Answer:
156.3 J
Explanation:
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
To find the energy delivered by the laser beam in 1 minute, we need to calculate the power of the laser beam and then multiply it by the duration.
E = Amplitude of the electric field = 25000 V/m
t = time interval = 1 minute = 60 sec
d = diameter of the spot = 2 mm = 2 x 10⁻³ m
Area of the spot is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
Energy delivered to the spot is given as
U = (0.5)ε₀ E² c A t
Inserting the values
U = (0.5) (8.85 x 10⁻¹²) (25000)² (3 x 10⁸) (3.14 x 10⁻⁶) (60)
U = 156.3 J
Why does a satellite in a circular orbit travel at a constant speed? A) The net force acting on the satellite is toward the center of the path.
B) There is a component of the net force acting in the direction of the motion of the satellite.
C) There is a force acting opposite to the direction of the motion of the satellite.
D) There is no component of the net force in the radial direction.
E) The gravitational force acting on the satellite is balanced by the centrifugal force acting on the satellite.
Explanation:
The motion of satellite is an example of uniform circular motion. In this type of motion, the velocity of object varies at each and every point but its speed is constant. The satellite in a circular orbit travel at a constant speed.
The gravitational force of earth is balanced by the centripetal force such that,
[tex]F_g=F_c[/tex]
A satellite travel at a constant speed because there is no force acting in the direction of motion of satellite i.e. there is no resistive force that opposite the motion of the satellite.
A proton travels at a speed 0.25 x 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.975 m. What is the field strength, in tesla?
Answer:
The magnetic field strength of the proton is 0.026 Tesla.
Explanation:
It is given that,
Speed of the proton, [tex]v=0.25\times 10^7\ m/s[/tex]
The radius of circular path, r = 0.975 m
It is moving perpendicular to a magnetic field such that the magnetic force is balancing the centripetal force.
[tex]qvB\ sin90=\dfrac{mv^2}{r}[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
q = charge on proton
[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 0.25\times 10^7\ m/s}{1.6\times 10^{-19}\ C\times 0.975\ m}[/tex]
B = 0.026 Tesla
So, the magnetic field strength of the proton is 0.026 Tesla.
A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic friction between the grass and the sled is 0.30. How much force must he apply to the sled?
180N
Explanation:Using Newton's law of motion;
∑F = m x a --------------------(i)
Where;
∑F = Resultant force
m = mass of the object (sled in this case)
a = acceleration of the sled
Calculate the resultant force;
Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;
i. Applied force ([tex]F_{A}[/tex]) in one direction and;
ii. Frictional force ([tex]F_{R}[/tex]) in the other direction to oppose motion
Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;
∑F = [tex]F_{A}[/tex] - [tex]F_{R}[/tex] -----------------------(i)
Frictional force [tex]F_{R}[/tex] is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e
[tex]F_{R}[/tex] = μ x W
Where;
W = mass(m) x gravity(g)
W = m x g
=> [tex]F_{R}[/tex] = μmg
Substitute [tex]F_{R}[/tex] into equation (ii)
∑F = [tex]F_{A}[/tex] - μmg
Substitute ∑F into equation (i)
[tex]F_{A}[/tex] - μmg = ma -------------------(iii)
Since the motion is at constant speed, it means acceleration is zero (0)
Substitute a = 0 into equation (iii) to give;
[tex]F_{A}[/tex] - μmg = 0
=> [tex]F_{A}[/tex] = μmg
Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;
=> [tex]F_{A}[/tex] = 0.3 x 60 x 10
=> [tex]F_{A}[/tex] = 180N
This means that the applied force should be 180N
The amount of force the football player must apply to the sled is 176.4 Newton.
Given the following data:
Mass of sled = 60 kgCoefficient of kinetic friction = 0.30We know that acceleration due to gravity (g) on Earth is equal to 9.8 [tex]m/s^2[/tex]
To find how much force the football player must apply to the sled:
Mathematically, the force of kinetic friction is given by the formula;
[tex]Fk = umg[/tex]
Where;
Fk represents the force of kinetic friction.μ represents the coefficient of friction.m represents the mass.g is the acceleration due to gravity.Substituting the given parameters into the formula, we have;
[tex]Fk = 0.30\times60\times9.8[/tex]
Force, Fk = 176.4 Newton.
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A 2.99-m-long2.99-m-long rod, as measured in its rest frame, speeds by you longitudinally at 6.49×107 m/s6.49×107 m/s . You measure its length as it passes. By how many millimeters do you determine the rod has contracted?
Answer:
The contraction in the rod is 71 mm.
Explanation:
Given that,
original length L'= 2.99 m
Speed [tex]v= 6.49\times10^{7}\ m/s[/tex]
We need to calculate the length
Using expression for length contraction
[tex]L'=\gamma L[/tex]
[tex]L=\dfrac{L'}{\gamma}[/tex]
Where,
[tex]\gamma=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]
[tex]L=\sqrt{1-\dfrac{v^2}{c^2}}L'[/tex]
Where, v = speed of observer
c = speed of the light
Put the value into the formula
[tex]L=\sqrt{1-\dfrac{(6.49\times10^{7})^2}{(3\times10^{8})^2}}\times2.99[/tex]
[tex]L=2.919\ m[/tex]
The expression for the contraction in the rod
[tex]d =L'-L[/tex]
[tex]d=2.99-2.919 [/tex]
[tex]d=0.071[/tex]
[tex]d= 71\ mm[/tex]
Hence, The contraction in the rod is 71 mm.
A 100.0 g ice cube at -10 degrees Celsius is placed in an aluminum cup whose initial temperature is 70 degrees Celsius. The system come to an equilibrium at 20 degrees Celsius. What is the mass of the cup?
Answer: 135 grams
Explanation:
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of ice = 100 g
[tex]m_2[/tex] = mass of aluminium cup =? g
[tex]T_{final}[/tex] = final temperature =[tex]20^0C[/tex]
[tex]T_1[/tex] = temperature of ice = [tex]-10^oC[/tex]
[tex]T_2[/tex] = temperature of aluminium cup= [tex]70^oC[/tex]
[tex]c_1[/tex] = specific heat of ice= [tex]2.03J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of aluminium cup = [tex] 0.902 J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex][100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)][/tex]
[tex]m_2=135g[/tex]
Therefore, the mass of the aluminium cup was 135 g.
By calculating the heat transfer between a 100 g ice cube and an aluminum cup, the mass of the cup is found to be approximately 978.3 g.
To solve this, we will use principles of thermal equilibrium and specific heat capacities. Let's go through the steps:
Step 1: Calculate the heat required to warm the ice to 0 degrees Celsius.
The specific heat capacity of ice is 2.1 J/g°C. The formula for heat is:
Q = m * c * ΔT
m = 100.0 g (mass of ice)
c = 2.1 J/g°C (specific heat capacity of ice)
ΔT = (0°C - (-10°C)) = 10°C
Q₁ = 100 g * 2.1 J/g°C * 10°C = 2100 J
Step 2: Calculate the heat required to melt the ice at 0 degrees Celsius.
The enthalpy of fusion of ice ,as we know ,is 334 J/g.
Q₂ = m * L
m = 100.0 g
L = 334 J/g
Q₂ = 100 g * 334 J/g = 33400 J
Step 3: Calculate the heat required to warm the melted ice from 0°C to 20°C.
The specific heat capacity of water is 4.18 J/g°C.
Q₃ = m * c * ΔT
m = 100.0 g
c = 4.18 J/g°C
ΔT = (20°C - 0°C) = 20°C
Q₃ = 100 g * 4.18 J/g°C * 20°C = 8360 J
Step 4: Calculate the total heat gained by the ice.
[tex]Q_{total[/tex] = Q₁ + Q₂ + Q₃ = 2100 J + 33400 J + 8360 J = 43860 J
Step 5: Calculate the heat lost by the aluminum cup.
The specific heat capacity of aluminum is 0.897 J/g°C. Since we need the heat lost, we use:
[tex]Q_{lost[/tex] = m * c * ΔT
[tex]Q_{lost[/tex] = 43860 J
c = 0.897 J/g°C
ΔT = (70°C - 20°C) = 50°C
Rearranging for m:
m = [tex]Q_{lost[/tex] / (c * ΔT)
m = 43860 J / (0.897 J/g°C * 50°C) = 978.3 g
Therefore, the mass of the aluminum cup is approximately 978.3 g.
What is the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east?
Answer:
The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.
Explanation:
Given that,
Electric field strength = 1500 N/C
We need to calculate the electric force
Using formula of electric field
[tex]F = Eq[/tex]
E = electric field strength
q = charge of electron
Electron has negative charge.
Put the value into the formula
[tex]F=1500\times(-1.6\times10^{-19})[/tex]
[tex]F=-2.4\times10^{-16}\ N[/tex]
Negative sign shows the opposite direction of the field
Hence, The magnitude of the electric force on an electron in a uniform electric is [tex]2.4\times10^{-16}\ N[/tex] to the west.
The magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.
What is the magnitude of the electric force?We know that electric force is given by the formula,
[tex]F = E \times q[/tex]
It is given that the electric field, E = 1500 N/C,
We also know that an electron is negatively charged and has a charge of 1.60217662 × 10⁻¹⁹ C.
[tex]F = E \times q\\\\F = 1500 \times 1.6 \times 10^{-19}\\\\F = 2.4 \times 10^{-16}\rm\ N[/tex]
Hence, the magnitude of the electric force on an electron in a uniform electric field of strength 1500 N/C that points due east is 2.4x10⁻¹⁶ C.
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A 64.0 kg skater moving initially at 2.81 m/s on rough horizontal ice comes to rest uniformly in 3.93 s due to friction from the ice. What force does friction exert on the skater?
Answer:
Force of friction, F = 45.76 N
Explanation:
t is given that,
Mass of the skater, m = 64 kg
Initial velocity of the skater, u = 2.81 m/s
Finally it comes to rest, v = 0
Time, t = 3.93 s
We need to find the force of friction. According to seconds law of motion as :
F = m × a
[tex]F=m\times \dfrac{v-u}{t}[/tex]
[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]
F = −45.76 N
So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.
You and your surfing buddy are waiting to catch a wave a few hundred meters off the beach. The waves are conveniently sinusoidal, and you notice that when you're on the top of one wave and moving toward your friend, she is exactly halfway between you and the trough of the wave. 1.50 seconds later, your friend is at the top of the wave. You estimate the horizontal distance between you and your friend at 8.00 m. (a) What is the frequency of the waves?
Answer:
(a): The frequency of the waves is f= 0.16 Hz
Explanation:
T/4= 1.5 s
T= 6 sec
f= 1/T
f= 0.16 Hz (a)
(a) The frequency of the waves is ¹/₆ Hz ≈ 0.167 Hz
(b) The speed of the waves is 5¹/₃ m/s ≈ 5.33 m/s
[tex]\texttt{ }[/tex]
Further explanationLet's recall the speed of wave and intensity of wave formula as follows:
[tex]\large {\boxed {v = \lambda f}}[/tex]
f = frequency of wave ( Hz )
v = speed of wave ( m/s )
λ = wavelength ( m )
[tex]\texttt{ }[/tex]
[tex]\large {\boxed {I = 2 \pi^2 A^2 f^2 \rho v}}[/tex]
I = intensity of wave ( W/m² )
A = amplitude of wave ( m )
f = frequeny of wave ( Hz )
ρ = density of medium ( kg/m³ )
v = speed of wave ( m/s )
Let's tackle the problem!
[tex]\texttt{ }[/tex]
Given:
time taken = t = 1.50 seconds
distance covered = d = 8.00 m
Asked:
(a) frequency of the waves = ?
(b) speed of the waves = ?
Solution:
Question (a):[tex]t = \frac{1}{4}T[/tex]
[tex]1.50 = \frac{1}{4}T[/tex]
[tex]T = 4 \times 1.50[/tex]
[tex]T = 6 \texttt{ seconds}[/tex]
[tex]\texttt{ }[/tex]
[tex]f = \frac{1}{T}[/tex]
[tex]f = \frac{1}{6} \texttt{ Hz}[/tex]
[tex]\texttt{ }[/tex]
Question (b):[tex]d = \frac{1}{4}\lambda[/tex]
[tex]8.00 = \frac{1}{4}\lambda[/tex]
[tex]\lambda = 8.00 \times 4[/tex]
[tex]\lambda = 32 \texttt{ m}[/tex]
[tex]\texttt{ }[/tex]
[tex]v = \lambda f[/tex]
[tex]v = 32 \times \frac{1}{6}[/tex]
[tex]v = 5\frac{1}{3} \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
Learn moreDoppler Effect : https://brainly.com/question/3841958Example of Doppler Effect : https://brainly.com/question/810552Sound Waves Cannot Travel In Space. : https://brainly.com/question/546436Frequency of The Beats - Doppler Effect : https://brainly.com/question/12367463[tex]\texttt{ }[/tex]
Answer detailsGrade: College
Subject: Physics
Chapter: Sound Waves
Determine explicitly which is faster, 75 miles per hour or 75 m/s? Express all your results in MKS units and explain your conclusions from the numbers, and explain using words. Show all work.
Answer:
75 m/s is faster
Explanation:
MKS units stands for meter kilogram seconds
75 miles per hour = 75 mph
1 mile = 1609.34 meters
1 hour = 60×60 = 3600 seconds
1 mph = 1609.34/3600 = 0.44704 m/s
75 mph = 75×0.44707 = 33.52792 m/s
Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.
An electric potential difference of 10,00 V is applied to a 4.5 x 10^-3 C test charge. How much work is done on the test charge?
Answer:
The work done on the test charge is 4.5 joules
Explanation:
It is given that,
Electric potential difference, [tex]V=1000\ V[/tex]
Charged particle, [tex]q=4.5\times 10^{-3}\ C[/tex]
Work done on the test charge is given by the product of test charge and potential difference i.e.
W = q × V
[tex]W=4.5\times 10^{-3}\ C\times 1000\ V[/tex]
W = 4.5 Joules
So, the work done on the test charge is 4.5 joules. Hence, this is the required solution.
A projectile is fired at an upward angle of 55° from the top of a 120 m cliff with a speed of 150 m/s. What will be its speed when it strikes the ground below? (USE CONSERVATION OF ENERGY and neglect air resistance.) Express your answer to three significant figures and include the appropriate units.
Answer:
The speed when it strikes the ground below is V= 157.64 m/s < -56.92º .
Explanation:
V= 150m/s
α= 55º
hi= 120m
Vy= V*sinα
Vy= 122.87 m/s
Vx= V * cos α
Vx= 86.03 m/s
h= hi + Vy * t - g*t²/2
clearing t we get the total flying time of the projectile:
t(total fly)= 26.01 sec
0= Vy - g*t
clearing t we get the maximum height time:
t(max height)= 12.53 sec
to get the fall time:
t(fall)= t(total fly) - t(max height)
t(fall)= 13.48 sec
Vy'= g* t(fall)
Vy'= 132.1 m/s
V'= √(Vx² +Vy'²)
V'= 157.64 m/s
α'= tg⁻¹ (Vy'/Vx)
α'= -56.92º
By using the conservation of energy principle, we find the speed of the projectile when it hits the ground to be roughly 165 m/s (rounded to 3 significant figures).
Explanation:
Given that a projectile is fired upward at an angle of 55° from the top of a 120 m cliff at a speed of 150 m/s, it is asked what its speed will be when it strikes the ground. To answer such a question, we employ the principle of conservation of energy, which states that the total mechanical energy (kinetic energy + potential energy) of an isolated system remains constant if non-conservative forces like air resistance are negligible.
Now, the energy of the projectile at the top of the cliff is equal to its kinetic energy (as it's launched) and its potential energy (due to its height). When the projectile hits the ground, all its potential energy will be converted into kinetic energy as the body has descended from the height, which the body will possess as its speed.
Let's calculate:
Initial Energy = Final Energy
1/2 * mass * (initial speed)² + mass * g * height = 1/2 * mass * (final speed)²
Solving for final speed, -(initial speed)² - 2gh = - (final speed)²
finalSpeed = sqrt((initial speed)² + 2*g*h)
Given that the initial speed = 150m/s, g (acceleration due to gravity) = 9.81m/s², and the height = 120m, the final speed would amount to 164.833 m/s. Therefore, when rounding to 3 significant figures, the speed when it hits the ground will be 165 m/s.
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A system is initially at conditions of
P= 200 kPa and V= 0.1 m^3
Heat is now added to the system, and the system expands in a constant pressure process. At the end of the process the volume has doubled.
(a) Calculate the work transferred during the process.
(b) During the process the internal energy of the system increases by 200 kJ; kinetic and potential energy remains constant. Determine the net heat transfer during the process.
Answer:
a)W=20 KJ
b) ΔQ= 220 KJ
Explanation:
Given:
V₁=0.1 m^3, P₁=200 kPa and heat is added to the system such that system expands under constant pressure.
Therefore V₂= 2V₁= 0.2 m^3
a) Work transfer W= P(V₂-V₁)= [tex]200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules[/tex]
W=20 KJ
b) internal energy change ΔU= 200 KJ
from first law we know that ΔQ(net heat transfer)= ΔU + W
ΔQ= [tex]200\times10^3 +2\times10^4[/tex]
ΔQ=[tex]22\times10^4 J[/tex]
ΔQ= 220 KJ
At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.8 h. How long does it take for the water to fall a distance 0.250d from its highest level
Answer:
2.13 h
Explanation:
Simple harmonic motion is:
x = A sin(2π/T t + φ) + B
where A is the amplitude, T is the period, φ is the phase shift, and B is the midline.
This can also be written in terms of cosine:
x = A cos(2π/T t + φ) + B
Here, A = d/2, T = 12.8, φ = 0, and B = d/2. I'll use cosine so that the highest level is at t=0.
x = d/2 cos(2π/12.8 t) + d/2
When x = d - 0.250 d = 0.750 d:
0.750 d = d/2 cos(2π/12.8 t) + d/2
0.250 d = d/2 cos(2π/12.8 t)
0.500 = cos(2π/12.8 t)
π/3 = 2π/12.8 t
12.8/6 = t
t = 2.13
It takes 2.13 hours to fall 0.250 d from the highest level.
The time it takes for the water to fall a distance 0.250d from its highest level in the tide cycle can be calculated by determining a quarter of the period, which is 3.2 hours.
The time it takes for the water to fall a distance 0.250d from its highest level can be calculated using the concept of simple harmonic motion. Since the period of the tide is 12.8 hours, to find the time for the water to fall 0.250d, we need to determine the fraction of the period corresponding to this distance.
Given that the complete cycle from highest to lowest level is 12.8 hours, the time to fall 0.250d would be one-quarter of that period, which equals 3.2 hours.
A clarinetist, setting out for a performance, grabs his 3.450 kg3.450 kg clarinet case (including the clarinet) from the top of the piano and carries it through the air with an upward force of 38.92 N.38.92 N. Find the case's vertical acceleration. Indicate an upward acceleration as positive and a downward one as negative.
Answer:
1.5 m/s²
Explanation:
m = mass of the clarinet case = 3.450 kg
F = upward force on the clarinet through the air = 38.92 N
W = weight of the clarinet case in down direction
weight of the clarinet case is given as
W = mg
W = (3.450) (9.8)
W = 33.81 N
a = acceleration of the case
Force equation for the motion of the case is given as
F - W = ma
38.92 - 33.81 = (3.450) a
a = 1.5 m/s²
A 1.83 kg1.83 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.5220.522 and the coefficient of kinetic friction is 0.2830.283 . How much force is needed to begin moving the book?
Answer:
9.4 N
Explanation:
m = mass of the book = 1.83 kg
[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.522
[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.283
[tex]f_{s}[/tex] = Static frictional force
F = force needed to make the book move
force needed to make the book move is same as the magnitude of maximum static frictional force applied by the desk on the book
Static frictional force is given as
[tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg
Hence, the force need to move the book is given as
F = [tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg
F = [tex]\mu _{s}[/tex] mg
F = (0.522) (1.83 x 9.8)
F = 9.4 N
You are given three resistors with the following resistances: R1 = 6.32 Ω, R2 = 8.13 Ω, and R3 = 2.29 Ω. What is the largest equivalent resistance you can build using these three resistors?
Answer:
The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω
Explanation:
Adding Resistances in serie is the way to build de largest equivalent value possible.
Rt= R1+R2+R3
Rt= 6.32 + 8.13 + 2.29
Rt= 16.74Ω
What quantities determine the resistance of a piece of material? Choose all that apply.
The length of the piece of material
The cross-sectional area of the piece of material
The type of material
The voltage across the material
The current flowing through the piece of material
Answer:
Option (a), (b) and (c)
Explanation:
The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.
The formula for the resistance is given by
R = ρ x l / A
Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.
So, It depends on the length, area and the type of material.
Resistance of a material is determined by its length, cross-sectional area, and the type of material; voltage and current do not determine resistance but are influenced by it.
Explanation:The Quantities Determining the Resistance of a Material
The resistance of a piece of material is determined by several key factors, specifically:
The length of the piece of material: Resistance is directly proportional to the length. The longer the material, the higher the resistance as there will be more collisions between the charges and atoms within the material.The cross-sectional area of the piece of material: Resistance is inversely proportional to the cross-sectional area. Larger areas allow more current to flow, reducing resistance.The type of material: Different materials have different intrinsic resistivity properties, which affect their resistance.It is important to note that the voltage across the material and the current flowing through the material are not factors that determine resistance. These two quantities are actually influenced by the resistance according to Ohm's law, which states that Voltage (V) equals Current (I) times Resistance (R), or V = IR.
Two children push on opposite sides of a door during play. Both push horizontally and perpendicular to the door. One child pushes with a force of 17.5 N at a distance of 0.600 m from the hinges, and the second child pushes at a distance of 0.450 m. What force must the second child exert to keep the door from moving? Assume friction is negligible.
Answer:
The second child must exert a force of magnitude 23.3N to keep the door from moving.
Explanation:
We have to find the moment that the first child exerts and then match it to that exercised by the second child.
F1= 17.5N
d1= 0.6m
F2= ?
d2= 0.45m
M= F * d
M1= 17.5N * 0.6m
M1= 10.5 N.m
M1=M2
M2= F2 * 0.45m
10.5 N.m= F2 * 0.45m
10.5 N.m/0.45m = F2
F2=23.3 N
The force that the second child must exert to keep the door from moving is 23.33 N.
What is a balanced force?A balanced force occurs when an object subjected to different forces are at equilibrium.
Torque applied to the doorF1r1 = F2r2
(17.5 x 0.6) = F2(0.45)
F2 = 23.33 N
Thus, the force that the second child must exert to keep the door from moving is 23.33 N.
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If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1 R3 . If the resistances are measured in ohms as R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R. (Round your answer to three decimal places.) Ω
To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.
Explanation:To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.
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The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.
Explanation:The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:
1/R = 1/R1 + 1/R2 + 1/R3
To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.
After calculating 1/R using the given resistances:
1/R = 1/100 + 1/25 + 1/10
1/R = 0.01 + 0.04 + 0.1 = 0.15
Therefore, R = 1 / 0.15 = 6.667 Ω
We calculate the maximum possible errors in resistances as:
Error in R1 (eR1) = 0.005 * 100 Ω = 0.5 ΩError in R2 (eR2) = 0.005 * 25 Ω = 0.125 ΩError in R3 (eR3) = 0.005 * 10 Ω = 0.05 ΩUsing the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:
eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |
Plugging in the values:
eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |
eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)
The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.
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What is your weight on Venus, assuming that the acceleration due to gravity on Venus is 8.875 m/s^2 and your mass is 50 kg.
Answer:
Weight on Venus = 443.75 N
Explanation:
Weight of a body is the product of mass and acceleration due to gravity.
So we have
Weight = Mass x Acceleration due to gravity
W = mg
Mass, m = 50 kg
Acceleration due to gravity, g = 8.875 m/s²
W = 50 x 8.875 = 443.75 N
Weight on Venus = 443.75 N
Answer:
443.75 N
Explanation:
Weight is the force with which a planet can attract anybody towards its centre.
Weight = mass of body × acceleration due to gravity on that planet
Weight = 50 × 8.875 = 443.75 N
A person hums into the top of a well and finds that standing waves are established at frequencies of 140, 196, and 252 Hz. The frequency of 140 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?
Answer:
Depth of well 3.06m
Explanation:
We know that for a pipe closed at one end the frequencies are in ratios if 1:2:3:5:7.... to the fundamental frequency
In our case the given frequencies are in the ratio of
a)[tex]\frac{140}{196}=\frac{5}{7}[/tex]
b) [tex]\frac{196}{252}=\frac{7}{9}[/tex]
Thus the fundamental frequency can be calculated as [tex]140Hz=5n[/tex]
[tex]\therefore n=\frac{140}{5}=28Hz[/tex]
Now we know that
[tex]\lambda_{1}=4l\\\\\frac{v}{f}=4l\\\\l=\frac{v}{4f}[/tex]
Applying values we get
[tex]L=3.06m[/tex]
By identifying 56 Hz as the fundamental frequency and using the formula for the resonance of open-ended tubes, the depth of the well is calculated to be approximately 3.06 meters using the speed of sound as 343 m/s.
Explanation:The student has observed standing waves at frequencies of 140 Hz, 196 Hz, and 252 Hz in a well. These frequencies represent the natural resonant frequencies or harmonics of the well. To find the depth of the well, we need to consider these frequencies as harmonics of a sound wave corresponding to the lengths of the air column in the well where each harmonic creates a standing wave.
Given the multiple frequencies, there is a constant difference of 56 Hz (196 Hz - 140 Hz, 252 Hz - 196 Hz) between consecutive frequencies. This indicates that 56 Hz is the fundamental frequency of the harmonics. We can represent these frequencies as [tex]F1 = 56 Hz \(n=1\), F2 = 112 Hz \(n=2\)[/tex]ere the observed frequencies correspond to F3, F4, and F5.
To determine the length of the well (L), we use the formula for the resonance of open-ended tubes, [tex]\(L = \frac{v}{2f}\)[/tex]d of sound (343 m/s) and 'f' is the frequency (56 Hz in this case). Therefore, [tex]L = \frac{343 m/s}{2 \times 56 Hz} = 3.0625 m[/tex]ly 3.06 meters.
A red ball of mass 5.57 kg is suspended between two walls on a string, as illustrated here. The weight of the ball causes the string to sag, making an angle \theta=θ= 19.4 degrees as indicated. What is the resulting tension F T T in the string, in Newtons?
Answer:
164.33 N
Explanation:
Given:
The mass of the ball, m = 5.57 kg
Angle made by the sag = 19.4°
it is required to find the tension [tex]F_T[/tex] in the string
Note: Refer the attached figure
The tension in the string will be caused by the weight of the wall
thus,
Weight of the ball, W = 5.57 kg × 9.8 m/s² = 54.586 N
Now the resolving the tension [tex]F_T[/tex] in the string into components as shown in the figure attached, the weight of the ball will be balanced by the sin component of the tension
thus,
W = [tex]F_T[/tex]sinΘ
or
54.586 N = [tex]F_T[/tex] sin 19.4°
or
[tex]F_T[/tex] = 164.33 N
Hence, the tension in the string will be 164.33 N
A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tangentially at the rim (a) Determine the rotational kinetic energy oh the hoop (b) What is instantaneous change rate of the kinetic energy?
Answer:
The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
Explanation:
Given that,
Mass = 2 kg
Radius = 0.5 m
Angular speed = 3 rad/s
Force = 10 N
(I). We need to calculate the rotational kinetic energy
Using formula of kinetic energy
[tex]K.E =\dfrac{1}{2}\timesI\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}\times mr^2\times\omega^2[/tex]
[tex]K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2[/tex]
[tex]K.E=2.25\ J[/tex]
(II). We need to calculate the instantaneous change rate of the kinetic energy
Using formula of kinetic energy
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
On differentiating
[tex]\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}[/tex]
[tex]\dfrac{K.T}{dt}=mva[/tex]....(I)
Using newton's second law
[tex]F = ma[/tex]
[tex]a= \dfrac{F}{m}[/tex]
[tex]a=\dfrac{10}{2}[/tex]
[tex]a=5 m/s^2[/tex]
Put the value of a in equation (I)
[tex]\dfrac{K.E}{dt}=mva[/tex]
[tex]\dfrac{K.E}{dt}=mr\omega a[/tex]
[tex]\dfrac{K.E}{dt}=2\times0.5\times3\times5[/tex]
[tex]\dfrac{K.E}{dt}=15\ J/s[/tex]
Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.
The rotational kinetic energy of the hoop is 2.25 J, and the instantaneous change rate of the kinetic energy is 15 W.
(a) To determine the rotational kinetic energy (RKE) of the hoop, we use the formula for RKE, which is given by:
[tex]\[ RKE = \frac{1}{2} I \omega^2 \][/tex]
where I is the moment of inertia of the hoop and [tex]\( \omega \)[/tex] is the angular speed. The moment of inertia I for a hoop (or a thin-walled cylinder) rotating about its center is:
[tex]\[ I = m r^2 \][/tex]
where m is the mass of the hoop and r is the radius. Given that m = 2kg and ( r = 0.5 ) m, we can calculate I:
[tex]\[ I = 2 \times (0.5)^2 = 0.5 \text{ kg} \cdot \text{m}^2 \][/tex]
Now, we know [tex]\( \omega = 3 \)[/tex] rad/s. We can substitute the values into the RKE formula:
[tex]\[ RKE = \frac{1}{2} \times 0.5 \times (3)^2 = \frac{1}{2} \times 0.5 \times 9 = 2.25 \text{ J} \][/tex]
So, the rotational kinetic energy of the hoop is 2.25 J.
(b) The instantaneous change rate of the kinetic energy is equal to the power input to the system. Power P due to a torque [tex]\( \tau \)[/tex] is given by:
[tex]\[ P = \tau \omega \][/tex]
The torque [tex]\( \tau \)[/tex] can be calculated from the force F applied tangentially at the rim of the hoop:
[tex]\[ \tau = F \times r \][/tex]
Given that ( F = 10 ) N and ( r = 0.5 ) m, we find:
[tex]\[ \tau = 10 \times 0.5 = 5 \text{ N} \cdot \text{m} \][/tex]
Now, we can calculate the power:
[tex]\[ P = \tau \omega = 5 \times 3 = 15 \text{ W} \][/tex]
Therefore, the instantaneous change rate of the kinetic energy is 15 W.
Two small plastic spheres each have a mass of 1.1 g and a charge of -50.0 nC . They are placed 2.1 cm apart (center to center). A. What is the magnitude of the electric force on each sphere? B. By what factor is the electric force on a sphere larger than its weight?
Answer:
Part a)
F = 0.051 N
Part b)
Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.
Explanation:
Part a)
Electrostatic force between two charged spherical balls is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we will have
[tex]q_1 = q_2 = 50 nC[/tex]
here the distance between the center of two balls is given as
[tex]r = 2.1 cm = 0.021 m[/tex]
now we will have
[tex]F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}[/tex]
[tex]F = 0.051 N[/tex]
Part b)
Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.
A place kicker applies an average force of 2400 N to a football of .040 kg. The force is applied at an angle of 20.0 degrees from the horizontal. Contact time is .010 sec. a. Find the velocity of the ball upon leaving the foot. b. Assuming no air resistance find the time to reach the goal posts 40.0 m away. c. The posts are 4.00 m high. Is the kick good? By how much?
Answer:
a) The velocity of the ball upon leaving the foot = 600 m/s
b) Time to reach the goal posts 40.0 m away = 0.07 seconds
c) The kick won't e going inside goal post, it is higher by 10.34m.
Explanation:
a) Rate of change of momentum = Force
[tex]\frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s[/tex]
The velocity of the ball upon leaving the foot = 600 m/s
b) Horizontal velocity = 600 cos20 = 563.82 m/s
Horizontal displacement = 40 m
Time
[tex]t=\frac{40}{563.82}=0.07s[/tex]
Time to reach the goal posts 40.0 m away = 0.07 seconds
c) Vertical velocity = 600 sin20 = 205.21 m/s
Time to reach the goal posts 40.0 m away = 0.07 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m
Height of ball = 14.34 m
Height of post = 4 m
Difference in height = 14.34 - 4 = 10.34 m
The kick won't e going inside goal post, it is higher by 10.34m.
A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the lowest frequency associated with this portion of the spectrum?
Answer:
the lowest frequency is [tex]7.5\times 10^{14} Hz[/tex]
Explanation:
In the question it is given that wavelength(L) in the range of 200μm to 400μm.
let ν be frequency of wave v velocity = 3\times 10^8
velocity v= Lν
therefore ν=[tex]\frac{v}{L}[/tex]
frequency ν be lopwest when L will be heighest
ν(lowest)=[tex]\frac{3\times 10^8}{400\times 10^-9}[/tex]
ν=[tex]7.5\times 10^{14} Hz[/tex]
A snowboarder is at the top of a 90 foot tall slope making an angle of 40 degrees with the horizontal direction. If the snowboarder weighs 170 pounds and the coefficient of friction between the snow and snowboarder is 0.2, what is the snowboarders acceleration down the hill?
Answer:[tex]4.802 m/s^2[/tex]
Explanation:
height of slope(h) =90 foot
[tex]\theta =40[/tex]
weight of snowboarder=170 pounds\approx 77.1107 kg
[tex]\mu =0.2[/tex]
as the snowboarder is sliding down the slope therefore
Now net acceleration of snowboarder is
[tex]a_{net}=gsin\theta -\mu \cdot gcos\theta[/tex]
[tex]a_{net }=9.81\times sin\left ( 40\right )-0.2\times 9.81\times cos\left ( 40\right )[/tex]
[tex]a_{net }=4.802 m/s^2[/tex]
A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 0.163 m, while 1.95 × 104 J of heat is removed from the gas. If the piston has a radius of 0.272 m, calculate the change in internal energy of the system.
Answer:
The change in internal energy of the system is -17746.78 J
Explanation:
Given that,
Pressure [tex]P=4.63\times10^{4}\ Pa[/tex]
Remove heat [tex]\Delta U= -1.95\times10^{4}\ J[/tex]
Radius = 0.272 m
Distance d = 0.163 m
We need to calculate the internal energy
Using thermodynamics first equation
[tex]dU=Q-W[/tex]...(I)
Where, dU = internal energy
Q = heat
W = work done
Put the value of W in equation (I)
[tex]dU=Q-PdV[/tex]
Where, W = PdV
Put the value in the equation
[tex]dU=-1.95\times10^{4}-(4.63\times10^{4}\times3.14\times(0.272)^2\times(-0.163))[/tex]
[tex]dU=-17746.78\ J[/tex]
Hence, The change in internal energy of the system is -17746.78 J
A container of gas is at a pressure of 3.7 × 105 Pa How much work is done by the gas if its volume expands by 1.6 m3?
Answer:
Work is done by the gas = 5.92 x 10⁵ J = 592 kJ
Explanation:
Work done at fixed pressure, W = PΔV
Pressure, P = 3.7 x 10⁵ Pa
Change in volume, ΔV = 1.6 m³
Substituting the values of pressure and change in volume we will get
Work done at fixed pressure, W = PΔV = 3.7 x 10⁵ x 1.6 = 5.92 x 10⁵ J
Work is done by the gas = 5.92 x 10⁵ J = 592 kJ
A wheel starts from rest and rotates with constant angular acceleration and reaches an anaular speed of 12 rad/s in 3 s. Find (a) the magnitude of the angular acceleration of the wheel and (b) the angle (in radian) through which it rotates in this time.
Answer:
(a) 4 rad/s^2
(b) 18 rad
Explanation:
w0 = 0, w = 12 rad/s, t = 3 s
(a) Let α be the angular acceleration.
w = w0 + α t
12 = 0 + 3 α
α = 4 rad/s^2
(b) Let θ be the angle rotated
θ = w0 t + 1/2 α t^2
θ = 0 + 0.5 x 4 x 9
θ = 18 rad