Answer:
Distance of closest approach, [tex]r=1.91\times 10^{-14}\ m[/tex]
Explanation:
It is given that,
Charge on proton, [tex]q_p=e[/tex]
Charge on alpha particle, [tex]q_a=2e[/tex]
Mass of proton, [tex]m_p=1.67\times 10^{-27}\ kg[/tex]
Mass of alpha particle, [tex]m_a=4m_p=6.68\times 10^{-27}\ kg[/tex]
The distance of closest approach for two charged particle is given by :
[tex]r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}[/tex]
[tex]r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}[/tex]
[tex]r=1.91\times 10^{-14}\ m[/tex]
So, their distance of closest approach, as measured between their centers [tex]1.91\times 10^{-14}\ m[/tex]. Hence, this is the required solution.
An electric field of 4.0 μV/m is induced at a point 2.0 cm from the axis of a long solenoid (radius = 3.0 cm, 800 turns/m). At what rate is the current in the solenoid changing at this instant?
Answer:
The rate of current in the solenoid is 0.398 A/s
Explanation:
Given that,
Electric field [tex]E = 4.0\ \mu V/m[/tex]
Distance = 2.0 cm
Radius = 3.0 cm
Number of turns per unit length = 800
We need to calculate the rate of current
Using formula of electric field for solenoid
[tex]E = \dfrac{x}{2}\mu_{0}n\dfrac{dI}{dt}[/tex]
Where, x = distance
n = number of turns per unit length
E = electric field
r = radius
Put the value into the formula
[tex]4.0\times10^{-6}=\dfrac{2.0\times10^{-2}}{2}\times4\pi\times10^{-7}\times800\times\dfrac{dI}{dt}[/tex]
[tex]\dfrac{dI}{dt}=\dfrac{4.0\times10^{-6}\times2}{2.0\times10^{-2}\times4\pi\times10^{-7}\times800}[/tex]
[tex]\dfrac{dI}{dt}=0.397\ A/s[/tex]
Hence, The rate of current in the solenoid is 0.398 A/s.
Which of these situations would NOT qualify as a perfectly inelastic collision? A. A frog that jumps onto and sits on a lily pad.
B. A bowling ball that knocks pins out of its way as it rolls down a lane
C. A fish, swimming in one direction, that comes upon and swallows another fish swimming towards it
D. A piece of bubble gum that flies through the air and sticks to a golf ball at rest
Answer:
Option (B)
Explanation:
In a perfectly inelastic collision, the two bodies collide each other after the collision.
A. As the frog and the lily pad stick together so it is a perfectly inelastic collision.
B. The bowling ball and the pin do not stick together so it is not an example of perfectly inelastic collision.
C. A fish swallows the another, so it is a perfectly inelastic collision.
D. Bubble gum sticks to the golf ball so it is a perfectly inelastic collision.
While doing an experiment you measure the length of an object to be 32.5cm. The measuring device that you are using measures to a millimeter. How can you decrease the error associated with your measurements
Answer and Explanation:
By measuring in millimeter we can decrease the associated error with the measurement because when we measure in smaller unit the measurement is more precise and accurate rather than when we measure in larger unit so when we measure 325 millimeter instead of 32.5 cm then there is chance of less error produced.
An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV = ΔVmax sin(2πft) and the source frequency is 16.9 Hz, at what time t will the current flowing in this circuit be 55.0% of the peak current?
0.005366 seconds is the time will the current flowing in this circuit be 55.0% of the peak current.
In a simple AC circuit with a resistor connected to an AC voltage source, the current flowing through the circuit can be calculated using Ohm's law:
I(t)=ΔV(t)/R
Where: I(t) is the current at time
ΔV(t) is the instantaneous voltage at time t
R is the resistance of the resistor.
Given that the source voltage is [tex]\bigtriangleup V=\bigtriangleup V_{max}sin(2 \pi ft)[/tex], and you want to find the time t at which the current is 55% of the peak current, you need to find the time when [tex]I(t)=0.55\times I_{peak}.[/tex]
Find the peak current [tex]I_{peak}[/tex].
The peak current corresponds to the current when the source voltage is at its peak value [tex]\bigtriangleup V_{max}[/tex] .
Using Ohm's law:
[tex]I_{peak}=\frac{\bigtriangleup V_{max}}{R}[/tex]
The time t when the current I(t) is 55% of the peak current:
[tex]I(t)=0.55 \times I_{peak}[/tex]
Substitute the expressions for I(t) and [tex]I_{peak}[/tex].
[tex]\frac{\bigtriangleup V(t)}{R}=0.55 \times \frac{\bigtriangleup V_{max}}{R}[/tex]
2πft=arcsin(0.55)
t=arcsin(0.55)/2πf
t= 0.5716/2×3.14×16.9
t=0.5716/106.132
t=0.0053s
Hence, at approximately 0.005366 seconds after the start of the AC cycle (or 5.366 ms 5.366ms), the current flowing in the circuit will be 55.0% of the peak current.
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During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket has an angular speed of 12.0 rad/s. If the distance between the top of the racket and the shoulder is 1.30 m, find the magnitude of the total acceleration of the top of the racket.
Final answer:
The magnitude of the total acceleration of the top of the racket during the serve, considering both tangential and radial components, is approximately 270.5 m/s^2.
Explanation:
To find the magnitude of the total acceleration of the top of the racket, we need to consider both tangential and radial (centripetal) acceleration components. The tangential acceleration (at) is directly provided by the angular acceleration (α), and is calculated by multiplying the angular acceleration by the radius (r), so at = α × r. Here, α = 150 rad/s2 and r = 1.30 m, giving at = 195 m/s2. The radial acceleration (ar), also known as centripetal acceleration, depends on the angular speed (ω) and the radius (r), and is calculated with the formula ar = ω2 × r. Given ω = 12.0 rad/s and r = 1.30 m, we find ar = 187.2 m/s2. Finally, the total acceleration (a) is the square root of the sum of the squares of at and ar, resulting in a = √(at2 + ar2). This gives us a = √(1952 + 187.22) m/s2, and by calculating, we find the magnitude of the total acceleration to be approximately 270.5 m/s2.
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per kg of steam for the process? Calculate the amount of entropy produced, in kJ/K per kg of steam. What is the magnitude of the maximum theoretical work that could be obtained from the given initial state to the same final pressure?
Answer:
work is 130.5 kJ/kg
entropy change is 1.655 kJ/kg-k
maximum theoretical work is 689.4 kJ/kg
Explanation:
piston cylinder assembly
100 bar, 360°C to 1 bar, 160°C
to find out
work and amount of entropy and magnitude
solution
first we calculate work i.e heat transfer - work = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C .................1
so first we get some value from steam table with the help of 100 bar @360°C and 1 bar @ 160°C
specific volume = 0.0233 m³/kg
specific enthalpy = 2961 kJ/kg
specific internal energy = 2728 kJ/kg
specific entropy = 6.004 kJ/kg-k
and respectively
specific volume = 1.9838 m³/kg
specific enthalpy = 2795.8 kJ/kg
specific internal energy = 2597.5 kJ/kg
specific entropy = 7.659 kJ/kg-k
now from equation 1 we know heat transfer q = 0
so - w = specific internal energy @1 bar, 160°C - specific internal energy @ 100 bar, 360°C
work = 2728 - 2597.5
work is 130.5 kJ/kg
and entropy change formula is i.e.
entropy change = specific entropy ( 100 bar @360°C) - specific entropy ( 1 bar @160°C )
put these value we get
entropy change = 7.659 - 6.004
entropy change is 1.655 kJ/kg-k
and we know maximum theoretical work = isentropic work
from steam table we know specific internal energy is 2038.3 kJ/kg
maximum theoretical work = specific internal energy - 2038.3
maximum theoretical work = 2728 - 2038.3
maximum theoretical work is 689.4 kJ/kg
The work and entropy produced in the adiabatic expansion of steam can be calculated using the First Law of Thermodynamics and thermodynamic principles. Maximum theoretical work corresponds to an isentropic process, which is an ideal, reversible adiabatic process that occurs without an increase in entropy.
Explanation:The question regards the calculation of work and entropy of an adiabatic expansion process of steam in a piston-cylinder assembly. We'd need specific heat capacity, initial and end states temperatures data and the thermodynamic equations for adiabatic processes. The First Law of Thermodynamics equation dEint = CyndT will be used to calculate internal energy changes. Final work, W, can be calculated using the equation W = pΔV. Entropy change can be obtained using the formula ΔS=Q/T, where Q represents heat transfer, which is 0 for adiabatic processes.
As for the maximum theoretical work obtainable, this corresponds to a reversible adiabatic process or an isentropic process - entropy remains constant. This requires a quasi-static or reversible deformation. The work obtained in this case is at its maximum because there are no irreversibilities or entropy generation, which are responsible for decreasing the performance of real processes and engines.
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What will be the new volume when 128 mL of gas at 20.0°C is heated to 40.0°C while pressure remains unchanged? What will be the new volume when 128 mL of gas at 20.0°C is heated to 40.0°C while pressure remains unchanged? 64.0 mL 256 mL 120. mL 137 mL 128 mL
The answer will be 137 mL
A fire hose ejects a stream of water at an angle of 35.0° above the horizontal. The water leaves the nozzle with a speed of 25.0 m/s. Assuming that the water behaves like a projectile, how far from a building should the fire hose be located to hit the highest possible fire?
To find the distance from the building where the water will hit the highest possible fire, we analyze the horizontal and vertical components of the water's motion. Using the given information, we calculate the horizontal distance using the horizontal velocity and time, and the vertical distance using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can determine the horizontal distance.
Explanation:A fire hose ejects a stream of water at an angle of 35.0° above the horizontal with a speed of 25.0 m/s. To find the distance from the building where the water will hit the highest possible fire, we can analyze the horizontal and vertical components of the water's motion. The horizontal distance can be calculated using the horizontal velocity and time, while the vertical distance can be determined using the vertical velocity and time. By finding the time at which the water reaches its maximum height, we can calculate the horizontal distance.
Using the given information, we can determine that the initial horizontal velocity is 25.0 m/s * cos(35.0°) = 20.4 m/s, and the vertical velocity is 25.0 m/s * sin(35.0°) = 14.3 m/s. The time it takes for the water to reach its maximum height (where the vertical velocity becomes zero) can be found using the equation vy = v0y - gt, where vy is the vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time. Solving for t, we get t = v0y / g. Plugging in the values, we have t = 14.3 m/s / 9.8 m/s^2 = 1.46 seconds. Finally, we can find the horizontal distance by multiplying the initial horizontal velocity by the time: x = v0x * t = 20.4 m/s * 1.46 s = 29.8 m. Therefore, the fire hose should be located approximately 29.8 meters away from the building to hit the highest possible fire.
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The fire hose should be located approximately 59.8 meters from the building from the building to hit a fire at the highest point the water stream can reach.
Projectile Motion in Fire Hose Stream
To determine how far from a building a fire hose should be located to hit the highest possible fire, we need to analyze the projectile motion of the water stream ejected.
The water stream is ejected at an angle of 35.0° above the horizontal with an initial speed of 25.0 m/s. We will assume the water reaches its highest point where its vertical component of velocity becomes zero.
Step-by-Step Calculation
Horizontal and Vertical Components of Velocity:A 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20m apart, on frictionless level ice. When the hunter pulls the polar bear to him, the polar bear will move: A. 1:0m B. 3:3m C. 10m D. 12m E. 17m
Answer:
When the hunter pulls the polar bear to him, the polar bear will move 3.3 meters.
Explanation:
It is given that, a 640-N hunter gets a rope around a 3200-N polar bear. They are stationary, 20 m apart, on friction less level ice.
The given system is an isolated system which is initially at rest and they both will meet at the center of mass. Then finding the center of mass of the system as :
[tex]X_{com}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}[/tex]
Multiplying "g" to the numerator and denominator both as :
[tex]X_{com}=\dfrac{gm_1x_1+m_2x_2g}{m_1g+m_2g}[/tex]
[tex]X_{com}=\dfrac{640(0)+3200\times 20}{640+3200}=16.67\ m[/tex]
i.e. 640 N hunter moves at a distance of 16.67 meters and when the hunter pulls the polar bear to him, the polar bear will move, 20 m - 16.67 m = 3.3 meters. Hence, this is the required solution.
While you watch a parade, a band on a float passes you. You detect the frequency of a note played on a flute to be 356 Hz when the float is coming toward you and 348 Hz after the float passes you. Part A At what speed is the float traveling
Answer:
Float is travelling at speed of 3.86 m/s
Explanation:
As per Doppler's effect we know that when source and observer moves relative to each other then the frequency is different from actual frequency
so here when source moves closer to the observer
[tex]f_1 = f_o(\frac{v}{v - v_s})[/tex]
now we have
[tex]356 = f_o(\frac{v}{v - v_s})[/tex]
again when source moves away from the observer then we have
[tex]f_2 = f_o(\frac{v}{v + v_s})[/tex]
[tex]348 = f_o(\frac{v}{v + v_s})[/tex]
now divide above two equations
[tex]\frac{356}{348} = \frac{v + v_s}{v - v_s}[/tex]
[tex]1.023 = \frac{v + v_s}{v - v_s}[/tex]
[tex]1.023( v - v_s) = (v + v_s)[/tex]
here we know that the speed of sound in air is 340 m/s
so we have
[tex]1.023(340 - v_s) = (340 + v_s)[/tex]
[tex]7.816 = 2.023 v_s[/tex]
now we have
[tex]v_s = 3.86 m/s[/tex]
The speed of the float in the parade is determined using the observed frequencies of the flute when the band is approaching and moving away informed by the Doppler Effect physics theory. You generate equations for both scenarios and solve for the speed of the source, using the average speed of sound in the air.
Explanation:This problem relates to the Doppler Effect, which explains why the frequency of a sound changes when the source of the sound is moving in relation to the observer. The formula to calculate a Doppler Shift when the source is moving away (as in the float in the parade) is given by: f' = f [(v + v0) / v] and when the source is approaching is given by: f' = f [(v - v0) / v]. Where, f' is the observed frequency, f is the source frequency, v is the sound speed (340 m/s), and v0 is the speed of the source.
In the problem, it's given that the observed frequencies when the band is approaching and moving away are 356 Hz and 348 Hz, respectively. By setting up two equations using the Doppler Shift formula for both observed frequencies, we can solve for v0. In this case, we use the average speed of sound in the air, which is about 340 m/s.
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A 0.012 kg ping pong ball placed 1.8 m away from 0.3 kg billbard ball. calculate gravitational force between two objects
Answer:
7.4 x 10^-14 N
Explanation:
m = 0.012 kg, M = 0.3 kg, r = 1.8 m
Let the gravitational force be F.
F = G M m / r^2
F = (6.67 x 10^-11 x 0.3 x 0.012) / (1.8)^2
F = 7.4 x 10^-14 N
A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24. What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
Answer:
The distance is 11 m.
Explanation:
Given that,
Friction coefficient = 0.24
Time = 3.0 s
Initial velocity = 0
We need to calculate the acceleration
Using newton's second law
[tex]F = ma[/tex]...(I)
Using formula of friction force
[tex]F= \mu m g[/tex]....(II)
Put the value of F in the equation (II) from equation (I)
[tex]ma=\mu mg[/tex]....(III)
[tex]a = \mu g[/tex]
Put the value in the equation (III)
[tex]a=0.24\times9.8[/tex]
[tex]a=2.352\ m/s^2[/tex]
We need to calculate the distance,
Using equation of motion
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
[tex]s=0+\dfrac{1}{2}2.352\times(3.0)^2[/tex]
[tex]s=10.584\ m\ approx\ 11\ m[/tex]
Hence, The distance is 11 m.
The truck can travel a maximum distance of 10.584 meters in 3.0 seconds without the box sliding.
To determine the maximum distance the truck can travel in 3.0 s without the box sliding, we start by calculating the maximum acceleration the box can experience before it starts to slide.
This is given by the static friction force, which is the product of the coefficient of static friction and the normal force (N). Assuming the box’s weight is W = mg, the normal force is N = mg.
The maximum static friction force (f) is:
f = μ N = μ mg
Given μs = 0.24, we need to ensure:
f > ma
Therefore, 0.24mg = ma, solving for a gives us:
a = 0.24g = 0.24 × 9.8 m/s² ≈ 2.352 m/s²
Next, we need to calculate the maximum distance using the kinematic equation for constant acceleration:
d = ut + (1/2)at²
Where:
u = initial velocity = 0a = 2.352 m/s²t = 3.0 sSubstituting these values, we get:
d = 0 + (1/2) × 2.352 m/s² × (3.0 s)² = 0.5 × 2.352 × 9 = 10.584 m
Thus, the maximum distance the truck can travel without the box sliding is approximately 10.584 meters.
An electric heater is rated at 1430 W, a toaster at 890 W, and an electric grill at 1760 W. The three appliances are connected in parallel to a common 120 V circuit. How much total current does this circuit draw?
Answer:
34.01 A
Explanation:
[tex]P_{heater}[/tex] = Power of the electric heater = 1430 W
[tex]P_{toaster}[/tex] = Power of the toaster = 890 W
[tex]P_{grill}[/tex] = Power of the electric grill = 1760 W
V = Voltage of the battery connected in parallel to appliances = 120 volts
Using ohm's law, current drawn by heater is given as
[tex]i_{heater} = \frac{P_{heater}}{V}[/tex]
[tex]i_{heater} = \frac{1430}{120}[/tex]
[tex]i_{heater} [/tex] = 11.92 A
Using ohm's law, current drawn by toaster is given as
[tex]i_{toaster} = \frac{P_{toaster}}{V}[/tex]
[tex]i_{toaster} = \frac{890}{120}[/tex]
[tex]i_{toaster} [/tex] = 7.42 A
Using ohm's law, current drawn by grill is given as
[tex]i_{grill} = \frac{P_{grill}}{V}[/tex]
[tex]i_{grill} = \frac{1760}{120}[/tex]
[tex]i_{grill} [/tex] = 14.67 A
Total circuit drawn is given as
[tex]i_{total} [/tex] = [tex]i_{heater} [/tex] + [tex]i_{toaster} [/tex] + [tex]i_{grill} [/tex]
[tex]i_{total} [/tex] = 11.92 + 7.42 + 14.67
[tex]i_{total} [/tex] = 34.01 A
Define electric potential and inductance using sentences an elementary student could understand. Do not use math
Answer and Explanation:
1). Electric Potential:
Electric Potential can be defined as the amount of work required to move a unit charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.
2). Inductance:
It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.
We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.
Answer:
this is my basic understanding o
. Electric Potential:
Electric Potential can be defined as the amount of work required to move a unit charge from infinity or reference point to a particular or specific point inside the electric field (without any accelerated motion of charge particle). The other terms for electric potential are 'electrostatic potential, potential difference(if a charge is moved from one point to another inside the field) or electric field potential.
2). Inductance:
It can be define as that property of electric circuits due to which it opposes any change in the voltage under the influence of changing magnetic field.
We can also say that the property of an electric conductor due to which it opposes any change in electric current passing through it. Any change in electric current results in the production of magnetic field which when varied results in the generation of Electro Motive Force (EMF) which according to Lenz law is produced in a way that it opposes its cause of production.
Explanation:
A 36-in-diameter pipeline carries oil (SG = 0.89) at 1 million barrels per day (bbl/day). The friction head loss is 13 ft/1000 ft of pipe. It is planned to place pumping stations every 10 mi along the pipe. Estimate the horsepower which must be delivered to the oil by each pump.
Answer:
144540.16735 hp
Explanation:
Diameter of pipe = 36 in
Specific gravity of oil = 0.89
ρ = Density of oil = 0.89×62.2407
Q = Flow rate of oil = 1×10⁶ bbl/day = 1×10⁶×5.61458/86400 = 64.9836033565 ft³/s
hf = 13 ft/1000ft = 0.013 ft/ft
for 10 miles
hf = 0.013×10×5280 = 686.4 ft
g = Acceleration due to gravity = 32.17405 ft/s²
Power
P = ρQghf
⇒P = 0.89×62.2407×64.9836033565×32.17405×686.4
⇒P = 79497097.342 ft lbf/s
1 ft lbf/s = 0.0018181817 hp
79497097.342 ft lbf/s = 144540.16735 hp
∴ Horsepower which must be delivered to the oil by each pump is 144540.16735 hp
Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay model, which is initially motionless and has a mass of 0.350 kg. Both being soft clay, they naturally stick together. What is their final velocity?
Answer:
0.278 m/s
Explanation:
We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.
So we can write:
[tex]mu=(m+M)v[/tex]
where
m = 0.200 kg is the mass of the koala bear
u = 0.750 m/s is the initial velocity of the koala bear
M = 0.350 kg is the mass of the other clay model
v is their final combined velocity
Solving the equation for v, we get
[tex]v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s[/tex]
An electron moving with a speed of 1.50*10^7 m/sec is projected at right angles into a uniform magnetic field of flux density 6.50*10^-8 w/m^2. Calculate quantitatively the new path of the electron.
Answer:
the path of the moving charge will be circular path now
Radius = 1312.5 m
Explanation:
Force on a moving charge due to constant magnetic field is given by
[tex]\vec F = q(\vec v \times \vec B)[/tex]
since here force on the moving charge is always perpendicular to the velocity always as it is vector product of velocity and magnetic field so here magnitude of the speed is always constant
Also the force is since perpendicular to the velocity always
so here the path of the moving charge will be circular path now
now to find out the radius of this circular path
[tex]F = \frac{mv^2}{R}[/tex]
[tex]qvBsin90 = \frac{mv^2}{R}[/tex]
[tex]R = \frac{mv}{qB}[/tex]
[tex]R = \frac{9.1\times 10^{-31}(1.50 \times 10^7)}{1.6 \times 10^{-19}(6.50 \times 10^{-8})}[/tex]
[tex]R = 1312.5 m[/tex]
a football of mass 0.38 kg is traveling towards a boy at 2.8ms-1. the boy kicks the ball with a force of 21N, reversing its direction. if the time of imapct between the football and boy’s foot is 250ms. what is the new speed of trhe ball?
Answer:
Final speed in reverse direction will be 11 m/s
Explanation:
As per impulse momentum theorem we know that
Change in momentum of the object is product of force and time
so we have
[tex]P_f - P_i = F\Delta t[/tex]
now we have
[tex]P_f = mv_f[/tex]
[tex]P_i = mv_i[/tex]
[tex]\Delta t = 250 ms[/tex]
F = 21 N
now we have
[tex]0.38(v_f) - 0.38(-2.8) = 21(250\times 10^{-3})[/tex]
[tex]v_f = 13.82 - 2.8[/tex]
[tex]v_f = 11 m/s[/tex]
What is the average speed of an object that travels 50 meters in 2 seconds? Select one: a. 100 m/s b. 25 m/s c. 50 m/s d. 2 m/s
Answer:
Average speed, v = 25 m/s
Explanation:
It is given that,
Distance travelled by the object, d = 50 meters
Time taken, t = 2 seconds
Average speed is defined as the total distance divided by total time taken i.e.
[tex]v=\dfrac{d}{t}[/tex]
[tex]v=\dfrac{50\ m}{2\ s}[/tex]
v = 25 m/s
So, the sped of the object is 25 m/s. Hence, this is the required solution.
The average speed of an object that travels 50 meters in 2 seconds is calculated by dividing the distance by time, resulting in 25 m/s.
The average speed of an object is calculated by dividing the total distance travelled by the total time taken to travel that distance. In this case, the object travels 50 meters in 2 seconds, so the average speed is calculated as follows:
Average speed = Total distance \/ Total time = 50 meters \2 seconds = 25 meters per second (m/s).
Therefore, the correct answer to the question is b. 25 m/s.
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg. The child grabs and clings to a bar that is 1.45 m from the center of the merry‑go‑round, causing the angular velocity of the merry‑go‑round to abruptly drop from 55.0 rpm to 17.0 rpm. What is the moment of inertia of the merry‑go‑round with respect to its central axis?
Answer:
[tex]I_1 = 32.5 kg m^2[/tex]
Explanation:
Here as we know that total angular momentum is always conserved for child + round system
so here by angular momentum conservation we have
[tex]I_1 \omega_1 = (I_1 + I_2)\omega_2[/tex]
here we have
[tex]I_2 = mR^2[/tex] (inertia of boy)
[tex]I_2 = (34.5)(1.45^2)[/tex]
[tex]I_2 = 72.5 kg m^2[/tex]
now we have
[tex]I_1(2\pi 55) = (I_1 + 72.5)(2 \pi 17)[/tex]
[tex]I_1 = (I_1 + 72.5)(0.31)[/tex]
[tex]I_1(1 - 0.31) = 22.4[/tex]
[tex]I_1 = 32.5 kg m^2[/tex]
The total moment of inertia of the merry-go-round with the child can be calculated by considering the child as a point mass at a given distance from the axis of rotation. The decrease in the angular velocity of the merry-go-round when the child jumps on is due to the increase in the moment of inertia. The moment of inertia of the merry-go-round itself is close to the total moment of inertia because it has far more mass distributed away from the axis.
Explanation:The subject is based on the physics concept of moment of inertia, which depends not only on the object's mass but also on its distribution relative to the axis of rotation. The moment of inertia initially decreases when the child, initially at rest, grabs onto the already spinning merry-go-round. This abrupt decrease in angular velocity is because of the increase in moment of inertia caused by the child.
To find the total moment of inertia (I), we first calculate the child’s moment of inertia (Ic) by considering the child as a point mass at a distance of 1.45 m from the axis of rotation. The formula for this is Ic = mR², where m is the child's mass (34.5 kg) and R is the distance from the center (1.45 m).
Then, we calculate for the angular velocity change from 55 rpm to 17 rpm. The goal is to find the moment of inertia of the merry-go-round itself, which should be close to the total moment of inertia because it has much more mass distributed away from the axis than the child does.
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Suppose that you time this pendulum, and it completes exactly 68.0 complete cycles in 100 s. [Hint: think carefully about all given distances.] 5. (5 pts) Which choice best represents the value of g in the vicinity of this pendulum?
Answer:
Explanation:
Number of oscillations in 100 s is 68.
time taken to complete one oscillation = 100 / 68 s = 1.47 s
Use the formula for the time period of an oscillating body
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex]
[tex]T^{2} = 4\pi^{2} \frac{L}{g}}[/tex]
In this equation substitute the value of T amd L and then find the value of g.
A man is holding a 6.0-kg (weight = 59 N) dumbbell at arm's length, a distance of 0.56 mfrom his shoulder. What is the torque on the shoulder joint from the weight of the dumbbell if thearm is held at 15° above the horizontal? On the picture, draw the lever arm for this force
Answer:
[tex]\vec \tau = 31.9 N[/tex]
Explanation:
As we know that torque due to a force is given by the formula
[tex]\vec \tau = \vec r \times \vec F[/tex]
here we know that force is exerted due to weight of the mass hold in his hand
so we have
F = mg = 59 N
Now Lever arm is the perpendicular distance on the line of action of force from the axis about which the system is rotated
so here we can say
[tex]r = Lcos\theta [/tex]
[tex]r = 0.56 cos15 = 0.54 m[/tex]
now we have
[tex]\vec \tau = (0.54)(59)[/tex]
[tex]\vec \tau = 31.9 Nm[/tex]
Calculate the rate of heat transfer by radiation from a car radiator at 111 °C into a 45 °C environment, if the radiator has an emissivity of 0.75 and a 1.6 m^2 surface area.
Answer:
783.63 Watt
Explanation:
T = 111 degree C = 111 + 273 = 384 K
T0 = 45 degree C = 45 + 273 = 318 K
A = 1.6 m^2
e = 0.75
According to the Stefan's law, the energy radiated per second is given by
[tex]E = \sigma eA\left ( T^{4}- T_{0}^{4}\right )[/tex]
[tex]E = 5.67\times 10^{-8}\times 0.75\times 1.6 ( 384^{4}- 318^{4}\right ))[/tex]
E = 783.63 Watt
A composite wall 8 m long and 4 m high consists of 4 cm of brick, 8 cm of concrete, 6 cm of fiberglass and 10 cm of corkboard. The inside temperature is 18°C, and the outside temperature is 40°C. How much heat is flow into the house through this wall in 10 hours?
Answer:
Heat flow through walls in 10 hours is
[tex]Q = 6.1 \times 10^6 J[/tex]
Explanation:
Thermal conductivity of all the materials is as follows
1). Concrete = 0.8
2). Brick = 0.6
3). Fibreglass = 0.04
4). Corkboard = 0.04
Area of the wall is given as
[tex]A = (8 \times 4) = 32 m^2[/tex]
now the thermal resistance due to each wall is given as
[tex]R_1 = \frac{0.04}{0.6(32)} = 2.08 \times 10^{-3}[/tex]
[tex]R_2 = \frac{0.08}{0.8(32)} = 3.125 \times 10^{-3}[/tex]
[tex]R_3 = \frac{0.06}{0.04(32)} = 46.87 \times 10^{-3}[/tex]
[tex]R_2 = \frac{0.10}{0.04(32)} = 78.13 \times 10^{-3}[/tex]
Now total thermal resistance of all walls
[tex]R = R_1 + R_2 + R_3 + R_4[/tex]
[tex]R = (2.08 + 3.125 + 46.87 + 78.13) \times 10^{-3}[/tex]
[tex]R = 0.130[/tex]
now rate of heat transfer per second is given as
[tex]\frac{dQ}{dt} = \frac{T_1 - T_2}{R}[/tex]
[tex]Q = \frac{40 - 18}{0.130}(10 \times 3600)[/tex]
[tex]Q = 6.1 \times 10^6 J[/tex]
J-s. If your 1400-kg car is parked in an 8.54-m-long garage, what is the uncertainty in its velocity? cm/s the tolerance is +/-2%
Answer:
[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]
Explanation:
As per Heisenberg's uncertainty principle we know that
[tex]\Delta P \times \Delta x = \frac{h}{4\pi}[/tex]
so here we have
[tex]\Delta P = m\Delta v[/tex]
[tex]\Delta x = 8.54 m[/tex]
now from above equation we have
[tex]m\Delta v \times (8.54) = \frac{h}{4\pi}[/tex]
[tex]1400(\Delta v) \times (8.54) = \frac{6.626 \times 10^{-34}}{4\pi}[/tex]
[tex]\Delta v = 4.41 \times 10^{-39} m/s[/tex]
[tex]\Delta v = 4.41 \times 10^{-37} cm/s[/tex]
Light that is polarized along the vertical direction is incident on a sheet of polarizing material. Only 92% of the intensity of the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet. What angle does the transmission axis of the second sheet make with the vertical?
The transmission axis of the second sheet makes a right angle (90 degrees) with the vertical given that no light passes through it. This is based on Malus's Law for polarization.
Explanation:The phenomenon described in your question is polarization, which is a characteristic of transverse waves such as light. The sheet of polarizing material allows only the light polarized in the same direction as the sheet's axis of transmission to pass through it. Following Malus's Law for polarization, we can determine that if no light passes through the second polarizer, its transmission axis must be at a right angle (90 degrees) to the initial polarization direction, which is vertical in this case.
The Malus's Law is an equation that describe the intensity of the light after passing through a polarizer as: I = I0 cos^2 θ where I0 is the initial intensity of the light, θ is the angle between the light polarization direction and the transmission axis of the polarizer. If I = 0, then cos^2 θ = 0, which indicates that θ is 90 degrees. This concept comes under the area of wave optics in Physics.
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The transmission axis of the second sheet of polarizing material makes a 90-degree angle with the vertical because it must be perpendicular to the first sheet's axis for no light to pass through. Hence, with the first polarizer vertical, the second must be horizontal.
Explanation:The question asks about the angle that the transmission axis of the second sheet of polarizing material makes with the vertical, given that no light passes through. According to the principles of light polarization, if no light passes through a second sheet of polarizing material, it means the transmission axis of the second sheet is perpendicular to the axis of the first one.
Since the first sheet polarizes light vertically, this implies the transmission axis of the second sheet must be horizontal. Therefore, as per common geometric considerations where vertical and horizontal lines are perpendicular, the angle between the vertical (first filter axis) and horizontal (second filter axis) is 90 degrees.
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A projectile is fired vertically with an initial velocity of 192 m/s. Calculate the maximum altitude h reached by the projectile and the time t after firing for it to return to the ground. Neglect air resistance and take the gravitational acceleration to be constant at 9.81 m/s2.
Answers:
h = m
t = s
Answer:
a) Maximum height reached = 1878.90 m
b) Time of flight = 39.14 seconds.
Explanation:
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g [tex]m/s^2[/tex] and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile , [tex]t=\frac{2usin\theta }{g}[/tex]
Vertical motion (Maximum height reached, H) :
We have equation of motion, [tex]v^2=u^2+2as[/tex], where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
[tex]0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}[/tex]
In the give problem we have u = 192 m/s, θ = 90° we need to find H and t.
a) [tex]H=\frac{u^2sin^2\theta}{2g}=\frac{192^2\times sin^290}{2\times 9.81}=1878.90m[/tex]
Maximum height reached = 1878.90 m
b) [tex]t=\frac{2usin\theta }{g}=\frac{2\times 192\times sin90}{9.81}=39.14s[/tex]
Time of flight = 39.14 seconds.
Two workers are sliding 390 kg crate across the floor. One worker pushes forward on the crate with a force of 450 N while the other pulls in the same direction with a force of 200 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?
Answer:
The coefficient of kinetic friction [tex]\mu= 0.16989[/tex]
Explanation:
From Newton's second law
[tex]\sum\overset{\rightarrow}{F}=m\cdot\overset{\rightarrow}{a}[/tex]
If the velocity is constant, that means the summation of all forces must be equal to zero. Draw the free-body diagram to obtain the sums of forces in x and y. It must include the Friction Force, in the opposite direction of the displacement, the weight ([tex]W=mg=390*9.81=3825.9N[/tex]), the Normal Force, which is the is the consequence of Newton's third law and the forces from the two workers.
The sum in y is:
[tex]\sum F_{y}=F_{N}-3825.9=0[/tex]
Solving for the [tex]F_{N}[/tex]:
[tex]F_{N}=$ $3825.\,\allowbreak9N[/tex]
The sum in x is:
[tex]\sum F_{x}=450+200-F_{f}=0[/tex]
Solving for the [tex]F_{f}[/tex]:
[tex]$F_{f}=650.0N[/tex]
The formula of the magnitude of the Friction force is
[tex]F_{f}=\mu F_{N}[/tex]
That means the coefficient of friction is:
[tex]\mu=\frac{F_{f}}{F_{N}}=\frac{650.0}{3825.\,\allowbreak9}=\allowbreak0.16989[/tex]
What is the magnitude (in Newtons) of the force a 796-uC charge exerts on a 481-nC charge 23.1-cm away? 2 pts Question 5 A proton is released in a uniform electric field, and it experiences an electric force of 6.9x10-15-N toward the south. What is the magnitude (in N/C) of the electric field?
Answer:
(a) 64.58 N
(b) 43125 N/C
Explanation:
(a) q1 = 796 uC = 796 x 10^-6 C, q2 = 481 nC = 481 x 10^-9 C,
r = 23.1 cm = 0.231 m
Force, F = k q1 q2 / r^2
F = (9 x 10^9 x 796 x 10^-6 x 481 x 10^-9) / (0.231)^2
F = 64.58 N
(b) F = 6.9 x 10^-15 N
E = F / q
E = (6.9 x 10^-15) / (1.6 x 10^-19) = 43125 N/C
A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total mass of the wagon, rider, and rock is 93.1 kg. The mass of the rock is 0.292 kg. Initially the wagon is rolling forward at a speed of 0.456 m/s. Then the person throws the rock with a speed of 15.4 m/s. Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown (a) directly forward in one case and (b) directly backward in another.
Answer:
a) 0.41 m/s
b) 0.51 m/s
Explanation:
(a)
M = total mass of wagon, rider and the rock = 93.1 kg
V = initial velocity of wagon = 0.456 m/s
m = mass of the rock = 0.292 kg
v = velocity of rock after throw = 15.4 m/s
V' = velocity of wagon after rock is thrown
Using conservation of momentum
M V = m v + (M - m) V'
(93.1) (0.456) = (0.292) (15.4) + (93.1 - 0.292) V'
V' = 0.41 m/s
b)
M = total mass of wagon, rider and the rock = 93.1 kg
V = initial velocity of wagon = 0.456 m/s
m = mass of the rock = 0.292 kg
v = velocity of rock after throw = - 15.4 m/s
V' = velocity of wagon after rock is thrown
Using conservation of momentum
M V = m v + (M - m) V'
(93.1) (0.456) = (0.292) (- 15.4) + (93.1 - 0.292) V'
V' = 0.51 m/s