Answer:
R = 6.67 Ohm's
Explanation:
Resistance is a property of a material that measures the opposition to the flow of current through the material. It is measured in Ohm's.
R = (ρl) ÷ A
where; R is the resistance of the material, A is its cross sectional area, l is its length and ρ is its resistivity.
For the first nichrome wire, resistance is 5 Ohm's.
i.e R = (ρl) ÷ A = 5 Ohm's
For the second nichrome wire, length = 4l, and area of cross section = 3A. The resistance of the second wire can be determined by;
R = (4ρl) ÷ 3A
= [tex]\frac{4}{3}[/tex] {(ρl) ÷ A}
= [tex]\frac{4}{3}[/tex] × 5
R = 6.67 Ohm's
Resistance of the second nichrome wire is 6.67 Ohm's.
g A The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x=x1. Find Ff, the magnitude of the average frictional force that acts on the box. (Since you don't know the coefficient of friction, don't include it in your answer.) Express the frictional force in terms of m, v0, and x1. View Available Hint(s) Ff = nothing Part B After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1.
Answer:
fr = m v₀² / 2 (x₁-x₀)
Explanation:
a) For this exercise we use Newton's second law
X axis
- fr = ma
Y Axis
N-W = 0
N=W
let's look for acceleration with expressions of kinematics
v² = v₀² - 2 a Δx
at the point where stop v = 0
a = v₀² / 2 Δx
let's replace
-fr = m (- v₀² / 2 (x₁-x₀))
fr = m v₀² / 2 (x₁-x₀)
b)they ask for the same
in this case part of rest
v₁² = 0 + 2 a Δx
a = v₁² / 2ΔX
we write Newton's second law
F - fr = m a
fr = F - ma
fr = F - m v₁² / 2Δx
Object A has a charge q on it, object B has a charge q on it, and Object C has a charge 2q on it. These charges are arranged, one each, at the vertices of an equilateral triangle. Which charge has the greatest magnitude electric force on it?a. Ab. Bc. Cd. All have equal magnitude forces on them.
Answer: Option d.
Explanation:
The force between charges can be expressed as:
F = k*q1*q2/r^2
where k is a constant, q1 and q2 are the charges and r is the distance between them.
We hare in a equilateral triangle, so al the distances are equal.
and the charges are:
qa = q
qb = q
qc = 2q
Now, for example, the force that experiments charge A is (in the y axis)
F = (k*qa*qb/r^2 + k*qa*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for charge B, the force is again, in the y axis.
F = (k*qb*qa/r^2 + k*qb*qc/r^2)*cos(30°) = cos(30°)*(q + 2q)*k*q/r^2 = cos(30°)*3*k*q^2/r^2
for particle C, we have:
F = (k*qc*qa/r^2 + k*qc*qb/r^2)*cos(30°) = cos(30°)*(q + q)*k*2q/r^2 = cos(30°)*4*k*q^2/r^2
Wher the cosine of 30° comes because we have a equilatiral triangle, where all the internal angles are 60°, so if we draw a line that cuts the angle by half (our y-axis) the angles to each side are 30°.
We can do a similar process for the forces in the x-axis, and we will reach the same conclusion:
Now, this means that the force that experiences the charge C is the biggest force, so the correct option is c.
g A small car travels up the hill with a speed of v = 0.2 s (m/s) where s is the distance measured from point A in meters. Determine the magnitude of the car’s acceleration when it reaches s = 50 (m) where the road’s radius of curvature is r
Answer:
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is [tex]2\,\frac{m}{s^{2}}[/tex].
Explanation:
The acceleration can be obtained by using the following differential equation:
[tex]a = v \cdot \frac{dv}{ds}[/tex]
Where [tex]a[/tex], [tex]v[/tex] and [tex]s[/tex] are the acceleration, speed and distance masured in meters per square second, meters per second and meters, respectively.
Given that [tex]v = 0.2\cdot s[/tex], its first derivative is:
[tex]\frac{dv}{ds} = 0.2[/tex]
The following expression is obtained by replacing each term:
[tex]a = 0.2\cdot 0.2\cdot s[/tex]
[tex]a = 0.04\cdot s[/tex]
The magnitude of the acceleration of the car when [tex]s = 50\,m[/tex] is:
[tex]a = 0.04\cdot (50\,m)[/tex]
[tex]a = 2\,\frac{m}{s^{2}}[/tex]
What is the main difference between work power and energy
Answer:
Work is the energy required to move an object from one point to another. while power is the energy transferred per unit time.
Energy can also be defined as the ability to do work.
Rock can melt at a depth of about
below Earth's surface.
O A. 50 m
B. 500 km
C. 100 km
O D. 2000 km
Rock can melt at a depth of about below Earth's surface 100 km
What role does surface physics play in the world?On the other hand, interface physics offers a wide variety of spectroscopic & microscopic techniques to characterize that deposition & structure creation process upon that sub-nanometer size and, therefore, to pave the way for effective manufacturing techniques.
How does Surface Chemistry work?It is the study of both the chemical processes that take place at the meeting point of two surfaces, such as solid-liquid, solid-gas, sturdy, liquid-gas, etc. Surface engineering refers to a few surface chemistry applications.
To know more about surface visit:
https://brainly.com/question/31464585
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Jack and Jill went up the hill to fetch a pail of water. Jack, who’s mass is 75 kg, 1.5 times heavier than Jill’s mass, fell down and broke his crown after climbing a 15 m high hill. Jillcame tumbling after covering the same distance as Jack in 1/3rd of the time.Required:a. Who did the most work climbing up the hill? b. Who applied the most power?
Answer:
a) Jack does more work uphill
b) Numerically, we can see that Jill applied the most power downhill
Explanation:
Jack's mass = 75 kg
Jill's mass = [tex]1.5x = 75[/tex]
Jill's mass = [tex]x = \frac{75}{1.5}[/tex] = 50 kg
distance up hill = 15 m
a) work done by Jack uphill = mgh
where g = acceleration due to gravity= 9.81 m/s^2
work = 75 x 9.81 x 15 = 11036.25 J
similarly,
Jill's work uphill = 50 x 9.81 x 15 = 7357.5 J
this shows that Jack does more work climbing up the hill
b) assuming Jack's time downhill to be t,
then Jill's time = [tex]\frac{t}{3}[/tex]
we recall that power is the rate in which work id done, i.e
P = [tex]\frac{work}{time}[/tex]
For Jack, power = [tex]\frac{11036.25}{t}[/tex]
For Jill, power = [tex]\frac{3*7357.5}{t}[/tex] = [tex]\frac{22072.5}{t}[/tex]
Numerically, we can see that Jill applied the most power downhill
Which three terms are needed to describe the energy a BASE jumper has as
she falls toward the ground?
O A. Potential
B. Electromagnetic
C. Gravitational
D. Kinetic
Answer:
I’m saying kinetic gravitational and electromagnetic and I will comment on this if I got it right
Explanation:.
A 0.13 kg ball is moving at 6.6 m/s when it is hit by a bat, causing it to reverse direction and having a speed of 10.3 m/s, What is the change in the magnitude of the momentum of the ball
Answer:
Change in momentum = 2.197 kgm/s
Explanation:
Momentum = MV
Initial momentum = MU
Final momentum = MV
Computation:
⇒ Change in momentum = MV - MU
⇒ Change in momentum = M (V - U)
⇒ Change in momentum = 0.13(-10.3 - 6.6)
⇒ Change in momentum = 0.13(16.9)
⇒ Change in momentum = 2.197 kgm/s
g 0 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 m along the surface before stopping. Part A What was the initial speed of the bullet? Express your answer with the appropriate units. v = nothing
Answer:
vb = 298 m/s
Explanation:
Given:-
- The mass of the bullet, mb = 5.00 g
- The mass of the wooden block, mw = 1.2 kg
- The coefficient of kinetic friction between block and horizontal surface, [tex]u_k = 0.2[/tex]
- The bullet and block combined moves a distance after impact, s = 0.390
Solution:-
- We will first consider the motion of the block and bullet combined after the bullet is embedded into the block.
- We are told that the wooden block is rested on a horizontal floor with coefficient of kinetic friction ( uk ).
- When the bullet is embedded into the block. Both combined will move with a velocity ( V ). Both will eventually loose all of their kinetic energy by doing work against the friction.
- We will apply the work - energy principle to the system ( block + bullet ) as follows:
[tex]W = dE_k[/tex]
Where,
W: Work done against friction
dEk: The change in kinetic energy of the system
- The work-done against friction is the product of frictional force ( Ff ) and the displacement over which the system travels ( s ).
- The frictional force ( Ff ) is proportional to the contact force ( N ) between the system and the surface.
- Apply static balance on the system in the direction normal to the surface as follows:
[tex]N - ( m_b + m_w )*g = 0\\\\N = ( m_b + m_w )*g[/tex]
- The frictional force is defined as:
[tex]F_f = u_k*N\\\\F_f = u_k* ( m_b + m_w ) * g[/tex]
- The work done against friction ( W ) is defined as:
[tex]W = F_f*s\\\\W = u_k*(m_b + m_w )*g*s[/tex]
Where,
g: the gravitational acceleration constant = 9.81 m/s^2
- Now use the energy balance to determine the velocity of the system after impact ( V ):
[tex]u_k*( m_b + m_w )*g*s = 0.5*( m_b + m_w )*V^2\\\\V = \sqrt{2*u_k*g*s } \\\\V = \sqrt{2*0.2*9.81*0.39 }\\\\V = 1.23707 m/s[/tex]
- Now we will again consider an independent system of bullet and the wooden block resting on the horizontal surface.
- The bullet is fired with velocity ( vb ) towards the wooden block. The system can be considered to be isolated and all other fictitious effects can be ignored. This validates the use of conservation of linear momentum for this system.
- The conservation of linear momentum denotes:
[tex]P_i = P_f[/tex]
Where,
Pi: the inital momentum of the system
Pf: the final momentum of the system
- Initially the block was at rest and bullet had a velocity ( vb ) and after striking the bullet is embedded into the block and moves with a velocity ( V ).
[tex]m_b*v_b = ( m_b + m_w )*V\\\\v_b = \frac{ ( m_b + m_w )}{m_b}*V\\\\v_b = \frac{ ( 0.005 + 1.2 )}{0.005}*(1.23707)\\\\v_b = 298 m/s[/tex]
Answer: The initial speed of the bullet is equivalent to the speed of the bullet just before the impact as vb = 298 m/s. This is under the assumption that forces like ( air resistance or gravitational or impulse) have negligible effect.
At rest, a car's horn sounds at a frequency of 365 Hz. The horn is sounded while the car is moving down the street. A bicyclist moving in the same direction with one-third the car's speed hears a frequency of 357 Hz. What is the speed of the car?
Answer:
10.15m/s
Explanation:
The change in the frequency of sound (or any other wave) when the source of the sound and the receiver or observer of the sound move towards (or away from) each other is explained by the Doppler effect which is given by the following equation:
f₁ = [(v ± v₁) / (v ± v₂)] f ----------------------(i)
Where;
f₁ = frequency received by the observer or receiver
v = speed of sound in air
v₁ = velocity of the observer
v₂ = velocity of the source
f = original frequency of the sound
From the question, the observer is the bicyclist and the source is the car driver. Therefore;
f₁ = frequency received by the observer (bicyclist) = 357Hz
v = speed of sound in air = 330m/s
v₁ = velocity of the observer(bicyclist) = (1 / 3) v₂ = 0.33v₂
v₂ = velocity of the source (driver)
f = original frequency of the sound = 365Hz
Note: The speed of the observer is positive if he moves towards the source and negative if he moves away from the source. Also, the speed of the source is positive if it moves away from the listener and negative otherwise.
From the question, the cyclist and the driver are moving in the same direction. But then, we do not know which one is at the front. Therefore, two scenarios are possible.
i. The bicyclist is at the front. In this case, v₁ and v₂ are negative.
Substitute these values into equation (i) as follows;
357 = [(330 - 0.33v₂) / (330 - v₂)] * 365
(357 / 365) = [(330 - 0.33v₂) / (330 - v₂)]
0.98 = [(330 - 0.33v₂) / (330 - v₂)]
0.98 (330 - v₂) = (330 - 0.33v₂)
323.4 - 0.98v₂ = 330 - 0.33v₂
323.4 - 330 = (0.98 - 0.33)v₂
-6.6 = 0.65v₂
v₂ = -10.15
The value of v₂ is not supposed to be negative since we already plugged in the right value polarity into the equation.
iI. The bicyclist is behind. In this case, v₁ and v₂ are positive.
Substitute these values into equation (i) as follows;
357 = [(330 + 0.33v₂) / (330 + v₂)] * 365
(357 / 365) = [(330 + 0.33v₂) / (330 + v₂)]
0.98 = [(330 + 0.33v₂) / (330 + v₂)]
0.98 (330 + v₂) = (330 + 0.33v₂)
323.4 + 0.98v₂ = 330 + 0.33v₂
323.4 - 330 = (0.33 - 0.98)v₂
-6.6 = -0.65v₂
v₂ = 10.15
The value of v₂ is positive and that is a valid solution.
Therefore, the speed of the car is 10.15m/s
A car is designed to get its energy from a rotating flywheel with a radius of 1.50 m and a mass of 430 kg. Before a trip, the flywheel is attached to an electric motor, which brings the flywheel's rotational speed up to 5,200 rev/min.
Required:
a. Find the kinetic energy stored in the flywheel.
b. If the flywheel is to supply energy to the car as would a 15.0-hp motor, find the length of time the car could run before the flywheel would have to be brought back up to speed.
Answer:
a
[tex]KE = 7.17 *10^{7} \ J[/tex]
b
[tex]t = 6411.09 \ s[/tex]
Explanation:
From the question we are told that
The radius of the flywheel is [tex]r = 1.50 \ m[/tex]
The mass of the flywheel is [tex]m = 430 \ kg[/tex]
The rotational speed of the flywheel is [tex]w = 5,200 \ rev/min = 5200 * \frac{2 \pi }{60} =544.61 \ rad/sec[/tex]
The power supplied by the motor is [tex]P = 15.0 hp = 15 * 746 = 11190 \ W[/tex]
Generally the moment of inertia of the flywheel is mathematically represented as
[tex]I = \frac{1}{2} mr^2[/tex]
substituting values
[tex]I = \frac{1}{2} ( 430)(1.50)^2[/tex]
[tex]I = 483.75 \ kgm^2[/tex]
The kinetic energy that is been stored is
[tex]KE = \frac{1}{2} * I * w^2[/tex]
substituting values
[tex]KE = \frac{1}{2} * 483.75 * (544.61)^2[/tex]
[tex]KE = 7.17 *10^{7} \ J[/tex]
Generally power is mathematically represented as
[tex]P = \frac{KE}{t}[/tex]
=> [tex]t = \frac{KE}{P}[/tex]
substituting the value
[tex]t = \frac{7.17 *10^{7}}{11190}[/tex]
[tex]t = 6411.09 \ s[/tex]
What is the particle arrangement in a liquid
Answer:
the particle arrangement in liquid are close together with no regular arrangement
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 26.5 m/s (about 59 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Answer:
The centripetal acceleration of the car will be 12.32 m/s² .
Explanation:
Given that
radius ,R= 57 m
Velocity , V=26.5 m/s
We know that centripetal acceleration given as follows
[tex]a_c=\dfrac{V^2}{R}[/tex]
Now by putting the values in the above equation we get
[tex]a_c=\dfrac{26.5^2}{57}=12.32\ m/s^2[/tex]
Therefore the centripetal acceleration of the car will be 12.32 m/s² .
how much weight can a man lift in the jupiter if he can lift 100kg on the earth.calculate
Answer:
2479 NewtonSolution,
Mass=100 kg
Acceleration due to gravity(g)=24.79 m/s^2
Now,.
[tex]weight = m \times g \\ \: \: \: \: \: \: \: \: \: \: = 100 \times 24.79 \\ \: \: \: \: \: \: = 2479 \: newton[/tex]
hope this helps ..
Good luck on your assignment..
One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand to 3 times its original
volume, its final pressure is 1.07 x 10^6 Pa. What is its final temperature:
Answer:
The final temperature is T2= 5.35°C
Explanation:
Apply the Gay-lussacs's law we have
[tex]\frac{P1}{T1} = \frac{P2}{T2}[/tex]
P1, initial pressure= 5.00 x 10^6 Pa
T1, initiation temperature= 25.°C
P2, final pressure= 1.07 x 10^6 Pa
T2, final temperature= ?
[tex]\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\[/tex]
Cross multiplying and making T2 subject of formula we have
[tex]T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35[/tex]
T2= 5.35°C
A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux through it is approximately Group of answer choices
Answer:
126 mWb
Explanation:
Given that:
length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.
We assume that the magnetic field in the solenoid is constant.
The magnetic flux is given as:
[tex]\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb[/tex]
1. Consider the ball example in the introduction when a ball is dropped from 3 meters. After the ball bounces, it raises to a height of 2 meters. The mass of the ball is 0.5 kg a. Calculate the speed of the ball right before the bounce. b. How much energy was converted into heat after the ball bounced off the ground
Answer:
(a) 7.67 m/s.
(b) 4.9 J
Explanation:
(a) From the law of conservation of energy,
P.E = K.E
mgh = 1/2(mv²)
therefore,
v = √(2gh)....................... Equation 1
Where v = speed of the ball before bounce, g = acceleration due to gravity, h = height from which the ball was dropped.
Given: h = 3 m, g = 9.8 m/s²
Substitute into equation 1
v = √(2×9.8×3)
v = √(58.8)
v = 7.67 m/s.
(b) Energy of the ball before the bounce = mgh = 0.5×9.8×3 = 14.7 J
Energy of the ball after the bounce = mgh' = 0.5(9.8)(2) = 9.8 J
Amount of energy converted to heat = 14.7-9.8 = 4.9 J
An ideal transformer has LaTeX: N_1 = 1000 N 1 = 1000 (number of windings on the primary side), and LaTeX: N_2 = 8000 N 2 = 8000 (number of windings on the secondary side). If the rms voltage on the primary side is LaTeX: V_{rms}=100V V r m s = 100 V , what is the rms voltage on the secondary side? Give your answer in terms of Volts (rms), without entering the units.
Answer:
V₂ = 800 Volts (rms)
Explanation:
The turns ratio of an ideal transformer is given by the following formula:
N₁/N₂ = V₁/V₂
where,
N₁ = No. of turns in primary coil of the transformer = 1000 N
N₂ = No. of turns in secondary coil of the transformer = 8000 N
V₁ = rms Voltage on primary side of the transformer = 100 V
V₂ = rms Voltage on secondary side of the transformer = ?
Therefore, using these values in equation, we get:
1000 N/8000 N = 100 V/V₂
V₂ = (100)(8) Volts
V₂ = 800 Volts (rms)
A shell (a large bullet) is shot with an initial speed of 20 m/s, 60 degrees above the horizontal. At the top of the trajectory, the bullet explodes into two fragments of equal mass. One fragment has a speed of zero just after the explosion and simply drops straight down. How far from the gun does the other fragment land, assuming that the ground is level and that the air drag is negligible.
Answer:
17.656 m
Explanation:
Initial speed u = 20 m/s
angle of projection α = 60°
at the top of the trajectory, one fragment has a speed of zero and drops to the ground.
we should note that the top of the trajectory will coincide with halfway the horizontal range of the the projectile travel. This is because the projectile follows an upward arc up till it reaches its maximum height from the ground, before descending down by following a similar arc downwards.
To find the range of the projectile, we use the equation
R = [tex]\frac{u^{2}sin2\alpha }{g}[/tex]
where g = acceleration due to gravity = 9.81 m/s^2
Sin 2α = 2 x (sin α) x (cos α)
when α = 60°,
Sin 2α = 2 x sin 60° x cos 60° = 2 x 0.866 x 0.5
Sin 2α = 0.866
therefore,
R = [tex]\frac{20^{2}*0.866 }{9.81}[/tex] = 35.31 m
since the other fraction with zero velocity drops a top of trajectory, distance between the two fragments assuming level ground and zero air drag, will be 35.31/2 = 17.656 m
A 200 turn coil is in a uniform magnetic field that is decreasing at the rate 0.20 T/s. The coil is perpendicular to the field and its dimensions are 0.20 m by 0.40 m. What is the magnitude of the induced emf in the coil
Answer:
emf = 3.2V
Explanation:
In order to calculate the magnitude of the induced emf in the coil you use the following formula:
[tex]emf=-N\frac{d\Phi_B}{dt}[/tex] (1)
N: turns of the coil = 200
ФB: magnetic flux = A*B
A: area of the coil = (0.20m)(0.40m) = 0.08m²
B: magnitude of the magnetic field
You take into account that the area of the coil is constant, while magnetic field changes on time. Then, the equation (1) becomes:
[tex]emf=-NA\frac{dB}{dt}[/tex] (2)
dB/dt = rate of change of the magnetic field = -0.20T/s (it is decreasing)
You replace the values of all parameters in the equation (2):
[tex]emf=-200(0.08m^2)(-0.20T/s)=3.2V[/tex]
The induced emf in the coil is 3.2V
Strontium decays by beta decay part of the nuclear equation is shown below fill in the blank with a number? 90/38Sr -> 0/-1e 90/blankY
Answer : The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Explanation :
Beta decay : It is defined as the process in which beta particle is emitted. In this process, a neutron gets converted to a proton and an electron.
The released beta particle is also known as electron.
The beta decay reaction is:
[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]
The chemical equation for the beta decay process of [tex]_{38}^{90}\textrm{Sr}[/tex] follows:
[tex]_{38}^{90}\textrm{Sr}\rightarrow _{39}^{90}\textrm{Y}+_{-1}^0\beta[/tex]
Answer:
the blank is 39
Explanation: a p e x
A surface is bombarded by particles, each of mass small 'm', which have velocity
normal to the surface. On average, n particles strike unit area of the surface each second
and rebound elastically. What is the pressure on the surface?
A. nmv
B. 2nmv
C. nmv²
D. 1/2nmv²
Answer:
B. 2nmv
Explanation:
Pressure is force over area.
P = F / A
Force is mass times acceleration.
F = ma
Acceleration is change in velocity over change in time.
a = Δv / Δt
Therefore:
F = m Δv / Δt
P = m Δv / (A Δt)
The total mass is nm.
The change in velocity is Δv = v − (-v) = 2v.
A = 1 and Δt = 1.
Plugging in:
P = (nm) (2v) / (1 × 1)
P = 2nmv
A woman is standing in the ocean, and she notices that after a wave crest passes by, five more crests pass in a time of 29.4 s. The distance between two successive crests is 31.4 m. What is the wave's (a) period, (b) frequency, (c) wavelength, and (d) speed
Answer:
(a) 5.88 s
(b) 0.17 Hz
(c) 31.4 m
(d) 5.338 m/s
Explanation:
From the question,
(a) Period = time between successive crest = t/n............ Equation 1
where t = time, n = number of wave crest
Given: t = 29.4, n = 5
therefore,
Period (T) = 29.4/5 = 5.88 s.
(b) Frequency = 1/period
Frequency = 1/5.88
Frequency = 0.17 Hz.
(c) wavelength = distance between two successive crest = 31.4 m
(d) using,
v = λf............... Equation 2
Where v = speed, f = frequency, λ = wavelength
given: f = 0.17 Hz, λ = 31.4 m
Substitute into equation 2
v = 0.17(31.4)
v = 5.338 m/s
2) (Ch. 14.1, Equation 14.1) a. Write an equation for the oscillation period in terms of frequency. Define all SI units. b. If an oscillator completes 10 cycles in 5.0 seconds, what are the period and the frequency of the oscillation
Answer:
a) T = 2 π√ (m / k) c) T = 2 s , f = 0.5 Hz
Explanation:
a) In an oscillatory system the angular velocity is
w = √ (k / m)
where w is the angular velocity, k the constant of the spring and m the applied mass
the angular velocity is related to the frequency and the period
w = 2π f
f = 1 / T
w = 2π / T
we substitute
2π / T = √ k / m
T = 2 π√ (m / k)
b) The units If kilograms (m = [kg]) are for the masses, the spring constant is in Newton per meters (k = [N / m], angular velocity in (w = [rad / s]
c) they ask us the period if we have 10 cycles in 5 s
T = 10/5
T = 2 s
f = 1 / T
f = 0.5 Hz
4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strikes the ground at a horizontal distance of 160 m from the base of the building 5.0 s after being thrown. Determine the speed with which the rock was thrown.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Answer:
55.42 m/s
Explanation:
Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by
u_x = d/t
Where
d = 160 m is the distance covered
t = 5.0 s is the time taken
Substituting, we get
u_x =160/5 = 32 m/s.
Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -9.8 m/s^2 (downward). Therefore, the vertical distance covered is given by the
[tex]S=u_yt+\frac{1}{2}at^2[/tex]
where
S = -100 m is the vertical displacement
u_y is the initial vertical velocity
Replacing t = 5.0 s and solving the equation for u_y, we find
-100 = u_y(5) + (-9.81)(5)^2/2
u_y = 45.25 m/s
Therefore, the speed with which the rock was thrown u
[tex]u= \sqrt{u_x^2+u_y^2} \\=\sqrt{32^2+45.25^2}\\ = 55.42 m/s[/tex]
The figure shows an arrangement of four charged particles, with θ = 20.0° and d1 = 3.00 cm, which is the distance from the origin to a charge q1. Charge q1 is unknown, but q2= +7.00×10‒19 C and q3 = q4 = ‒2.00×10‒19 C. If there is no nett electrostatic force on q1 due to the other charges (the nett electrostatic force on q1 is zero), calculate the distance from the origin to q2, given by d2, in cm. Assume that all forces apart from the electrostatic forces in the system are negligible
Answer:
[tex]d_2=3.16cm[/tex]
Explanation:
So, in order to solve this problem, we must start by building a diagram of the problem itself. (See attached picture) And together with the diagram, we must build a free body diagram, which will include the forces that are being applied on the given charged particle together with their directions.
In this case we only care about the x-direction of the force, since the y-forces cancel each other. So if we do a sum of forces on the x-direction, we get the following:
[tex]\sum{F_{x}}=0[/tex]
so:
[Tex]-F_{12}+F_{13x}+F_{14x}=0[/tex]
Since [tex]F_{13x}=F_{14x}[/tex] we can simplify the equation as:
[tex]-F_{12}+2F_{13x}=0[/tex]
we can now solve this for [tex]F_{12}[/tex] so we get:
[tex]F_{12}=2F_{13x}[/tex]
Now we can substitute with the electrostatic force formula, so we get:
[tex]k_{e}\frac{q_{1}q_{2}}{r_{12}^{2}}=2k_{e}\frac{q_{1}q_{3}}{r_{13}^{2}}cos \theta[/tex]
We can cancel [tex]k_{e}[/tex] and [tex]q_{1}[/tex]
so the simplified equation is:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{r_{13}^{2}}cos \theta[/tex]
From the given diagram we know that:
[tex]cos \theta = \frac{d_{1}}{r_{13}}[/tex]
so when solving for [tex]r_{13}[/tex] we get:
[tex]r_{13}=\frac{d_{1}}{cos\theta}[/tex]
and if we square both sides of the equation, we get:
[tex]r_{13}^{2}=\frac{d_{1}^{2}}{cos^{2}\theta}[/tex]
and we can substitute this into our equation:
[tex]\frac{q_{2}}{r_{12}^{2}}=2\frac{q_{3}}{d_{1}^{2}}cos^{3} \theta[/tex]
so we can now solve this for [tex]r_{12}[/tex] so we get:
[tex]r_{12}=\sqrt{\frac{d_{1}^{2}q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
which can be rewritten as:
[tex]r_{12}=d_{1}\sqrt{\frac{q_{2}}{2q_{3}cos^{3}\theta}}[/tex]
and now we can substitute values.
[tex]r_{12}=(3cm)\sqrt{\frac{7x10^{-19}C}{2(2x10^{-19}C)cos^{3}(20^{o})}}[/tex]
which solves to:
[tex]r_{12}=6.16cm[/tex]
now, we must find [tex]d_{2}[/tex] by using the following equation:
[tex]r_{12}=d_{1}+d_{2}[/tex]
when solving for [tex]d_{2}[/tex] we get:
[tex]d_{2}=r_{12}-d_{1}[/tex]
when substituting we get:
[tex]d_{2}=6.16cm-3cm[/tex]
so:
[tex]d_{2}=3.16cm[/tex]
In the 25 ftft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of intensity 2500 W/m^2 at the floor of the facility. (This simulates the intensity of sunlight near the planet Venus.)
Required:
Find the average radiation pressure (in pascals and in atmospheres) on
a. A totally absorbing section of the floor.
b. A totally reflecting section of the floor.
c. Find the average momentum density (momentum per unit volume) in the light at the floor.
Answer:
a) 8.33 x [tex]10^{-6}[/tex] Pa or 8.22 x [tex]10^{-11}[/tex] atm
b) 1.66 x [tex]10^{-5}[/tex] Pa or 1.63 x [tex]10^{-10}[/tex] atm
c) 2.77 x [tex]10^{-14}[/tex] kg/m^2-s
Explanation:
Intensity of light = 2500 W/m^2
area = 25 ft^2
a) average radiation pressure on a totally absorbing section of the floor[tex]Pav = \frac{I}{c}[/tex]
where I is the intensity of the light
c is the speed of light = [tex]3*10^{8} m/s[/tex]
[tex]Pav = \frac{2500}{3*10^{8} }[/tex] = 8.33 x [tex]10^{-6}[/tex] Pa
1 pa = [tex]9.87*10^{-6}[/tex]
8.33 x [tex]10^{-6}[/tex] Pa = 8.22 x [tex]10^{-11}[/tex] atm
b) average radiation for a totally radiating section of the floor
[tex]Pav = \frac{2I}{c}[/tex]
this means that the pressure for a totally radiating section is twice the average pressure of the totally absorbing section
therefore,
Pav = 2 x 8.33 x [tex]10^{-6}[/tex] = 1.66 x [tex]10^{-5}[/tex] Pa
or
Pav in atm = 2 x 8.22 x [tex]10^{-11}[/tex] = 1.63 x [tex]10^{-10}[/tex] atm
c) average momentum per unit volume is
[tex]m = \frac{I}{c^{2} }[/tex]
[tex]m = \frac{2500}{(3*10^{8}) ^{2} }[/tex] = 2.77 x [tex]10^{-14}[/tex] kg/m^2-s
Calculate the electric potential due to a dipole whose dipole moment is 5.2×10−30 C⋅m at a point r = 2.8×10−9 m away. Suppose that r≫ℓ, where ℓ is the distance between the charges in the dipole.
Answer:
V = 8.01*10^-12 V
Explanation:
In order to calculate the electric potential produced by the dipole you use the following formula:
[tex]V=k\frac{p}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
p: dipole moment = 5.2×10−30 C⋅m
r: distance to the dipole = 2.8*10^-9m
You replace the values of the parameters:
[tex]V=(8.98*10^9Nm^2/C^2)\frac{5.2*10^{-30}Cm}{(2.8*10^{-9}m)}\\\\V=8.01*10^{-12}V[/tex]
The electric potential of the dipole is 8.01*10^-12V
The flowers of the bunchberry plant open with astonishing force and speed, causing the pollen grains to be ejected out of the flower in a mere 0.30 ms at an acceleration of 2.5 × 104 m. s2 If the acceleration is constant, what impulse is delivered to a pollen grain with a mass of 1.0 × 10−7g?
Answer:
I = 7.5*10^-10 kg m/s
Explanation:
In order to calculate the impulse you first take into account the following formula:
[tex]I=m\Delta v=m(v-v_o)[/tex] (1)
m: mass of the pollen grain = 1.0*10^-7g = 1.0*10^-10 kg
v: final speed of the pollen grain = ?
vo: initial speed of the pollen grain = 0 m/s
Next, you calculate the final speed of the pollen grain by using the information about the acceleration and time. You use the following formula:
[tex]v=v_o+a t[/tex] (2)
a: acceleration = 2.5*10^4 m/s^2
t: time = 0.30ms = 0.30*10^-3 s
[tex]v=0m/s+(2.5*10^4m/s^2)(0.30*10^{-3}s)=7.5\frac{m}{s}[/tex]
Next, you replcae this value of v in the equation (1) and calculate the impulse:
[tex]I=m(v-v_o)=(1.0*10^{-10}kg)(7.5m/s-0m/s)=7.5*10^{-10}kg.\frac{m}{s}[/tex]
The impulse delivered to the pollen grain is 7.5*10^-10 kg m/s
A car and a truck start from rest at the same instant, with the car initially at some distance behind the truck. The truck has a constant acceleration of 2.10 m/s2 and the car an acceleration of 3.40 m/s2. The automobile overtakes the truck after the truck has moved 60.0 m.Requried:a. How much time does it take the car to overtake the truck? b. How far was the car behind the truck initially? c. What is the speed of each when they are abreast? d. On a single graph, sketch the position of each vehicle as a function of time. Take x = 0 at the initial location of the truck.
Answer:
A = 7.56s
B = 37.16m
C = for car 25.704m/s and for truck = 15.876m/s
Explanation:
Hello,
This is an example of relative motion between two moving bodies and equation of motion are usually modified to solve problems involving relative motion.
In this question, we'll be making use of
x - x₁ = v₁t + ½at²
The above equation is a modification of
x = vt + ½at²
Where x = distance
v = velocity of the body
a = acceleration of the body
t = time
x - x₁ = v₁t + ½at²
Assuming the starting point of the bodies x₁ = 0 which is at rest, the car sped past the truck at 60m.
x - x₁ = v₁t + ½at²
x₁ = 0
v₁ = 0
x = 60
a = 2.10
t = ?
60 - 0 = 0 × t = ½ × 2.10t²
Solve for t
60 = 1.05t²
t² = 60 / 1.05
t² = 57.14
t = √(57.14)
t = 7.56s
The time it took the car to over take the truck is 7.56s
b).
From our previous equation,
x - x₁ = v₁t + ½at²
Same variable except
a = 3.40m/s² and t = 7.56s
60 - x₁ = 0 + ½ × 3.40 × 7.56²
60 - x₁ = 97.16
x₁ = 60 - 97.16
x₁ = -37.16m
The car was behind the truck by 37.16m or the truck was ahead of the car by 37.16m.
c).
The initial speed of the car ?
v = u + at
u = 0m/s
a = 3.40m/s²
v = 0 + 3.40 × 7.56
v = 25.704m/s
The initial speed of the truck = ?
v = u + at
u = 0m/s
a = 2.10m/s
v = 0 + 2.10 × 7.56
v = 15.876m/s
d).
Due to some circumstance, kindly check attached document for a sketch of the graph