Answer:
a. TRUTH
b. FALSE
c. TRUTH
d. FALSE
Explanation:
The emf (electromagnetic force) is generated in a loop or solenoid by the change in the magnetic flux in a closed conductor path (for example, a wire).
This can be noted in the following formula, which is known as the Lenz's law:
[tex]emf=-N\frac{d\Phi_B}{dt}=-N\frac{d(AB)}{dt}[/tex] (1)
Then, the change, in time, of the area of the conductor, or the change in the magnitude of the magnetic field, the induced emf acquires different values. Furthermore, the loops have a resistance, then, a current can be measured when an emf is induced.
Based on this information you have:
a. an induced emf is caused by a changing magnetic flux. TRUTH
b. an emf can only be induced in a conducting loop by moving the loop through an area that has a constant magnetic field. FALSE
c. an induced emf can be observed by measuring the current that is created. TRUTH
d. an induced emf and conventional induced current are in opposite directions. TRUTH (the minus sing in the equation (1) )
A student writes down several steps of scientific method. Put the steps in the best order
Answer:
Make a hypothesis, conduct an experiment, Analyze the experimental data..
PLEASE HELP !
Complete the following sentence. Choose the right answer from the given ones. The internal energy of the body can be changed A / B / C. A. only when the body is warmed or cooled B. when work is done on the body or heat flow C. only when the body does work
B
HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.
Answer:
Letter A. [tex]y=4.55 m[/tex]
Explanation:
Let's use the wave equation:
[tex]y=Asin(kx-\omega t)[/tex]
A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction (x=1.21 m)Therefore the displacement y will be:
[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]
[tex]y=4.55 m[/tex]
The answer is letter A.
I hope it helps you!
Answer:
Explanation:
Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
Amplitude (A) of the simple harmonic wave = 6.44 m
wave number (k) of the given wave = 2.34 m-1
Angular frequency (ω) of the given wave = 2.88 rad/s
Displacement x = 1.21 m and time t = 0.71 s
Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by
y = Asin(kx - ωt)
= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]
Y=6.44sin(0.7866 rad)
0.7866rad*(180 degrees/pi rad) =45.1
Y=6.44sin(45.1)
Y=4.55m
In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.
Answer:
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
Explanation:
The Free Body Diagram associated with the experiment is presented as attachment included below.
Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.
Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:
[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]
Where:
[tex]W[/tex] - Weight of the eraser, measured in newtons.
[tex]f[/tex] - Friction force, measured in newtons.
[tex]N[/tex] - Normal force, measured in newtons.
[tex]\theta[/tex] - Angle of the incline, measured in degrees.
The maximum allowable static friction force is:
[tex]f = \mu_{s} \cdot N[/tex]
Where:
[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
Likewise, the kinetic friction force is described by the following model:
[tex]f = \mu_{k} \cdot N[/tex]
Where:
[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.
[tex]N[/tex] - Normal force, measured in newtons.
And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:
[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]
[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]
But [tex]N = m\cdot g \cos \theta[/tex], so that:
[tex]\mu = \tan \theta[/tex]
Now, coefficients of static and kinetic friction are, respectively:
[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]
[tex]\mu_{s} \approx 0.716[/tex]
[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]
[tex]\mu_{k} \approx 0.596[/tex]
The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.
A particle with a charge of 5.1 μC is 3.02 cm from a particle with a charge of 2.51 μC . The potential energy of this two-particle system, relative to the potential energy at infinite separation, is
Answer:
U = 3.806 J
Explanation:
The potential energy between the two charges q1 and q2, is given by the following formula:
[tex]U=k\frac{q_1q_2}{r}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = 5.1*10^-6 C
q2 = 2.51*10^-6 C
r: distance of separation between particles = 3.02cm = 3.02*10^-2 m
You replace the values of all parameters in the equation (1):
[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]
The potential energy of the two particle system is 3.806 J
Two radio antennas A and B radiate in phase. Antenna B is a distance of 100 m to the right of antenna A. Consider point Q along the extension of the line connecting the antennas, a horizontal distance of 50.0 m to the right of antenna B. The frequency, and hence the wavelength, of the emitted waves can be varied.
Required:
a. What is the longest wavelength for which there will be destructive interference at point Q?
b. What is the longest wavelength for which there will be constructive interference at point Q?
Answer:
a. 200 m
b. 100 m
Explanation:
Solution:-
- We will first draw three points marked A,B and Q from left most to right most.
- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).
- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:
AQ - BQ = n*λ/2
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ/2
- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ/2
λ = 200 m .... Answer
- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:
AQ - BQ = n*λ
Where,
n: The order of wavelength
AQ: The distance between antenna A and point Q
BQ: The distance between antenna B and point Q
- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,
150 - 50 = n*λ
- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,
100 = λ
λ = 100 m .... Answer
An accident in a laboratory results in a room being contaminated by a radioisotope with a half life of 4.5 hours. If the radiation is measured to be 64 times the maximum permissible level, how much time must elapse before the room is safe to enter? The mass of Helium atom is 4.002602 u (where u = 1.66 x 10-27 kg) but the mass of 1 proton is 1.00730 u and 1 neutron is 1.00869 u. Calculate the binding energy per nucleon in MeV.
Answer:
a) t = 27.00 h
b) B = 6.84 MeV/nucleon
Explanation:
a) The time can be calculated using the following equation:
[tex] R = R_{0}e^{-\lambda*t} [/tex]
Where:
R: is the radiation measured at time t
R₀: is the initial radiation
λ: is the decay constant
t: is the time
The decay constant can be calculated as follows:
[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]
Where:
t(1/2): is the half life = 4.5 h
[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]
We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:
[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]
[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]
b) The binding energy (B) can be calculated using the following equation:
[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]
Where:
Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{p}[/tex]: is the proton mass = 1.00730 u
N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])
[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u
[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u
A = N + Z = 2 + 2 = 4
The binding energy of [tex]^{4}_{2}He[/tex] is:
[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]
Hence, the binding energy per nucleon is 6.84 MeV.
I hope it helps you!
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line
Answer:
Speed v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Explanation:
Given;
Mass transfer rate m = 5.37x10^-2 kg/s.
Density d = 739 kg/m3
radius of pipe r = 3.37x10^-3 m
We know that;
Density = mass/volume
Volume = mass/density
Volumetric flow rate V = mass transfer rate/density
V = m/d
V = 5.37x10^-2 kg/s ÷ 739 kg/m3
V = 0.00007266576454 m^3/s
V = 7.267 × 10^-5 m^3/s
V = cross sectional area × speed
V = Av
Area A = πr^2
V = πr^2 × v
v = V/πr^2
Substituting the given values;
v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))
v = 0.203678639672 × 10 m/s
v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
A 110-kg football player running at 8.00 m/s catches a 0.410-kg football that is traveling at 25.0 m/s. Assuming the football player catches the ball with his feet off the ground with both of them moving horizontally, calculate: the final velocity if the ball and player are going in the same directio
Answer:[tex]8.062\ m/s[/tex]
Explanation:
Given
masss of football player [tex]M=110\ kg[/tex]
Velocity of football player [tex]u_1=8\ m/s[/tex]
mass of football [tex]m=0.41\ kg[/tex]
velocity of football [tex]u_2=25\ m/s[/tex]
Final velocity will be given by applying conservation of linear momentum
After catching the ball Player and ball moves with same velocity
[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]
[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]
[tex]\Rightarrow 880+10.25=110.41\times v[/tex]
[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]
So, final velocity will be [tex]8.062\ m/s[/tex]
How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?
Answer:
Explanation:
There's a formula for this:
[tex]F = k*displacement[/tex]
F being force, k being the spring constant, and displacement being the change in x
We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
the heat capacity of 0.125Kg of water is measured to be 523j/k at a room temperature.Hence, calculate the heat capacity of water
(a) per unit mass
(b) per unit volume
Answer:
A. 4148 J/K/Kg
B. 4148 J/K/L
Explanation:
A. Heat capacity per unit mass is known as the specific heat capacity, c.
C = Heat capacity/mass(kg)
C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg
B. Volume of water = mass/density
Density of water = 1 Kg/L
Volume of water = 0.125 Kg/ 1Kg/L
Volume of water = 0.125 L
Heat capacity per unit volume = (523 J/K) / 0.125 L
Heat capacity per unit volume = 4148 J/K/L
A compact disk, which has a diameter of 12.0 cm, speeds up uniformly from zero to 4.30 rev/s in 3.05 s . Part A What is the tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed is 2.00 rev/s
Answer:
[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
Explanation:
Angular acceleration
[tex]\begin{aligned}
\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\
\omega_{i} &=0 \\
\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\
&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\
&=27.02 \mathrm{rad} / \mathrm{s} \\
\alpha &=\frac{(27.02-0)}{3.15} \\
&=8.57 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
a)Tangential acceleration
[tex]\begin{aligned}
a &=r \alpha \\
&=\frac{12}{2} \times 10^{-2} \times 8.57 \\
a &=0.51 \mathrm{m} / \mathrm{s}^{2}
\end{aligned}[/tex]
The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]
This question involves the concepts of the equations of motion for angular motion.
The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".
First, we will use the first equation of motion for the angular motion to find out the angular acceleration:
[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]
where,
[tex]\alpha[/tex] = angular acceleration = ?
[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s
[tex]\omega_i[/tex] = initial angular speed = 0 rad/s
t = time taken = 3.05 s
Therefore,
[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]
Now, the tangential acceleration can be given as follows:
[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]
a = 0.532 m/s²
Learn more about the angular motion here:
brainly.com/question/14979994?referrer=searchResults
The attached picture shows the angular equations of motion.
A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field B fills the region between the rails . Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :
A. v2m / ILB to yhe right
B. 3v2m /2 ILB to yhe left
C. 5v2m/ 2ILB to the right
D. v2m / 2ILB to the left
Answer:
F = ILB
Explanation:
To find the net force on the conducting bar you take into account the following expression:
[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]
I: current in the conducting bar
L: length of the bar
B: magnitude of the magnetic field
In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:
+i X +k = +j
The direction of the force is to the right and its magnitude is F = ILB
Espresso is a coffee beverage made by forcing steam through finely ground coffee beans. Modern espresso makers generate steam at very high pressures and temperatures, but in this problem we'll consider a low-tech espresso machine that only generates steam at 100?C and atomospheric pressure--not much good for making your favorite coffee beverage.The amount of heat Q needed to turn a mass m of room temperature ( T1) water into steam at 100?C ( T2) can be found using the specific heat c of water and the heat of vaporization Hv of water at 1 atmosphere of pressure.Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0?C into steam at 100?C. If c=4187J/(kg??C) and Hv=2,258kJ/kg, how much heat Q is absorbed by the water from the heating resistor inside the machine?Assume that this is a closed and isolated system.Express your answer in joules to three significant figures.Q = _________________ J
Answer:
Q = 3877 KJ
Explanation:
Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:
Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)
Q = Q₁ + Q₂ ----- equation (1)
Now, for Q₁:
Q₁ = m C ΔT
where,
m = Mass of Water = 1.5 kg
C = Specific Heat of Water = 4187 J/kg.°C
ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C
Therefore,
Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)
Q₁ = 490 x 10³ J =490 KJ
Now, for Q₂:
Q₂ = m H
where,
m = Mass of Water = 1.5 kg
H = Heat of Vaporization of Water = 2258 KJ/kg
Therefore,
Q₂ = (1.5 kg)(2258 KJ/kg)
Q₂ = 3387 KJ
Substituting the values in equation (1), we get:
Q = Q₁ + Q₂
Q = 490 KJ + 3387 KJ
Q = 3877 KJ
uring a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the wall. If the time the ball was in contact with the wall was 60.0 ms, what was the magnitude of the average force applied to the ball? 26.7 N 16.7 N 13.3 N 107 N 40.0 N
Answer:
107 N, option d
Explanation:
Given that
mass of the ball, m = 0.2 kg
initial velocity of the ball, u = 20 m/s
final velocity of the ball, v = -12 m/s
time taken, Δt = 60 ms
Solving this question makes us remember "Impulse Theorem"
It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"
Mathematically, it is represented as
FΔt = m(v - u), where
F = the average force
Δt = time taken
m = mass of the ball
v = final velocity of the ball
u = initial velocity of the ball
From the question we were given, if we substitute the values in it, we have
F = ?
Δt = 60 ms = 0.06s
m = 0.2 kg
v = -12 m/s
u = 20 m/s
F = 0.2(-12 - 20) / 0.06
F = (0.2 * -32) / 0.06
F = -6.4 / 0.06
F = -106.7 N
Thus, the magnitude is 107 N
A projectile is fired from ground level with an initial speed of 55.6 m/s at an angle of 41.2° above the horizontal. (a) Determine the time necessary for the projectile to reach its maximum height. (b) Determine the maximum height reached by the projectile. (c) Determine the horizontal and vertical components of the velocity vector at the maximum height. (d) Determine the horizontal and vertical components of the acceleration vector at the maximum heigh
Answer:
(a) t = 3.74 s
(b) H = 136.86 m
(c) Vₓ = 41.83 m/s, Vy = 0 m/s
(d) ax = 0 m/s², ay = 9.8 m/s²
Explanation:
(a)
Time to reach maximum height by the projectile is given as:
t = V₀ Sinθ/g
where,
V₀ = Launching Speed = 55.6 m/s
Angle with Horizontal = θ = 41.2°
g = 9.8 m/s²
Therefore,
t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)
t = 3.74 s
(b)
Maximum height reached by projectile is:
H = V₀² Sin²θ/g
H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)
H = 136.86 m
(c)
Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:
Vₓ = V₀ₓ = V₀ Cos θ
Vₓ = (55.6 m/s)(Cos 41.2°)
Vₓ = 41.83 m/s
Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.
Vy = 0 m/s
(d)
Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.
ax = 0 m/s²
The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:
ay = 9.8 m/s²
an object's resistance to any change in motion is the_________ of the object.
An object's resistance to any change in motion is the Inertia of the object.
Potential difference is measured in which units?
volts
amps
currents
watts
Answer:
Potential difference is measured in volts
Explanation:
The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.
Answer:
Your answer is A.) volts
Explanation:
On April 13, 2029 (Friday the 13th!), the asteroid 99942 mi Apophis will pass within 18600 mi of the earth-about 1/13 the distance to the moon! It has a density of 2600 kg/m^3, can be modeled as a sphere 320 m in diameter, and will be traveling at 12.6 km/s.
1)If, due to a small disturbance in its orbit, the asteroid were to hit the earth, how much kinetic energy would it deliver?
2)The largest nuclear bomb ever tested by the United States was the "Castle/Bravo" bomb, having a yield of 15 megatons of TNT. (A megaton of TNT releases 4.184x10^15 J of energy.) How many Castle/Bravo bombs would be equivalent to the energy of Apophis?
Answer:
Explanation:
Volume of asteroid = 4/3 x π x 160³
= 17.15 x 10⁶
mass = volume x density
= 17.15 x 10⁶ x 2600
= 445.9 x 10⁸ kg
kinetic energy
= 1/2 x 445.9 x 10⁸ x( 12.6 )² x 10⁶
= 35.4 x 10¹⁷ J .
2 )
energy of 15 megaton
= 4.184 x 10¹⁵ x 15 J
= 62.76 x 10¹⁵ J
No of bombs required
= 35.4 x 10¹⁷ / 62.76 x 10¹⁵
= 56.4 Bombs .
What is a substance?
Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new pipe which has a diameter half that of the first section of pipe. Determine the speed and pressure of the water in the new, reduced in size pipe.
Answer:
The velocity is [tex]v_2 = 6.8 \ m/s[/tex]
The pressure is [tex]P_2 = 204978 Pa[/tex]
Explanation:
From the question we are told that
The speed at which water is travelling through is [tex]v = 1.7 \ m/s[/tex]
The pressure is [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]
The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]
Where D is the diameter of first pipe
According to the principal of continuity we have that
[tex]A_1 v_1 = A_2 v_2[/tex]
Now [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as
[tex]A_1 = \pi \frac{D^2}{4}[/tex]
and [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as
[tex]A_2 = \pi \frac{d^2}{4}[/tex]
Recall [tex]d = \frac{D}{2}[/tex]
[tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]
[tex]A_2 = \frac{A_1}{4}[/tex]
So [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]
substituting value
[tex]1.7 = \frac{1}{4} * v_2[/tex]
[tex]v_2 = 4 * 1.7[/tex]
[tex]v_2 = 6.8 \ m/s[/tex]
According to Bernoulli's equation we have that
[tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]
substituting values
[tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]
[tex]P_2 = 204978 Pa[/tex]
Your electric drill rotates initially at 5.35 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.331 rad/s2 for 4.81 s. What is the drill's angular displacement during that time interval?
Answer:
The angular displacement is [tex]\theta = 29.6 \ rad[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w = 5.35 \ rad/s[/tex]
The angular acceleration is [tex]\alpha = 0.331 rad /s^2[/tex]
The time take is [tex]t = 4.81 \ s[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]
substituting values
[tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]
[tex]\theta = 29.6 \ rad[/tex]
At an accident scene on a level road, investigators measure a car’s skid mark (mass of car is M) to be of length d. It was a rainy day and the coefficient of friction was estimated to be μk.
A) Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.B) Why does the car's mass not matter?1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation.2) Since the work done by friction does not depend on mass.3) Since the change in kinetic energy and the work done by friction do not depend on mass.
Answer:
1) Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Explanation:
The kinetic friction works against the kinetic energy of the car and the car stops when these two equalises .
friction force = μk x R , μk is coefficient of kinetic friction and R is reaction from the ground.
= μk x mg
work done by friction
= force x displacement
= μk x mg x d
kinetic energy of car at the time of accident = 1/2 m v²
kinetic energy = work done by friction
1/2 m v² = μk x mg x d
d = v² / (2 μk x g)
v² = 2dμk g
v = √(2dμk g)
Since both the change in kinetic energy and the work done by friction are proportional to the mass. The mass cancels out of the equation
Someone please helppppppp!!!!!
A mass of 5.00 kg pulls down vertically on a string that is wound around a rod of radius 0.100 m and negligible moment of inertia. The rod is fixed in the center of a disk. The disk has mass 125 kg and radius 0.2 m. They turn freely about a fixed axis through the center. What is the angular acceleration of the rod, in radians/s 2
Answer:
0.981 rad/sec^2
Explanation:
mass that pulls on string = 5 kg
weight due to mass = 5 x 9.81 = 49.05 N
radius of rod = 0.1 m
torque produced by this force on the rod = force x radius
torque = 49.05 x 0.1 = 4.905 N-m
mass of disk = 125 kg
radius of disk = 0.2 m
moment of inertia of the disk I = m[tex]r^{2}[/tex]
I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2
from the equation, T = Iα
where T is torque
I is moment of inertia
α is angular acceleration
imputing values,
4.905 = 5α
α = 4.905/5 = 0.981 rad/sec^2
g The potential energy of a pair of hydrogen atoms separated by a large distance x is given by U(x)=−C6/x6, where C6 is a positive constant. Part A What is the force that one atom exerts on the other? Express your answer in terms of C6 and x. Fx = nothing Request Answer Part B Is this force attractive or repulsive? Is this force attractive or repulsive? attractive repulsive
Answer:
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
Attractive
Explanation:
Data provided in the question
The potential energy of a pair of hydrogen atoms given by [tex]\frac{C_6}{X_6}[/tex]
Based on the given information, the force that one atom exerts on the other is
Potential energy μ = [tex]\frac{C_6}{X_6}[/tex]
Force exerted by one atom upon another
[tex]F_x = \frac{\partial U}{\partial X} = \frac{\partial}{\partial X} (-\frac{C_6}{X^6})[/tex]
or
[tex]F_x = \frac{\partial}{\partial X} (\frac{C_6}{X^6})[/tex]
or
[tex]F_x = -\frac{6 C_6}{2^7}[/tex]
As we can see that the [tex]C_6[/tex] comes in positive and constant which represents that the force is negative that means the force is attractive in nature
Mark Watney (Matt Damon in the Martian movie) and Marvin the Martian (Looney Tunes cartoon character) are having an argument on the surface of Mars (negligible air resistance). They are testing out their new potato launcher that fires projectiles at a constant speed. Mark launches his potato at an angle of 60◦ and Marvin launches his identical potato at an angle of 30◦ . Without any calculations try to answer the following questions, and justify each answer.
(A) Which potato lands farther away from the launcher (potatoes are launched from ground level)?
(B) Which potato spends more time in the air before hitting the ground
(C) Which potato has a greater speed just before it hits the ground?
Answer:
A) The two potatoes cover the same horizontal distance from the launcher.
B) Mark's potato spends more time in the air than Marvin's potato before hitting the ground.
C) Marvin's potato hits the ground with a greater speed than Mark's potato
Explanation:
A) For projectile motion, the final horizontal distance of the projectile from where it was initially launched (its range) is given as
R = (u² sin 2θ)/g
where
u = initial velocity of the projectile
θ = angle above the horizontal at which the projectile was launched = 30°, 60°
g = acceleration due to gravity on Mars
Since, u and g are the same for Mark and Marvin, sin 2θ would determine which range is higher.
Sin (2×60°) = sin 120°
Sin (2×30°) = sin 60°
Sin 120° = Sin 60°
Hence, the two potatoes cover the same horizontal distance from the launcher.
B) Time spent in the air for a projectile is given as
T = (2u sin θ)/g
Again, since u and g are the same for Mark and Marvin on Mars, sin θ will give the required idea of whose potato spends more time in the air.
Sin 60° = 0.866
Sin 30° = 0.50
Sin 60° > Sin 30°
Hence, Mark's potato spends more time in the air than Marvin's potato.
C) The horizontal velocity for projectile motion is constant all through the motion and is equal to u cos θ
u cos 60° < u cos 30°
And the initial vertical velocity is u sin θ
Final vertical velocity
= (initial vertical velocity) - gt
g = acceleration due to gravity on Mars
T = time of flight
For Mark,
initial vertical velocity = u sin 60°, greater than Marvin's u sin 30°
And Mark's potato's time of flight is greater as established in (B) above.
But for Marvin
initial vertical velocity = u sin 30°, less than Mark's u sin 60°
And Marvin's potato's time of flight is lesser as established in (B) above
So, at the end of the day, the final vertical velocity is almost the same for both Mark's and Marvin's potatoes.
Hence, the horizontal component of the final velocity edges the final speed of the potatoes just before hitting the ground in Marvin's favour.
Hope this Helps!!!
A wire of length L is made up of two sections of two different materials connected in series. The first section of length L1 = 17.7 m is made of steel and the second section of length L2 = 28.5 m is made of iron. Both wires have the same radius of 5.30 ✕ 10−4 m. If the compound wire is subjected to a tension of 148 N, determine the time taken for a transverse pulse to move from one end of the wire to the other. The density of steel is 7.75 ✕ 103 kg/m3 and the density of iron is 7.86 ✕ 103 kg/m3.
Answer:
Explanation:
velocity of wave in a tense wire is given by the expression
[tex]v= \sqrt{\frac{T}{m} }[/tex]
v is velocity . T is tension and m is mass per unit length .
for steel wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.75 x 10³
= 683.57 x 10⁻⁵ kg/m
v = [tex]\sqrt{\frac{148}{683.57\times 10^{-5}} }[/tex]
= 1.47 x 10² m /s
= 147 m /s
for iron wire
m = π r² ρ where r is radius and ρ is density
= 3.14 x (5.3 x 10⁻⁴)²x7.86 x 10³
= 693.27 x 10⁻⁵ kg/m
[tex]v = \sqrt{\frac{148}{693.27\times 10^{-5}} }[/tex]
= 146 m /s
Time taken to move from one end to another
= 17.7 / 147 + 28.5 / 146
= .12 + .195
= .315 s .
Leah is moving in a spaceship at a constant velocity away from a group of stars. Which one of the following statements indicates a method by which she can determine her absolute velocity through space?
A) She can measure her increases in mass.
B) She can measure the contraction of her ship.
C) She can measure the vibration frequency of a quartz crystal.
D) She can measure the changes in total energy of her ship.
E) She can perform no measurement to determine this quantity.
Answer:
E) She can perform no measurement to determine this quantity.
Explanation:
A spacecraft is a machine used to fly in outer space.
According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.
As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.
