Answer:
15.6 m
Explanation:
d = diameter of the wheel = 43 cm = 0.43 m
r = radius of the wheel = (0.5) d = (0.5) (0.43) = 0.215 m
w₀ = initial angular velocity = 86 rpm = 9.01 rad/s
v₀ = initial linear speed = r w₀ = (0.215) (9.01) = 1.94 m/s
w = final angular velocity = 362 rpm = 37.91 rad/s
v = final linear speed = r w = (0.215) (37.91) = 8.15 m/s
t = time taken = 3.1 s
Using the equation
v = v₀ + at
8.15 = 1.94 + a (3.1)
a = 2 m/s²
d = distance traveled by a point at the edge of the wheel
Using the equation
d = v₀ t + (0.5) at²
d = (1.94) (3.1) + (0.5) (2) (3.1)²
d = 15.6 m
A 6.0-μF air capacitor is connected across a 100-V battery. After the battery fully charges the capacitor, the capacitor is immersed in transformer oil (dielectric constant = 4.5). How much additional charge flows from the battery, which remained connected during the process?
Final answer:
An additional charge of 2100 μC flows from the battery into the capacitor when it is immersed in transformer oil with a dielectric constant of 4.5.
Explanation:
To determine how much additional charge flows from the battery when a 6.0-μF air capacitor is immersed in transformer oil with a dielectric constant of 4.5, we need to examine the effect of the dielectric on the capacitor's capacitance.
Initially, the capacitance of the air capacitor Cair is 6.0 μF. The charge Qinitial on the capacitor when connected to a 100-V battery is given by:
Qinitial = Cair × Vbattery
Qinitial = 6.0 μF × 100 V
Qinitial = 600 μC
When the capacitor is immersed in oil, the capacitance increases due to the dielectric constant (κ) of the oil:
Coil = κ × Cair
Coil = 4.5 × 6.0 μF = 27.0 μF
Since the battery remains connected, the voltage across the capacitor stays at 100 V, so the new charge Qfinal becomes:
Qfinal = Coil × Vbattery
Qfinal = 27.0 μF × 100 V
Qfinal = 2700 μC
The additional charge ΔQ that flows from the battery is the difference between Qfinal and Qinitial:
ΔQ = Qfinal - Qinitial
ΔQ = 2700 μC - 600 μC
ΔQ = 2100 μC
Therefore, an additional charge of 2100 μC flows from the battery into the capacitor.
A particle with a charge of 33 μC moves with a speed of 74 m/s in the positive x direction. The magnetic field in this region of space has a component of 0.41 T in the positive y direction, and a component of 0.86 T in the positive z direction. Part A What is the magnitude of the magnetic force on the particle? Express your answer using two significant figures.
Explanation:
33×10^-6 ×74 ×(0.86 - 0.41)
A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is the magnification?
Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object
[tex]\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}[/tex]
Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula
[tex]\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}[/tex]
[tex]\dfrac{1}{u}=\dfrac{103}{4080}[/tex]
[tex]u =\dfrac{4080}{103}[/tex]
The magnification is
[tex]m = \dfrac{-v}{u}[/tex]
[tex]m=\dfrac{-240\times103}{4080}[/tex]
[tex]m = -6.05[/tex]
Hence, The magnification is -6.05.
An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?
Answer:
10 seconds
Explanation:
x = x₀ + v₀ t + ½ at²
250 = 0 + (0) t + ½ (5) t²
250 = 2.5 t²
t² = 100
t = 10
It takes 10 seconds to land from a height of 250 ft.
Answer:
10 seconds
Explanation:
It takes 10 seconds for an object to fall from an elevation of 250 ft.
250 = 0 + (0) t + ½ (5) t²
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, what must have been the reference pressure?
Given:
I = 30dB
P = 66 × [tex]10^{-9}[/tex] Pa
Solution:
Formula used:
I = [tex]20\log_{10}(\frac{P}{P_{o}})[/tex] (1)
where,
I = intensity of sound
P = absolute pressure
[tex]P_{o}[/tex] = reference pressure
Using Eqn (1), we get:
[tex]30 = 20\log _{10}\frac{66\times 10^{-9}}{P^{o}}[/tex]
[tex]P_{o}[/tex] = [tex]\frac{66\times 10^{-9}}{10^{1.5}}[/tex]
[tex]P_{o}[/tex] = 2.08 × [tex]10^{-9}[/tex] Pa
The reference pressure for a sound intensity level of 0 dB is always 20 micropascals, or 2 x 10^-5 Pa, regardless of the absolute pressure of the sound.
If a sound is 30 dB and its absolute pressure was 66 x 10-9 Pa, we need to find the reference pressure. The reference pressure is known as the threshold of hearing and corresponds to a sound intensity level of 0 dB. In acoustics, 0 dB is quantified relative to a reference which has been set at a sound pressure level of 20 micropascals, equivalent to 2 x 10-5 Pa. The question of what is the reference pressure can be answered easily: the reference pressure is always 20 micropascals or 2 x 10-5 Pa, because the decibel scale is logarithmic and based on this fixed reference.
A police car on the side of the road (at rest) uses a radar gun to catch speeders as they approach it. The frequency the radar gun emits is 8 x 109 Hz and the speed limit is 65 mi/hr. What is the difference in frequency of the emitted and returned radar wave
Final answer:
Using the Doppler Effect formulae for electromagnetic waves, the difference in frequency (Dopler shift) experienced by radar signals upon hitting and returning from a moving vehicle allows a radar gun to calculate the vehicle's speed. The principle involves the use of shift in frequency and the speed of light to measure the speed at which the vehicle is moving.
Explanation:
The question involves the Doppler Effect in physics, specifically its application in a radar speed trap. To find the frequency shift, we must use the Doppler Effect formulae for electromagnetic waves. However, it seems there might have been a mix-up with the frequencies provided in several example problems. Since those seem to be examples rather than the actual frequencies we are working with, let's focus on finding the principle behind calculating the speed of the vehicle based on a known frequency shift of a radar emission and its return signal.
The frequency shift (Δf) in the Doppler Effect for electromagnetic waves such as radar can be calculated by the formula: Δf = (2 * f * v) / c, where f is the original frequency emitted by the radar gun, v is the speed of the vehicle, and c is the speed of light. The factor of 2 is because the radar signal experiences a frequency shift once when it hits the moving vehicle and another shift when the echo returns. The radar gun's internal processors calculate the difference in frequency (the Dopler shift) to find the speed of the vehicle. For accurate measurement, the radar unit must be able to discern even small frequency shifts to effectively differentiate speeds with fine resolution.
A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?
Answer:
31.1 N
Explanation:
m = mass attached to string = 0.50 kg
r = radius of the vertical circle = 2.0 m
v = speed of the mass at the highest point = 12 m/s
T = force of the string on the mass attached.
At the highest point, force equation is given as
[tex]T + mg =\frac{mv^{2}}{r}[/tex]
Inserting the values
[tex]T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}[/tex]
T = 31.1 N
Answer: 41N
Explanation :
T= mv^2/R + mgcos θ
At the highest point on the circle θ=0
Cos 0 = 1
T= mv^2/R + mg
m = 0.5kg
Velocity at the highest point (amplitude)= 12m/s
T = 0.5× 12^2/2 + 0.5×10
0.5×144/2 +5
T = 0.5×72 + 5
T = 36+5
T = 41N
3. Which statement is not true of simple machines?
A They can change the direction in which you apply a force
B They can change the amount of force you exert.
CThey can change the amount of work done on an object.
D. They can change the distance over which you apply a force
Answer:
C They can change the amount of work done on an object. Controlling amount of work done would be down to user / more complex machines.
Explanation:
Simple machines help direct, increase, and affect distance force is applied.
Simple machines do not change the amount of work done but can alter the force applied and distance over which the force is exerted.
Simple machines do not change the amount of work done. Although they cannot alter the work done, they can change the amount of force applied and the distance over which the force is exerted. For instance, a simple machine like a lever can help reduce the force needed to lift an object.
An uncharged 5.0-µF capacitor and a resistor are connected in series to a 12-V battery and an open switch to form a simple RC circuit. The switch is closed at t = 0 s. The time constant of the circuit is 4.0 s. What is the charge on either plate after one time constant has elapsed?
For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:
Q(t) = Cℰ(1-e^{-t/(RC)})
Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.
The time constant of the circuit τ is the product of the resistance and capacitance:
τ = RC
Q(t) can be rewritten as:
Q(t) = Cℰ(1-e^{-t/τ})
We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Plug in these values and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
The charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
The charge stored on one of the capacitor's plates can be calculated by,
[tex]\bold{Q(t) = C\epsilon(1-e^{-t/(RC)})}[/tex]
Where,
Q(t) - the charge,
t - time,
ℰ- the battery's terminal voltage,
R - the resistor's resistance,
C - the capacitor's capacitance.
The time constant of the circuit τ is equal to the product of the resistance R and capacitance C :
τ = RC
The amount of charge Q(t) when t = τ. put the values,
Q(τ) = Cℰ(1-e^{-τ/τ})
Q(τ) = Cℰ(1-1/e)
Q(τ) = Cℰ(0.63)
Given values:
C = 5.0×10⁻⁶F
ℰ = 12V
Put the values in the formula and solve for Q(τ):
Q(τ) = (5.0×10⁻⁶)(12)(0.63)
Q(τ) = 3.8×10⁻⁵C
Therefore, the charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.
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A seismographic station receives S and P waves from an earthquake, separated in time by 17.8 s. Assume the waves have traveled over the same path at speeds of 4.50 km/s and 6.90 km/s. Find the distance from the seismograph to the focus of the quake.
Answer:
D = 230.2 Km
Explanation:
let distance between seismograph and focus of quake is D
From time distance formula we can calculate the time taken by the S wave
[tex]T_1 =\frac{D}{4.5}[/tex]
From time distance formula we can calculate the time taken by the P wave
[tex]T_2 =\frac{D}{6.90}[/tex]
It is given in equation both waves are seperated from each other by 17.8 sec
so we have
[tex]T_1 - T_2 = 17.8[/tex]sec
Putting both time value to get distance value
[tex]\frac{D}{4.5} - \frac{D}{6.90} = 17.8[/tex]
D = 230.2 Km
Calculate the force of gravity on the 1.2-kg mass if it were 1.9×107 m above earth's surface (that is, if it were four earth radii from earth's center).
Answer:
Force of gravity, F = 0.74 N
Mass of an object, m = 1.2 kg
Distance above earth's surface, [tex]d=1.9\times 10^7\ m[/tex]
Mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]
Radius of Earth, [tex]r=6.37\times 10^6\ m[/tex]
We need to find the force of gravity above the surface of Earth. It is given by :
[tex]F=G\dfrac{mM}{R^2}[/tex]
R = r + d
R = 25370000 m
[tex]F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}[/tex]
F = 0.74 N
So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.
The force of gravity exerted on the 1.2-kg mass located four Earth radii from Earth's center is 0.059 N. This result showcases the inverse-square law as the gravitational force is significantly less than the same mass at Earth’s surface (11.76 N).
Explanation:The force of gravity on an object depends on its mass and its distance from the center of the Earth. The general equation to calculate this force is F = GMm/r², where F is the gravitational force, G is the universal gravitational constant (6.67x10^-11 N(m/kg)²), M is the mass of the Earth (5.98x10^24 kg), m is the mass of the object (in this case, 1.2 kg), and r is the distance from the center of the Earth.
Put strictly, the stated distance must be added to Earth's radius (6.37x10^6 m) to find the total distance from Earth's center. So, r = Earth’s radius + the object’s height from the surface, which gives us r = 6.37x10^6 m + 1.9x10^7 m = 2.537x10^7 m.
Substituting these values into the equation gives F = (6.67x10^-11 N(m/kg)² * 5.98x10^24 kg * 1.2 kg) / (2.537x10^7 m)² = 0.059 N.To put this into perspective, the weight of the same mass (1.2 kg) on the surface of the Earth would be 1.2 kg * 9.8 m/s² (acceleration due to gravity at Earth’s surface) = 11.76 N. This demonstrates the inverse-square law of gravity, where the gravitational force decreases with the square of the distance from the source.
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A 300.0 W immersion heater is used to heat 0.250 kg of water from 10.0 °C to 70.0 °C. About how many seconds did this take? (There are 4186 Joules per kcal.)
Answer:
209.3 seconds
Explanation:
P = 300 W, m = 0.250 kg, T1 = 10 degree C, T2 = 70 degree C
c = 4186 J / kg C
Heat given to water = mass x specific heat of water x rise in temperature
H = 0.250 x 4186 x (70 - 10)
H = 62790 J
Power = Heat / Time
Time, t = heat / Power
t = 62790 / 300 = 209.3 seconds
It took approximately [tex]\( 209.3 \)[/tex] seconds to heat the water from [tex]\( 10.0^\circ \text{C} \)[/tex] to [tex]\( 70.0^\circ \text{C} \)[/tex].
To determine how many seconds it took to heat the water using the immersion heater, we need to calculate the amount of energy required and then use the power of the heater to find the time.
First, calculate the change in temperature of the water:
[tex]\[ \Delta T = 70.0^\circ \text{C} - 10.0^\circ \text{C} = 60.0^\circ \text{C} \][/tex]
Next, calculate the energy [tex]\( Q \)[/tex] required to heat the water using the specific heat capacity of water [tex]\( C = 4186 \text{ J/kg}^\circ \text{C} \)[/tex]:
[tex]\[ Q = mc\Delta T \][/tex]
where [tex]\( m \)[/tex] is the mass of water and [tex]\( c \)[/tex] is the specific heat capacity of water.
Given:
[tex]\[ m = 0.250 \text{ kg} \][/tex]
[tex]\[ c = 4186 \text{ J/kg}^\circ \text{C} \][/tex]
[tex]\[ \Delta T = 60.0^\circ \text{C} \][/tex]
[tex]\[ Q = 0.250 \times 4186 \times 60.0 \][/tex]
[tex]\[ Q = 62790 \text{ Joules} \][/tex]
Now, calculate the time t required using the power P of the heater:
[tex]\[ P = 300.0 \text{ W} \][/tex]
The time [tex]\( t \)[/tex] is given by:
[tex]\[ t = \frac{Q}{P} \][/tex]
[tex]\[ t = \frac{62790}{300.0} \][/tex]
[tex]\[ t = 209.3 \text{ seconds} \][/tex]
The sides of a square increase in length at a rate of 3 m/sec. a. At what rate is the area of the square changing when the sides are 14 m long? b. At what rate is the area of the square changing when the sides are 25 m long?
The area of a square is given by:
A = s²
A is the square's area
s is the length of one of the square's sides
Let us take the derivative of both sides of the equation with respect to time t in order to determine a formula for finding the rate of change of the square's area over time:
d[A]/dt = d[s²]/dt
The chain rule says to take the derivative of s² with respect to s then multiply the result by ds/dt
dA/dt = 2s(ds/dt)
A) Given values:
s = 14m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(14)(3)
dA/dt = 84m²/s
B) Given values:
s = 25m
ds/dt = 3m/s
Plug in these values and solve for dA/dt:
dA/dt = 2(25)(3)
dA/dt = 150m²/s
When the side of the square is 14 m, the rate at which the area is changing is 84 m²/s.
When the side of the square is 25 m, the rate at which the area is changing is 150 m²/s.
The given parameters;
rate at which the side of the square is increasing, [tex]\frac{dl}{dt}[/tex] = 3 m/sThe area of the square is calculated as;
A = L²
The change in the area is calculated as;
[tex]\frac{dA}{dt} = 2l\frac{dl}{dt}[/tex]
When the side of the square is 14 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 14 \times 3\\\\\frac{dA}{dt} = 84 \ m^2/s[/tex]
When the side of the square is 25 m, the rate at which the area is changing is calculated as;
[tex]\frac{dA}{dt} = 2l \frac{dl}{dt} \\\\\frac{dA}{dt} = 2 \times l \times \frac{dl}{dt}\\\\\frac{dA}{dt} = 2 \times 25 \times 3\\\\\frac{dA}{dt} = 150 \ m^2/s[/tex]
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In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will “stick” to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. What is the minimum rotational frequency, in rpm, for which the ride is safe?
Answer:
[tex]\omega = 2.56 rad/s[/tex]
Explanation:
As the cylinder rotates the centripetal force on all the passengers is due to normal force due to the wall
So here we can say
[tex]N = m\omega^2 R[/tex]
now when floor is removed all the passengers are safe because here friction force on the passenger is counter balanced by the weight of the passengers
so we can say
[tex]F_f = mg[/tex]
[tex]\mu_s F_n = mg[/tex]
[tex]\mu_s (m\omega^2 R) = mg[/tex]
[tex]\mu_s \omega^2 R = g[/tex]
[tex]\omega = \sqrt{\frac{g}{\mu_s R}}[/tex]
for minimum rotational speed we have
[tex]\omega = \sqrt{\frac{9.8}{0.60(2.5)}[/tex]
[tex]\omega = 2.56 rad/s[/tex]
The minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.
Explanation:To determine the minimum rotational frequency for which the ride is safe, we need to consider the static coefficient of friction between clothing and steel. Since the passengers are standing inside a hollow cylinder, they will experience a centrifugal force pushing them against the wall of the cylinder. To prevent sliding, the static friction force needs to be greater than or equal to the gravitational force pulling them downward. The formula to calculate the static friction force is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force.
For clothing against steel, the coefficient of static friction ranges from 0.60 to 1.0. Assuming the worst-case scenario with μs = 0.60, we can calculate the minimum rotational frequency:
Centrifugal force = m * g = m * ω^2 * R, where m is the mass of the passengers, g is the acceleration due to gravity, ω is the angular velocity in radians per second, and R is the radius of the cylinder.
Static friction force = μs * m * g
Equating these two forces, we get μs * m * g = m * ω^2 * R
Simplifying the equation, we find ω = sqrt(μs * g / R)
Converting the angular velocity to revolutions per minute (rpm), we have rpm = 60 * ω / (2 * π)
Substituting the values, the minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.
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A vibratory system in a vehicle is to be designed with the following parameters: k-295 N/m, C-2N-s/m, m-13 kg. Calculate the natural frequency of damped vibration
Answer:
[tex]w_{damped}= 4.76[/tex] s^-1
Explanation:
The mathematical relationship is
[tex]w_{damped}=w_{undamped} *\sqrt{1-(\frac{c}{2\sqrt{km}})^{2}}[/tex]
where:
c is the damper constant
k is the spring constant
m is the mass
ω_undamped is the natural frequency
ω_damped is the damped frequency
[tex]w_{undamped} =\sqrt{\frac{k}{m}}=4.79[/tex] s^-1
[tex]w_{damped}= 4.79 *\sqrt{1-(\frac{2}{2\sqrt{295*13}})^{2}}[/tex]
[tex]w_{damped}= 4.76[/tex] s^-1
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at x = 2.00 m is 1 000 V, what is the change in potential energy of a particle with a charge of + 2.80 x 10-3 C as it moves from x = 1.9 m to x = 2.1 m?
Answer:
[tex]\Delta U = 0.2072 J[/tex]
Explanation:
Potential difference between two points in constant electric field is given by the formula
[tex]\Delta V = E.\Delta x[/tex]
here we know that
[tex]E = 370 N/C[/tex]
also we know that
[tex]\Delta x = 2.1 - 1.9 = 0.2 m[/tex]
now we have
[tex]\Delta V = 370 (0.2) = 74 V[/tex]
now change in potential energy is given as
[tex]\Delta U = Q\Delta V[/tex]
[tex]\Delta U = (2.80 \times 10^{-3})(74)[/tex]
[tex]\Delta U = 0.2072 J[/tex]
An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. How much heat was generated by the current? (Ccopper = 0.093 kcal/kgC°)
Answer:
Heat generated by the current = 1547.89 J
Explanation:
We have equation for heat energy H = mCΔT
Mass of copper = 0.221 kg
Specific heat of copper = 0.093 kcal/kgC° = 389.112 J/kgC°
ΔT = 38 - 20 = 18°C
Substituting in H = mCΔT
H = 0.221 x 389.112 x 18 = 1547.89 J
Heat generated by the current = 1547.89 J
The heat generated by the electric current that heated a 221 g copper wire from 20.0 °C to 38.0 °C is calculated to be 1.542 kJ, using the specific heat capacity of copper and the formula Q = mcΔT.
An electric current heats a 221 g (0.221 kg) copper wire from 20.0 °C to 38.0 °C. To calculate the heat generated by the current, we use the formula for heat energy, Q = mcΔT, where m is the mass in kg, c is the specific heat capacity in kcal/kg°C, and ΔT is the change in temperature in °C.
Given:
m = 0.221 kg
c = 0.093 kcal/kg°C
ΔT = (38.0 - 20.0) °C = 18.0 °C
Substituting the values into the formula:
Q = 0.221 kg * 0.093 kcal/kg°C * 18.0 °C = 0.368658 kcal
To convert kcal to joules (since 1 kcal = 4.184 kJ),
Q = 0.368658 kcal * 4.184 kJ/kcal = 1.542 kJ
Therefore, the heat generated by the electric current is 1.542 kJ.
A 200-loop coil of cross sectional area 8.5 cm2 lies in the plane of the paper. Directed out of the plane of the paper is a magnetic field of 0.06 T. The field out of the paper decreases to 0.02 T in 12 milliseconds. What is the direction of the current induced?
Explanation:
It is given that,
Number of turns, N = 200
Area of cross section, A = 8.5 cm²
Magnetic field is directed out of the paper and is, B = 0.06 T
The magnetic field is out of the paper decreases to 0.02 T in 12 milliseconds. We need to find the direction of current induced. The induced emf is given by :
[tex]\epsilon=-N\dfrac{d\phi}{dt}[/tex]
Since, [tex]\epsilon=IR[/tex]
I is the induced current
[tex]I=-\dfrac{N}{R}\dfrac{d\phi}{dt}[/tex]
According to Lenz's law, the direction of induced current is such that it always opposes the change in current that causes it.
Here, the field is directed out of the plane of the paper, this gives the induced current in counterclockwise direction.
A superball and a clay ball are dropped from a height of 10cm above a tabletop. They have the same mass 0.05kg and the same size. The superball bounces off the table and rises back to the same height. The clay ball sticks to the table. For the superball in the previous question, if it was in contact with the table for 34.3ms, calculate the average force exerted on the ball by the table. Hint: First calculate the momentum before and after hitting the table. Don't forget the gravitational force.
Answer:4.08 N
Explanation:
Given data
superball dropped from a height of 10 cm
Mass of ball[tex]\left ( m\right )[/tex]=0.05kg
time of contact[tex]\left ( t\right )[/tex]=34.3[tex]\times 10^{-3}[/tex] s
Now we know impulse =[tex]Force\times time\ of\ contact[/tex]=Change in momentum
[tex]F_{average}\times t[/tex]=[tex]m\left ( v-(-v)\right )[/tex]
and velocity at the bottom is given by
v=[tex]\sqrt{2gh}[/tex]
[tex]F_{average}\times 34.3\times 10^{-3}[/tex]=[tex]0.05\left ( 1.4-(-1.4)\right )[/tex]
[tex]F_{average}[/tex]=4.081N
To find the average force exerted on the superball by the table, we calculate its change in momentum during the bounce and divide by the contact time. The result is an approximate average force of 4.08 N upward.
The question involves a superball and a clay ball, both with the same mass of 0.05 kg, dropped from a height of 10 cm. The superball bounces back to its original height, while the clay ball sticks to the table. We need to calculate the average force exerted on the superball by the table.
Step-by-Step Solution
Calculate the velocity of the superball just before hitting the table: Using the equation for free fall, v = √(2gh), where g is 9.8 m/s² (acceleration due to gravity) and h is 0.10 m.
v = √(2 * 9.8 * 0.10)
v = √(1.96)
v ≈ 1.40 m/s
Determine the change in momentum: Before impact, the momentum is [tex]P_{before[/tex] = m * v. Since the superball bounces back with the same speed, the momentum after impact is [tex]P_{after[/tex] = -m * v because the direction changes.
[tex]P_{before[/tex] = 0.05 kg * 1.40 m/s = 0.07 kg·m/s
[tex]P_{after[/tex] = 0.05 kg * (-1.40 m/s) = -0.07 kg·m/s
Calculate the impulse: Impulse is the change in momentum, so Impulse = [tex]P_{after[/tex] - [tex]P_{before[/tex].
Impulse = -0.07 kg·m/s - 0.07 kg·m/s
Impulse = -0.14 kg·m/s
Calculate the average force: The impulse-momentum theorem states that Impulse = [tex]F_{avg[/tex] * Δt, where [tex]F_{avg[/tex] is the average force and Δt is the time of contact (34.3 ms = 0.0343 s). Solving for [tex]F_{avg[/tex] ,
[tex]F_{avg[/tex] = Impulse / Δt
[tex]F_{avg[/tex] = -0.14 kg·m/s / 0.0343 s
[tex]F_{avg[/tex] ≈ -4.08 N
The magnitude of the average force exerted on the ball by the table is approximately 4.08 N in the upward direction.
Water is stored in an open stored in an open storage tank. The absolute pressure in the tank above the liquid is 1.0 atm. The water level is 8m above the base of the tank. What is the instantaneous velocity of a fluid jet when a 0.5 cm diameter orifice is open at point A. The opening is 0.8m above the base. consider v1=0 at the instant the opening is made (1)
Answer:
[tex]V_{A}= 11.88 m/s[/tex]
Explanation:
given data:
water level at point A = 8 m
diameter of orifice = 0.5 cm
velocity at point A = 0
h1 =0.8 m
h2 = 8 - h1 = 8 - 0.8 = 7.2 m
Applying Bernoulli theorem between point A and B
[tex]P_{o}+\rho _{water}gh_{2}+\frac{1}{2}\rho v_{B}^{2}+\rho _{water}gh_{1}=P_{o} +\frac{1}{2}\rho v_{A}^{2}+\rho _{water}gh_{1}[/tex]
[tex]V_{A}=\sqrt{2gh_{2}}[/tex]
[tex]V_{A}=\sqrt{2*9.81*7.2}[/tex]
[tex]V_{A}= 11.88 m/s[/tex]
An object of mass m travels along the parabola yequalsx squared with a constant speed of 5 units/sec. What is the force on the object due to its acceleration at left parenthesis 2 Superscript 1 divided by 2 Baseline comma 2 right parenthesis? (Remember Newton's law, Fequalsma.)
Explanation:
The object is moving along the parabola y = x² and is at the point (√2, 2). Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.
First, we need to find the radius of curvature. This is given by the equation:
R = [1 + (y')²]^(³/₂) / |y"|
y' = 2x and y" = 2:
R = [1 + (2x)²]^(³/₂) / |2|
R = (1 + 4x²)^(³/₂) / 2
At x = √2:
R = (1 + 4(√2)²)^(³/₂) / 2
R = (9)^(³/₂) / 2
R = 27 / 2
R = 13.5
So the centripetal force is:
F = m v² / r
F = m (5)² / 13.5
F = 1.85 m
The force on an object moving along a parabola at a point can be determined by calculating the acceleration at that point (derived from the velocity and its change), and then applying Newton's second law of motion (F=ma). The computations involve complex physics and calculus concepts.
Explanation:An object of mass m moves along a parabolic path y=x2 with constant speed, but the direction of its velocity is continuously changing, which should be considered as an acceleration and hence results in a force according to Newton's second law (F=ma).
The force on the object at a particular point (21/2 , 2) due to its acceleration can be determined by first calculating the acceleration at that point and then using Newton's second law. The details of these calculations involve some complex high school level physics and calculus concepts, but essentially involve calculating the derivative of the object's velocity with respect to time at the given point, then multiplying that by the object's mass.
Learn more about Newton's Second Law here:https://brainly.com/question/18769451
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The 0.41-kg cup of a James Bond anti-Martini-Maker is attached to a spring of spring constant 110 N/m. The cup is displaced horizontally from the equilibrium position and let go. What is the period of martini oscillation? A) 0.023 s B) 0.38 s C) 0.80s D) 100 s
Answer:
Option B is the correct answer.
Explanation:
Period of a spring is given by the expression
[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]
Here, spring constant, k = 110 N/m
Mass = 0.41 kg
Substituting,
[tex]T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.41}{110}}=0.38s[/tex]
Option B is the correct answer.
A car drives over the top of a hill that has a radius of 40 m. ? Part A What maximum speed can the car have without flying off the road at the top of the hill?
Answer:
Maximum speed = 19.81 m/s
Explanation:
Maximum speed can the car have without flying off the road at the top of the hill.
For this condition to occur we have
Centripetal force ≥ Weight of car.
[tex]\frac{mv^2}{r}\geq mg[/tex]
For maximum speed without flying we have
[tex]\frac{mv^2}{r}=mg\\\\\frac{v^2}{r}=g\\\\v=\sqrt{rg}=\sqrt{40\times 9.81}=19.81m/s[/tex]
Maximum speed = 19.81 m/s
The maximum speed of the car on top of hill is 19.8 m/s.
The given parameters;
radius of the hill, r = 40 m
The maximum speed of the car on top of hill is calculated as follows;
the centripetal force must be equal or greater than weight of the car.
[tex]F_c = mg\\\\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v = \sqrt{40 \times 9.8} \\\\v = 19.80 \ m/s[/tex]
Thus, the maximum speed of the car on top of hill is 19.8 m/s.
Learn more here:https://brainly.com/question/20905151
The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.67 from 8.19 Wb in a time of 0.0386 s. The average induced current in the coil is 275 A. What is the resistance of the wire?
A student performs this experiment and measures the bar to have a mass of 150g and length of 36cm. What is the moment of inertia of the bar?
Answer:
The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]
Explanation:
Given that,
mass of bar = 150 g
Length l = 36 cm
We need to calculate the moment of inertia of the bar
Using formula of moment inertia
[tex]I=\dfrac{1}{12}Ml^2[/tex]
Where,
M = mass of the bar
L = length of the bar
Put the value into the formula
[tex]I=\dfrac{1}{12}\times150\times10^-3\times36\times10^{-2}[/tex]
[tex]I=45\times10^{-4}\ kg-m^2[/tex]
Hence, The moment of inertia of the bar is [tex]45\times10^{-4}\ kg-m^2[/tex]
The moment of inertia of a bar with a mass of 150g and length of 36cm, rotating about its center, is approximately 0.00162 kg·m². The calculation uses the formula I = (1/12) * M * L². First, convert the mass and length to SI units and then substitute them into the formula.
To find the moment of inertia of a uniform bar with a mass of 150g and a length of 36cm, we can use the formula for a rod rotating about its center:
I = (1/12) * M * L²
Where I is the moment of inertia, M is the mass of the bar, and L is the length of the bar. Let's convert the mass to kilograms and the length to meters:Mass, M = 150g = 0.15kgLength, L = 36cm = 0.36mNow substitute these values into the formula:I = (1/12) * 0.15kg * (0.36m)²
I = (1/12) * 0.15kg * 0.1296m²
I ≈ 0.00162 kg·m²
Therefore, the moment of inertia of the bar rotating about its center is approximately 0.00162 kg·m².
The capacitance of a fully-charged capacitor is 11 F. Determine the capacitor's capacitance when it is half charged.
Answer:
The capacitance is 11 F for half and fully charged capacitor.
Explanation:
Capacitance of capacitor is given by the expression
[tex]C=\frac{\epsilon A}{d}[/tex]
Where ε is the vacuum permittivity, A is the area of plates and d is the separation between plates.
So capacitance does not depend upon charge and potential. So capacitance fully and half charged capacitors are same.
Here the capacitance is 11 F for half and fully charged capacitor.
Hi there!
[tex]\boxed{C = 11F}[/tex]
Even though C = Q/V, the capacitance is NEVER changed by the charge or voltage.
The only factors that change the capacitance of a capacitor are those related to its geometry or if a dielectric is inserted. We can look at some examples:
For a parallel plate capacitor:
[tex]C = \frac{\epsilon_0A}{d}[/tex]
C = Capacitance (F)
A = Area of plates (m²)
d = distance between plates (m)
For a spherical capacitor:
[tex]C = 4\pi \epsilon_0 (\frac{r_{outer}r_{inner}}{r_{outer} - r_{inner}}})[/tex]
Notice how the capacitance is strictly determined by its geometric properties. Therefore, changing the charge or voltage has no effect, so the capacitance will remain 11 F.
How many electrons does 1.00 kg of water contain?
Answer:
6 x (10)^26 electrons.
Explanation:
1 mole = 18 gr
1 gm =1/18 mole
1 kg = 1000/18 mole
Now , 1 mole of any compound = 6.022 x (10)^23 atoms.
Therefore, 1 kg of H20= (1000/18)*(6.022 x (10)^23) atoms
Roughly , 3.34 x (10)^25 molecules
And each molecule has 18 electrons
Therefore, 6 x (10)^26 electrons.
Thank you.
What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 12.0 ms?
Final answer:
The peak emf generated by a 1000-turn, 42.0 cm diameter coil initially perpendicular to the Earth's 5.00 × 10^-5 T magnetic field and rotated to be parallel in 12.0 ms is approximately 1.1 V.
Explanation:
Calculating the Peak EMF in a Rotating Coil
To calculate the peak emf generated by a rotating coil in a magnetic field, we can use Faraday's Law of electromagnetic induction. The formula derived from Faraday's Law for a coil with multiple turns is:
emf = -N * (change in magnetic flux)/change in time
The magnetic flux (Φ) is given by the product of the magnetic field (B), the area (A) of the coil, and the cosine of the angle (θ) between the magnetic field and the normal to the surface of the coil:
Φ = B * A * cos(θ)
The question states that the coil with 1000 turns and a 42.0 cm diameter is initially perpendicular to the Earth's magnetic field of 5.00 × 10-5 T, and then rotated to become parallel. This change goes from cos(90°), which is 0, to cos(0°), which is 1, over a time interval of 12.0 ms.
The area A of the coil is π * (radius)2, where radius is half the diameter. The radius is 42.0 cm / 2 = 21.0 cm = 0.21 m. Thus:
A = π * (0.21 m)2
Now, we plug the values into the equation to find the peak emf:
Peak emf = -(1000) * (5.00 × 10-5 T * π * (0.21 m)2 - 0) / (12.0 × 10-3 s)
After calculation:
Peak emf ≈ 1.1 V
This is the maximum emf induced in the coil during its rotation.
The peak emf generated by rotating the given coil in the Earth's magnetic field is approximately 3.63 V.
To find the peak emf generated by rotating a coil in a magnetic field, we use Faraday's Law of Induction. The peak emf (ε) can be calculated using the formula:
ε = NABω
where N is the number of turns, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity in radians per second. First, we need to find the area (A) of the coil:
A = πr²
Given the diameter of the coil is 42.0 cm, the radius (r) is 21.0 cm or 0.21 m. So,
A = π(0.21)² = 0.1385 m²
Next, we determine the angular velocity (ω) given one full rotation in 12.0 ms:
ω = 2π / T
where T is the period (12.0 ms or 0.012 s),
ω = 2π / 0.012 = 523.6 rad/s
Now, substituting the values into the emf formula:
ε = 1000 × 0.1385 m² × 5.00 × 10⁻⁵ T × 523.6 rad/s = 3.629 V
Thus, the peak emf generated is approximately 3.63 V.
A silicon wafer is used to attenuate the intensity from a laser emitting at a wavelength of 0.8 µm. If the laser output power is 100 mW what is the wafer thickness required to attenuate the power to 1 mW?
Answer:
[tex]t =4.605 *10^{-3}[/tex]
Explanation:
given data:
wavelength of emission =[tex]0.8 \mu m[/tex]
output power = 100 mW
We can deduce value of obsorption coefficient from the graph obsorption coefficient vs wavelength
for wavelength [tex]0.8 \mu m[/tex] the obsorption coefficient value is 10^{3}
intensity can be expressed as a function of thickness as following:
[tex]I(t) = I_{O} *e^{-\lambda *t}[/tex]
putting all value to get thickness
[tex]1*10^{-3} =100*10^{-3}e^{-10^{3}*t}[/tex]
[tex]0.01 = e^{10^{3}t}[/tex]
[tex]t =4.605 *10^{-3}[/tex]
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total resistance of the coil plus the resistor is 11.0 capital omega. (a) Find the current in the circuit after 0.5 ms.
Given:
L = 1 mH = [tex]1\times 10^{-3}[/tex] H
total Resistance, R = 11 [tex]\Omega[/tex]
current at t = 0 s,
[tex]I_{o}[/tex] = 2.8 A
Formula used:
[tex]I = I_{o}\times e^-{\frac{R}{L}t}[/tex]
Solution:
Using the given formula:
current after t = 0.5 ms = [tex]0.5\times 10^{-3} s[/tex]
for the inductive circuit:
[tex]I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}[/tex]
[tex]I = 2.8\times e^-5.5[/tex]
I =0.011 A