Answer:
the speed is 2
Explanation:
speed= distance/ time
speed= 50/25
= 2
hope this helps!
Two loops of wire are arranged so that a changing current in one will induce a current in the other. If the current in the first is increasing clockwise by one amp every second, the current in the second loop will
Answer:
The current in the second loop will stay constant
Explanation:
Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)
Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.
So, i₂R = -Mdi₁/dt
i₂ = -Mdi₁/dt/R
Since di₁/dt = + 1 A/s,
i₂ = -Mdi₁/dt/R
i₂ = -M × + 1 A/s/R
i₂ = -M/R
Since M and R are constant, this implies that i₂ = constant
So, the current in the second wire loop will stay constant.
A 1 500-kg car rounds an unbanked curve with a radius of 52 m at a speed of 12.0 m/s. What minimum coefficient of friction must exist between the road and tires to prevent the car from slipping
Explanation:
The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.
[tex]y:\:\:\:\:N - mg = 0[/tex]
[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]
or
[tex]m \dfrac{v^2}{r} = \mu mg[/tex]
Solving for [tex]\mu[/tex],
[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]
A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?
When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.
What is the Doppler formula?The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.
The frequency increase by the Doppler effect is represented by the formula
f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f
Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀ is 0.
Substituting the value into the equation will give us the velocity of the police car.
[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)
When the car is receding, the frequency of the receiving signal f = 4500 Hz.
[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)
Solving both equation, we get the velocity of a police car.
v = 33 m/s
Therefore, the velocity v of the police car is 33 m/s.
Learn more about Doppler equation.
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A curve in a road forms part of a horizontal circle. As a car goes around it at constant speed (12 m/s), the total force on the driver has a magnitude of 146 N. What is the total vector force (in N) on the driver if the speed is 18 m/s instead?
Answer:
a1 = v1^2 / R
a2 = v2^2 / R
a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4
F2 = 9/4 * F1 = 9/4 * 146 = N 329 N since F = m * a
A golf ball is dropped from rest from a height of 8.40 m. It hits the pavement, then bounces back up, rising just 5.60 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
Answer:
t1= 8.40/10 =.84 s
t2 = 5.60/10 = .56s
t3= 1.4/10 = .14s
total time = 1.54 sec
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30.08 as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp.
Answer:
2.55 m/s
Explanation:
A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.
Solution:
The work done by friction is given as:
[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]
The work done by gravity is:
[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]
Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above
Answer:
Explanation:no change in surface tension
An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.
Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.
The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.
As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
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Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has velocity 10 m/s and position s = 8m when t=0, determine its velocity and position when t= 3 s. Where the velocity become zero. Discuss briefly.
Explanation:
Given: a = -3v^2
By definition, the acceleration is the time derivative of velocity v:
[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]
Re-arranging the expression above, we get
[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]
Integrating this expression, we get
[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]
[tex] - \frac{1}{v} = - 3t + k[/tex]
Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as
[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]
or
[tex]v = \frac{10}{30t +1 } [/tex]
We also know that
[tex]v = \frac{ds}{dt} [/tex]
or
[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]
We can integrate this to get s:
[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]
Let u = 30t +1
du = 30dt
so
[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]
[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]
So we can now write s as
[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]
We know that when t = 0, s = 8 m, therefore k = 8 m.
[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]
Next, we need to find the position and velocity at t = 3 s. At t = 3 s,
[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]
[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]
Note: velocity approaches zero as t --> [tex]\infty [/tex]
13. What type of lens bends light outwards and away from a point?
concave
Answer:
No,it isn't concave. The correct answer is convex lens.
Explanation:
A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.
Convex lens is the answer.
See the attached diagram.
A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is
Answer:
a) T = 2π [tex]\sqrt{\frac{m}{k} }[/tex], b) T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
Explanation:
a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity
w² = k / m
angular velocity and period are related
w = 2π /T
we substitute
4π²/ T² = k / m
T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) We change the spring for another with k ’= 2 k, let's find the period
T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]
T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]
T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
2. Is this an open or a closed circuit?
open circuit
closed circuit
Answer:
Open Circuit
Explanation:
This Picture represents an open circuit.
Answer: Closed Circuit
Explanation:
Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.
5.00 × 109 J
10 × 107 J
50.0 × 108 J
1.00 × 108 J
Impervious surfaces prevent water from flowing through them.
True or false?
Answer: the answer is true
Explanation:
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
An object of mass 15 kg is at an elevation of 100 m relative to the surface of the Earth. What is the potential energy of the object, in kJ? If the object were initially at rest, to what velocity, in m/s, would you have to accelerate it for the kinetic energy to have the same value as the potential energy you calculated above? The acceleration of gravity is 9.8 m/s2.
Answer:
The potential energy of the object is 14.7 kJ
The velocity of the object is 44.27 m/s
Explanation:
Given;
mass of the given object, m = 15 kg
position of the object, h = 100 m
acceleration due to gravity, g = 9.8 m/s²
The potential energy of the object is calculated as;
P.E = mgh
P.E = 15 x 9.8 x 100
P.E = 14,700 J = 14.7 kJ
The velocity of the object if its kinetic energy must equal potential energy;
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 100)
v = 44.27 m/s
negative effects of restricting roles in a democratic society
Answer:
it could lead to riot
it could lead to uncertainty of life and property
Answer:
Positive: has everybody informed on whats going on and where. Unadulterated media could let the people know the truth about their governments.
Negative: adulterated media could give people wrong ideas and leading to more unnecessary problems. Media who prioritise money over society would always do what gets them more money, thus their information is unreliable. Since most people believe what the media says, media has the power to alter the truth within an instance
Electromagnetic waves from the sun carry what to the earth
Answer:
Solar radiation
Explanation:
Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.
==> Energy
==> Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays
==> They carry all these kinds of energy wherever they go. Not only to the Earth.
Four 8.5-kg spheres are located at the corners of a square of side 0.90 m.
A. Calculate the magnitude of the gravitational force exerted on one sphere by the other three.
B. Calculate the direction of the gravitational force exerted on one sphere by the other three.
Explanation:
the answer is in the above image
An 8.50 kg point mass and a 14.5 kg point mass are held in place 50.0 cm apart. A particle of mass (m) is released from a point between the two masses 12.0 cm from the 8.50 kg mass along the line connecting the two fixed masses.Find the magnitude of the acceleration of the particle.
Answer:
[tex]a=2.8*10^{-9}m/s[/tex]
Explanation:
From the question we are told that:
First Mass [tex]m=8.50kg[/tex]
2nd Mass [tex]m=14.5kg[/tex]
Distance
[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]
Generally the Newtons equation for Gravitational force is mathematically given by
[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]
Therefore
Initial force on m
[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]
Final force on m
[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]
Acceleration of m
[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]
[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]
[tex]a=2.8*10^{-9}m/s[/tex]
A physicist wants to estimate the rate of emissions of alpha particles from a certain source. He counts 400 emissions in 80 seconds. Estimate the rate, in emissions per second and find the uncertainty in this estimate.
Answer:
Emissions per second = 0.36
Explanation:
Please find the attached question
Solution
Given
Let X be the rate of background emission.
X = B/t
Where B = 36
And t = 100
X = 36/100 = 0.36
En un barrio un estudiante observa un roedor en medio de dos postes, los cuales están comunicados por un solo cable de fibra óptica, el cual calcula que tiene 50 m de longitud de poste a poste. Dicho roedor al estar apoyado pandea el cable 0.058 m. Si el estudiante intuye, que por sus clases de bilogía el roedor tiene aproximadamente 350gde masa, ¿Cuál será la tensión del cable debido a dicho suceso?
Answer:
7 Newton
Explanation:
Dado
Longitud de la cuerda = 50 m
El cable se dobla en 0,058 m.
Masa de roedor = 350 gramos = 0,35 kg
T = m * a + m × v2 / r
Sustituyendo los valores dados obtenemos
T = 0,35 (10 + 10)
T = 0,35 * 20
T = 7 Newton
A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2
Answer:
The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
Explanation:
Given the data in the question;
since the train starts from rest,
Initial velocity; u = 0 m/s
final velocity; v = 42 m/s
distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m
acceleration a = ?
From the third equation of motion;
v² = u² + 2as
we substitute in our values
( 42 )² = ( 0 )² + [ 2 × a × 5600 ]
1764 = 0 + [ 11200 × a ]
1764 = 11200 × a
a = 1764 / 11200
a = 0.1575 ≈ 0.16 m/s² { two decimal place }
Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²
Option c) 0.16 m/s² is the correct answer.
Water at 200 C has a bulk modulus of 2.2109 Pa, and the speed of sound in water at this temperature is 1480m/s. For 1000Hz sound waves in water at 200 C, what displacement amplitude is produced if the pressure amplitude is 310-2 Pa?
6. Show that the weight of an object on the moon is 1/6 its weight on earth.
Taking ratio of W & w. ≈ 6 . w = 1/6 W. Therefore , Weight of an object on the moon is 1/6 of its weight on the earth.
What is the law of conservation of energy
A 5.00-kg object is initially at rest. The object is acted on by a 9.00-N force toward the east for 3.00 s. No force acts on the object for the next 4.00 s. How far has the object moved during this 7.00 s interval?
Answer:
The total distance at 7 s is:
[tex]x_{tot}=27\: m[/tex]
Explanation:
Distance due to the force
We can use second Newton's law to find the acceleration.
[tex]F=ma[/tex]
[tex]a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}[/tex]
Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.
[tex]x_{1}=0.5at_{1}^{2}[/tex]
[tex]x_{1}=0.5(1.8)(3)^{2}[/tex]
[tex]x_{1}=8.1\: m[/tex]
In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.
First, we need to find the final velocity of the first interval
[tex]v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s[/tex]
So the second distance will be:
[tex]x_{2}=vt_{2}=5.4*4=21.6\: m[/tex]
Therefore, the total distance is:
[tex]x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m[/tex]
I hope it helps you!
The total distance at 7 s is: [tex]x_{total}=27\ m[/tex]
What is force?Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.
Distance due to the force
We can use second Newton's law to find the acceleration.
[tex]F=ma[/tex]
[tex]a=\dfrac{F}{m}=\dfrac{9}{5}=1.8\ \frac{m}{s^2}[/tex]
Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.
[tex]x_1=0.5at_1^2[/tex]
[tex]x_1=0.5(1.8)(3^2)[/tex]
[tex]x_1=8.1\ m[/tex]
In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.
First, we need to find the final velocity of the first interval
[tex]v=v_i+at_1=0+(1.8)(3)=5.4 \frac{m}{s}[/tex]
So the second distance will be:
[tex]x_2=vt_2=5.4\times 4=21.6\ m[/tex]
Therefore, the total distance is: [tex]x_{total}=27\ m[/tex]
To know more about force follow
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If the frequency of a sound wave is 250 hertz and its wavelength is 1.36 meters, what is the wave's velocity
OA. 250 meters/second
ОВ.
340 meters/second
O c.
200 meters/second
OD.
120 meters/second
Answer:
B
Explanation:
V=frequency*wavelength V=? W=1.36mF=250hertzAnswer:
B
Explanation:
Bc i say
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan
Answer:
F = 1010 Lb
the tension on the cable is greater than its resistance, which is why the plan is not viable
Explanation:
For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.
v = v₀ + a t
how the car comes out of rest v₀ = 0
a = v / t
let's reduce to the english system
v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s
let's calculate
a = 66/10
a = 6.6 ft / s²
now let's write Newton's second law
X axis
Fₓ = ma
with trigonometry
cos 20 = Fₓ / F
Fₓ = F cos 20
we substitute
F cos 20 = m a
F = m a / cos20
W = mg
F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]
let's calculate
F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20
F = 1010 Lb
Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.
Use the following information for questions 18 - 21. A 0.13 kg disk is rotating at an angular speed of 57 rad/s. The disk has a radius of 0.25 m. The disk speeds up for 3 s. After the 3 s have passed, the edge of the disk is under a centripetal force of 312.13 N. What is the centripetal acceleration of the disk at this time
Answer:
[tex]a=2401m/s^2[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.13kg[/tex]
Angular speed [tex]\omega=57rads/s[/tex]
Radius [tex]r=0.25m[/tex]
Time [tex]t=3s[/tex]
[tex]F=312.13N[/tex]
Generally the equation for centripetal acceleration is mathematically given by
[tex]a=\frac{f}{m}[/tex]
[tex]a=\frac{312.13}{0.13}[/tex]
[tex]a=2401m/s^2[/tex]
