The x-component and y-component for E1 are E1*cos(θ1) and E1*sin(θ1), respectively. Similarly, the x and y components for E2 are E2*cos(θ2) and E2*sin(θ2), respectively.
We then add together the respective x and y components: E(x total) = E1*cos(θ1) + E2*cos(θ2) and E(y total) = E1*sin(θ1) + E2*sin(θ2).
The overall magnitude of the resultant electric field can be calculated using Pythagoras' theorem, √[(E(x total))^2 + (E(y total))^2]. The direction of the total electric field can be evaluated using the tangent inverse relation θ total = tan^-1 [E(y total)/E(x total)].
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A 2.4-m-diameter merry-go-round with a mass of 270 kg is spinning at 20 rpm. John runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 34 kg . Part A Part complete What is the merry-go-round's angular speed, in rpm, after John jumps on?
Answer:
23.98 rpm
Explanation:
d = diameter of merry-go-round = 2.4 m
r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m
m = mass of merry-go-round = 270 kg
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²
M = mass of john = 34 kg
Moment of inertia of merry-go-round and john together after jump is given as
I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²
w = final angular speed
w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s
v = speed of john before jump
using conservation of angular momentum
Mvr + I w₀ = I' w
(34) (5) (1.2) + (194.4) (2.093) = (243.36) w
w = 2.51 rad/s
w = 23.98 rpm
The motor in normal operation carries a direct current of 0.750 A when connected to a 120 V power supply. The resistance of the motor windings is 19.6 Ω. Your supervisor asks you to determine by what factor the rate of change of internal energy in the windings will increase if the rotor seizes while it is operating and the supply voltage is not cut off.
Answer:
Factor by which the intrnal energy changes is 66.63
Explanation:
Given:
When the is motor operated normally, the current through the motor,I₁ =0.750A
Motor winding resistance, R = 19.6Ω
Thus, the Power dissipated at Normal condition,(P₁):
P₁=I₁²×R
P₁=(0.750)²×19.6= 11.025 W
Also,
Power = Rate of change of energy = [tex]\frac{dE}{dt}[/tex]
Now,
when the motor is connected to 120 V power supply, the current through the motor (I₂) =[tex]\frac{V}{R}[/tex]
or
I₂ =[tex]\frac{120}{19.6}A[/tex]
or
I₂ =6.122 A
Thus, the power dissipated due the current I₂ will be
P₂ = I₂² × R
⇒ P₂ = (6.122)² × 19.6 = 734.693 W
Hence, the factor by which the internal energy rate in resistor winding increases is = [tex]\frac{P_2}{P_1}=\frac{734.693}{11.025}=[/tex]66.63
The solution involves using Ohm's Law and the formula for power dissipation to determine the increase in the rate of change of internal energy in the motor windings when the rotor seizes and compare it to normal operating conditions.
Explanation:The student's question pertains to a direct current motor and involves calculating the rate of change of internal energy, back electromotive force (emf), and current under various scenarios. To address the student's question, one would need to apply Ohm's Law (V = IR, where V is voltage, I is current, R is resistance) and the power dissipation formula (P = I²R, where P is power).
Explanation of Solution
Initially, the motor's normal operation can be described by the values given: I = 0.750 A, V = 120 V, R = 19.6 Ω. The power dissipated (P) by the motor windings when the motor is operating normally would be P = I²R = (0.750 A)²(19.6 Ω).
However, if the rotor seizes, the motor will essentially become a static resistor and the current through it will increase. This is because without the back emf generated by the rotating rotor, the voltage across the motor windings will be the full supply voltage. The current in this case (let's call it I') can be calculated using Ohm's Law: I' = V/R = 120 V / 19.6 Ω. The new power dissipation (P') when the rotor seizes would be P' = I'²R. To find by what factor the rate of change of internal energy increases, we would need to compare P' to the original power dissipation P by calculating P'/P.
A police car has an 800-Hz siren. It is traveling at 35.0 m/s on a day when the speed of sound through air is 340 m/s. The car approaches and passes an observer who is standing along the roadside. What change of frequency does the observer hear?
Answer:
The observer experienced a change I'm frequency of 6971.42Hertz
Explanation:
In waves, the frequency of a wave is directly proportional to its velocity i.e F ∝ V
F = kV
k = F/V
F1/V1 = F2/V2 = k... 1
Given F1 = 800Hz at V1 = 35m/s
F2 = ? when V2 = 340m/s
Substituting this given datas into equation 1 to get the new frequency F2:
800/35 = F2/340
35F2 = 800×340
F2 = 800×340/35
F2 = 7771.42Hertz
Change of frequency = F2-F1
Change of frequency = 7771.42-800
Change of frequency = 6971.42Hertzz
The change of frequency does the observer hear due to police car siren is 6971.42 Hz.
What is frequency?Frequency of wave is the number of waves, which is passed thorough a particular point at a unit time.
For the cars approaching to the observer, the Doppler formula can be given as
[tex]\dfrac{f_1}{v_1}=\dfrac{f_2}{v_s}[/tex]
Here, [tex](v_s)[/tex] is the speed of the sound.
A police car has an 800-Hz siren. It is traveling at 35.0 m/s on a day when the speed of sound through air is 340 m/s.
As, the car approaches and passes an observer. Thus by the above formula,
[tex]\dfrac{800}{35}=\dfrac{f_2}{340}\\f_2=7771.41\rm Hz[/tex]
As the frequency of police car siren is 800-Hz. Thus, the change of the frequency observed by the observer is,
[tex]\Delta f=7771.42-8000\\\Delta f=6971.42\rm Hz[/tex]
Thus, change of frequency does the observer hear due to police car siren is 6971.42 Hz.
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In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge that are traveling perpendicular to the fields at 8.75 km s. The magnetic field available to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions.
Answer:
Part a)
E = 4812.5 N/C
Part b)
Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero
Explanation:
For uniform electric and magnetic field we will have a charge particle moving through it such that net force on it is zero
so here we have
[tex]qE = qvB[/tex]
magnetic force on moving charge will balanced by electrostatic force on moving charge
[tex]v = \frac{E}{B}[/tex]
now we have
[tex]v = 8.75 km/s[/tex]
B = 0.550 T
now we have
[tex]E = (8.75 \times 10^3)(0.550)[/tex]
[tex]E = 4812.5 N/C[/tex]
Part b)
Two field must be perpendicular to each other as well as perpendicular to the velocity of charge so that net force on the moving charge will be zero