Answer:
a). C = b/2 and C = b/4
b). [tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
c). m = 63.4 kg (approx.)
Explanation:
Ex. 2.4.4
The total force acting on mass m is [tex]$ F = F_{spring }= -kx $[/tex] , where x is the displacement from the equilibrium position.
The equation of motion is [tex]$ m {\overset{..}x} + kx = 0 $[/tex]
or [tex]$ {\overset{..}x}+ \frac{k}{m}x=0 $[/tex] or [tex]$ {\overset{..}x} + w_0^2 x = 0 $[/tex] , where [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex]
The solution is [tex]$ x = A \cos (w_0t + \phi) $[/tex] , where A and Ф are constants.
A is amplitude of motion
[tex]$ w_0$[/tex] is the angular frequency of motion
Ф is the phase angle.
Now, [tex]$ w_0 = 2 \pi f_0 = \sqrt{k/m} $[/tex]
or [tex]$ m = \frac{k}{4\pi f_0^2} $[/tex]
Given [tex]$ f_0 = 0.8 Hz , k = 4 N/m $[/tex]
a). [tex]$ m = \frac{4}{4(3.14)^2(0.8)^2} = 0.158\ kg$[/tex]
b). [tex]$ w_0^2 = k/m $[/tex]
or [tex]$ m = k/ w_0^2 = k / (2\pi f_0)^2 = k / 4 \pi^2 f_0^2 $[/tex]
Ex. 2.4.5
a). Total force acting on the mass m is [tex]$F = F_{spring}+f $[/tex]
[tex]$ = -kx-bv $[/tex]
The equation of motion is [tex]$ m {\overset{..}x}= -kx-b{\overset{.}x} $[/tex]
or [tex]$ w_0 = \sqrt{\frac{k}{m}} $[/tex] , angular frequency of the undamped oscillation.
γ = b/2m is called the damping coefficient (γ=C)
[tex]$ k = m w_0^2 = 4 \pi^2 m f_0^2 $[/tex]
for 1 kg weight (= 9.8 N), [tex]$ f_0$[/tex] = 1.1 Hz
k = 4 x (3.14)² x (9.8) x 1.1² = 4.6 x 10² N/m
For 2 kg weight (= 19.6 N), [tex]$ f_0$[/tex] = 0.8 Hz
k = 4 x 9.8596 x 2 x 9.8 x 0.8² = 5 x [tex]$ 10^7$[/tex] N/m
[tex]$ \gamma = \frac{b}{2m_1} = \frac{b}{2m_2} $[/tex]
or [tex]$ \gamma = \frac{b}{2 \times 1} = \frac{b}{2 \times 2} $[/tex]
γ = b/2 (for 1 kg) and γ = b/4 (for 2 kg)
C = b/2 and C = b/4
b). [tex]$ w_0^2 = \frac{k}{m} \Rightarrow \frac{k}{w_0^2} = \frac{k}{(2 \pi f_0)^2} = \frac{k}{4 \pi^2 f_0^2} $[/tex]
For two particle problem,
[tex]$ w'_0^2 = \sqrt{\frac{k(m_1+m_2)}{m_1 +m_2}} $[/tex]
[tex]$ \therefore T = 2 \pi \sqrt{\frac{m_1 +m_2}{k (m_1 + m_2)}} = 2 \pi \sqrt{ \mu/k}$[/tex]
where, μ is the reduced mass.
This time period is same for both the particles.
c). [tex]$ m =\frac{k}{4 \pi^2 f_0^2}$[/tex]
[tex]$ = \frac{5 \times 10^2}{4 \times 9.14^2 \times 0.2} = 63.4\ kg $[/tex] ( approx.)
A point charge Q moves on the x-axis in the positive direction with a speed of A point P is on the y-axis at The magnetic field produced at the point P, as the charge moves through the origin, is equal to What is the charge Q?
The question is missing. Here is the complete question.
A point charge Q moves on the x-axis in the positive direction with a speed of 160 m/s. A point P is on the y-axis at y = + 20mm. The magnetic field produced at point P, as the charge moves through the origin is equal to -0.6μT k. What is the charge Q? [tex]\mu_{0}=4.\pi.10^{-7}[/tex]T.m/A)
Answer: Q = [tex]-0.015.10^{-3}[/tex]C
Explanation: Magnetic Field (B) is a vector field, i.e., has magnitude and direction, and describes the distribution of magnetic force in the space around. It happens when an electrical charge is in movement.
Its magnitude is determined by the formula:
[tex]B = \frac{\mu_{0}}{4.\pi}\frac{Qvsin(\theta)}{r^{2}}[/tex]
where
[tex]\mu_{0}[/tex] is the vacuum permeability constant;
r is the distance between charge and a point you want to know the magnetic field;
θ is the angle between velocity and distance r;
For this question, magnetic field has that magnitude when charge is passing through the origin. So, angle between velocity and distance is 90°.
Calculating to determine charge:
[tex]-0.6.10^{-6} = \frac{4.\pi.10^{-7}}{4.\pi}\frac{Q.160.sin(90)}{(2.10^{-2})^{2}}[/tex]
[tex]-0.6.10^{-6} = 10^{-7}\frac{Q.160.1}{(2.10^{-2})^{2}}[/tex]
[tex]-2.4.10^{-10} = Q.160.10^{-7}[/tex]
[tex]Q = \frac{-2.4.10^{-10}}{160.10^{-7}}[/tex]
[tex]Q = -0.015.10^{-3}[/tex]
Charge Q is [tex]-0.015.10^{-3}[/tex]C or [tex]-0.015[/tex]mC.
Calculate the momentum of at 1500 kg car traveling at 6 m/s?
Answer:
Explanation:
momentum =mass*speed
p=1500 kg*6 m/s
p=9000 kgm/s
Julie drives 100 mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 38.0 mph and half the distance at 62.0 mph. On her return trip, she drives half the time at 38.0 mph and half the time at 62.0 mph.A) What is Julie's average speed on the way to grandmother's house?B) What is her average speed in the return trip?
Answer:
A.46.7mph
B.50mph
Explanation:
We know that from the data given
Julie drove 50 miles at a speed of 38 mph, and another 50 miles for 62mph. Thus , for the first 50 miles, she drove for the following time
T1= distance/speed
====> 50/38= 1.3hrs
the next 50 miles, she drove for
T2= 50/62
= 0.8hrs
So average speed is
Totaldistance/ total time
100/2.1
= 46.7mph
B. Because the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given which is
(50+50)/2 = 50mph
Explanation:
Answer:
46.95 mph
50 mph
Explanation:
first, we start by finding the average speed. The average speed if given by the relation
Average Speed = Total distance / Total Time
from the question, we know that
total distance is 100 mi
total time is time of the first distance + time of the second distance
t = (50mi / 38) + (50mi / 62)
t = 1.32 + 0.81
t = 2.13 hrs
Then, the average speed going to grandmothers house will be
Speed = 100mi / 2.13 hrs
Speed = 46.95 mph
on her return trip
speed on the first distance * half the time + speed on the second distance * half the time
x = V1 * t/2 + V2 * t/2
100 = 38 * t/2 + 62 * t/2
100 = 100 * t/2
t/2 = 100/100
t/2 = 1
t = 2 hrs
The average speed then will be, = 100 mi / 2 hrs = 50mph
the linear impulse and momentum equation is obtained by integrating the ______ with respect to time.
Answer:
and the forces acting on the particle as a function of time.
Explanation:
The principle of linear impulse and momentum is obtained by integrating the equation of motion with respect to time. hope this helps you :)
The linear impulse and momentum equation is obtained by integrating the force acting on the object with respect to time.
What is the linear impulse?The Linear Impulse is defined as the momentum exerted by a force during a time interval.
The linear impulse and momentum equation is obtained by integrating the force with respect to time.
From the equation of the impulse-momentum theorem:
∆p = J
where,
∆p is the change in momentum of the object,
J is the impulse applied to it.
The impulse is defined as the product of force and the time over which the force is applied.
Thus, we can write the impulse as:
J = ∫ F dt
where,
F is the force acting on the object,
dt is the time interval during which the force acts.
By substituting this expression for J in the momentum equation,
we get:
∆p = ∫ F dt
which is the linear impulse-momentum equation.
Learn more about the linear impulse here:
https://brainly.com/question/30252584
#SPJ6
Convert 37 mg to Dg (Dekagram)
0.37 dg
Explanation:1 dg ....... 100 mg
x dg .........37 mg
x = 1×37/100 = 37/100 = 0.37 dg
while you are in class, your iclicker falls off your desk to the floor. what forces are acting on your iclicker as it falls to the floor?
Answer:
Gravitational forces and drag forces
Explanation:
All objects that are above a certain distance above the ground has gravitational force acting on them. A falling body like the iclicker has gravitational force pulling it down with a magnitude proportional to its mass and the acceleration due to gravity. As the iclicker fall, it experiences air resistance due to the drag forces on it from the air molecules around. This drag force acts upwards so as to oppose the motion of the iclicker downwards, and the magnitude is minimal when compared to the gravitational force. The resultant force is therefore downwards
A sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A stationary observer hears a sound whose frequency is appreciably different from f because:________
a. interference effects set up a standing-wave pattern that alters the frequency
b. the wavelength established in the medium is not the same as it would be if the source were at rest.
c. the equation that relates velocity of propagation, frequency, and wavelength of a sound traveling through a medium does not apply in this situation
d. the sound wave travels through the medium with a velocity different from that which it would have if the source were at rest
Answer:
The correct answer is d
Explanation:
In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.
The expression for the Doppler effect of a moving source is
f ’= (v / (v- + v_s) f
From this expression we see that if the speed the sound source is different from zero feels a change in the frequency.
The correct answer is d
ANSWER ASAP PLEASEEE Kendra wonders what would happen if the oceans did not exist on Earth. Based on her understanding of the oceans and the water cycle she predicts that Earth would become warmer, with less rain, more clouds, and that living things would not survive. Which part of her prediction is incorrect answers; 1.That living things would not survive 2.That Earth would get warmer 3.That there would be less rain 4.That there would be more clouds
Answer:
that there would be more clouds.
Explanation:
one of the cycles is precipitation. without water there would be no rain and clouds are what causes the rain to come down
Answer:
it's D or 4 that there would be more clouds
Explanation:
hoped it helped!!
Is the statement "An object always moves in the direction of the net force acting on it" true or false
Answer:
False
Explanation:
This is because according to newtons second law which says the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. So take for example a net a net force in opposite direction will cause an object to slow down.
velocity vector here is not the same as acceleration vector
how high off the ground is a a book that has 120 J of potential energy and a 4-kg mass
A 40-turn coil has a diameter of 19 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]\epsilon = 7.57 \ V[/tex]
b
[tex]\epsilon =0.0757 \ V[/tex]
c
[tex]\epsilon =0.00699 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is N = 40 turns
The diameter is [tex]d = 19 \ cm = 0.19 \ m[/tex]
The initial magnetic field is [tex]B_i = 0.40 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
Generally the cross-sectional area of the coil is mathematically represented as
[tex]A = \pi \frac{d^2}{4}[/tex]
=> [tex]A = 3.142 * \frac{0.19^2}{4}[/tex]
=> [tex]A = 0.0284 \ m^2[/tex]
At [tex]\delta t = 0.60 \ s[/tex]
The induced emf is
[tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]
=> [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{0.06}[/tex]
=> [tex]\epsilon = 7.57 \ V[/tex]
At [tex]\delta t = 6 \ s[/tex]
The induced emf is
[tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]
=> [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{6}[/tex]
=> [tex]\epsilon =0.0757 \ V[/tex]
At [tex]\delta t = 65 \ s[/tex]
The induced emf is
[tex]\epsilon = - N * \frac{[B_f - B_i ] * A }{\delta t }[/tex]
=> [tex]\epsilon = - 40 * \frac{[- 0.40 ] * 0.0284 }{65}[/tex]
=> [tex]\epsilon =0.00699 \ V[/tex]
A car travelling at a constant speed of 70km/h passes a stationary police car. The police car immediately goes on the chase accelerating uniformly to reach a speed of 85km/h in 10s and continues at that speed until it overtakes the other car. Calculate the time taken for the police car to overtake the car?
Answer:
18.24 seconds
Explanation:
First you convert the km/h to m/s, 70km/h=(175/9)m/s,85km/h=(425/18)m/s.
You know it took 10 seconds for the police to reach 85 km/h. Calculate the distance that the car is ahead of the police (175/9)*10=1750/9m. Then by divide 1750/9 with 425/18, you will get the value 8.24. Add the 10 seconds with the 8.24 you will get 18.24 sec which is the total time.
A Georgia water worker accidentally strikes a water pipe with the end of his pickaxe while trying to dig a whole. If the pickaxe strikes with a force of 3,600N and the end of the pickaxe measures 15mm by 13mm, how much pressure is exerted on the pipe by the pickaxe
Answer:
Pressure, [tex]P=1.84\times 10^7\ Pa[/tex]
Explanation:
Given that,
Force with which pickaxe strikes is 3600 N
The end of the pickaxe measure 15mm by 13mm.
We need to find the pressure exerted on the pipe by the pickaxe. Pressure exerted is equal to the force acting per unit area. So,
[tex]P=\dfrac{F}{A}[/tex]
A = 15 mm × 13 mm = 195 mm² = 0.000195 m²
So,
[tex]P=\dfrac{3600\ N}{0.000195\ m^2}\\\\P=18461538.46\ Pa\\\\\text{or}\\\\P=1.84\times 10^7\ Pa[/tex]
So, the pressure of [tex]1.84\times 10^7\ Pa[/tex] is exerted on the pipe by the pickaxe.If you push a 20-kg box with a net force of 10 N, what is the acceleration of the box? A) 20 m/s^2 B) 2 m/s^2 C) 0.5 m/s^2 INTEGRATED SCIENCE to the person who keeps deleting my questions ':(
Answer:
C) 0.5 m/s^2
Explanation:
[tex]Mass = 20kg\\Force = 10 N\\Acceleration = ?\\\\Force = Mass \times Acceleration \\\\Acceleration = \frac{Force}{Mass}\\\\ Acceleration = \frac{10}{20}\\ \\Acceleration = 1/2 = 0.5m/s^2[/tex]
A 8700-kgkg boxcar traveling at 14 m/sm/s strikes a second boxcar at rest. The two stick together and move off with a speed of 5.0 m/sm/s.What is the mass of thesecond car?
Answer:
m₂ = 15660 kg
Explanation:
Given that,
Before collision,
Mass of box car 1, m₁ = 8700 kg
Speed of box car 1, u₁ = 14 m/s
Speed of box car 2, u₂ = 0 (at rest)
After collision,
The two stick together and move off with a speed of 5 m/s
Let m₂ is the mass of the second car. As cars stick together, it is a case of inelastic collision. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\8700\times 14+0=(8700+m_2)5\\\\121800=43500+5m_2\\\\121800-43500=5m_2\\\\m_2=\dfrac{78300}{5}\\\\m_2=15660\ kg[/tex]
So, the mass of the second car is 15660 kg.
Answer:
15000 kg
Explanation:
m1v1+m2v2=(m1m2)Vf
8700(15)=(8700+m)5.5
130500=8700+m(5.5)
130500/5.5=8700+m
130500-8700=m
m=15027.3 kg
It takes the Earth 24 hours to make a complete rotation around its axis. 33% Part (a) What is the period of rotation of the Earth in seconds? Grade Summary Deductions0 Potential 100 sin cotan) coso asin) atano acotan) sinh0 tan) acoso Attempts remaining per antemp4) detailed vie cosh0 tanhO cotanhO ODegrees O Radians Hint I give up Hints:-% deduction per hint. Hints remaining:- Feedback: dohction per fcedback ▲ 33% Part (b) What is the angular velocity of the Earth in rad/s? 33% Part (c) Given that Earth has a radius of 6.4 x 106 m at its equator, what is the linear velocity at Earth's surface?
Answer:
Explanation:
a )
one rotation per 24 hours
Time period of rotation of the earth T = time / no of rotation
T = 24 x 60 x 60 s / 1
= 86400 s .
b )
angular velocity = angle of rotation in radian / time
in one rotation , angle made at the centre = 2π radian
= 2 x 3.14 radian
angular velocity ω = 2π / T
= 2 x 3.14 / 86400 radian / s
= 72.68 x 10⁻⁶ radian / s
c )
Relation between linear and angular velocity is as follows
v = ω x R where R is radius of the earth and v is linear velocity .
linear velocity = 72.68 x 10⁻⁶ x 6.4 x 10⁶ m /s
= 465.152 m /s
=
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_______ → + 0 e1 + 13C6
Answer:
Nitrogen 13
Explanation:
We need fill in the blank the nuclide symbol for the missing particle in the following nuclear equation.
[tex]X\rightarrow ^0e_1+^{13}C_6[/tex]
In this reaction, the emission of positron takes place. The atomic number in the LHS and the RHS should be same. So,
[tex]^{13}_7 N \rightarrow ^0e_1+^{13}C_6[/tex]
So, the LHS have nitrogen 13, a radioactive isotope.
What is the charge delivered by a current with an average of 0.54 A over 28.3 minutes?
Answer:
916.9C
Explanation:
Using Q= It
Q= 0.54A x 1698s
= 916.9C
The charge delivered by a current with an average of 0.54 A over 28.3 minutes is 916.9C.
What is Electric Charge?Electric charge is defined as the physical property of matter that causes that matter to experience a force when placed in an electromagnetic field that may be positive or negative such that charges repel each other and unlike charges attract each other. An object with the absence of net charge is said to be neutral.
The SI unit of electric charge is the coulomb which is a derived SI unit and is represented by the symbol C.
Electric Charge is represented as
Q= It
where, Q= Electric charge
I= Electric current
t= time
Q= 0.54A x 1698s
= 916.9C
Thus, the charge delivered by a current with an average of 0.54 A over 28.3 minutes is 916.9C.
Learn more about Electric charge, here:
https://brainly.com/question/20342415
#SPJ2
You have a 100-kg object sitting on a frictionless tabletop. The object is connected to a spring with k = 1000 N/m and a natural length of 1 m with the other end of the spring connected to the wall. You pull the object 50 cm from the equilibrium position of the spring and hold it in place, and then release it.
Required:
a. How hard were you pulling on the object to hold it in place?
b. How much work did you do to move the object to that spot?
c. How close to the wall will the object get?
d. What is the fastest that the object moves and where is that location?
Answer:
a) 500 N
b) 250 J
c) 0.87 m
d) 1.58 m/s, at 0.6 m from the wall
Explanation:
The mass of the object m = 100 kg
the spring constant k = 1000 N/m
length of the the spring = 1 m
extension of the string = 50 cm = 0.5 m
a) Force used to pull the mass is gotten from Hooke's law equation
F = -kx
where F is the force used to pull = ?
k is the spring constant = 1000 N/m
x is the extension = 0.5 m
substituting, we have
F = 1000 x 0.5 = 500 N this force is used to pull the mass
b) The work done in moving the mass = Fx
==> 500 x 0.5 = 250 J
c) The energy stored up in the spring U = [tex]\frac{1}{2}kx^{2}[/tex]
U = [tex]\frac{1}{2}*1000*0.5^{2}[/tex] = 125 J
energy available for the mass from its equilibrium position = 250 - 125 = 125 J
this energy is equivalent to the work done by the spring on the mass by moving it closer to the wall
Work W = (weigh of the mass) x distance moved
weight = mg
where m is the mass = 100 kg
g is acceleration due to gravity = 9.81 m/s^2
substituting, we have
W = mgd
where d is the distance the mass moves closer to the wall
W = 100 x 9.81 x d
but W = 125 J
125 = 981d
d = 125/981 = 0.13 m
closeness to the wall = L - d
where L is the natural length of the spring = 1 m
closeness to the wall = 1 - 0.13 = 0.87 m
d) The maximum kinetic energy of the object will be halfway between the extended length and the final resting place.
extended length = 1 + 0.5 m = 1.5 m
distance from resting place = 1.5 - 0.87 = 0.63 m from the wall
At this point, all the mechanical energy on the mass and spring system is converted to kinetic energy of motion.
KE = [tex]\frac{1}{2}mv^{2}[/tex]
substituting,
125 = [tex]\frac{1}{2}*100*v^{2}[/tex] = [tex]50v^{2}[/tex]
[tex]v^{2}[/tex] = 125/50 = 2.5
v = [tex]\sqrt{2.5}[/tex] = 1.58 m/s
Give five difference between manometer and barometer
Answer:
Manometer is a measures to the atmospheric measure.
Barometer is a measure to the atmospheric pressure.
Explanation:
Manometer is there are two types (1) closed tube (2) open tube, is filled with liquid water and mercury and the tube.Manometer do not need equal size of the tubes to the open ended manometers.Manometers is simple to attach on this filled to the high pressure gas in the bulb,this gas pressure put on the tube.Manometer is the attach the gas and vacuum gas to the open of the air, its high pressure gas is to the one of end.Manometer are filled the mercury to the heavy liquid manometer filled the liquid water and oil with the mercury.Barometer is to the close ended tube and the glass tube is the open and that a vacuum is the other end.Barometer can the u shaped and the closed ended tubes to the barometers.Barometer is the mercury closed end and vacuum is the tallest point in the liquid.Barometer is also used to be a table or left standing and also be digital and portable traditional glass, may find the digital clock .Barometer are filled the mercury to the high level liquid must be extreme tall in change in atmospheric pressure.In a Young experiment two slits are speared by 6(um), the third dark fringe is formed at an angle 5.6°. The distance between slits and viewing screen is 2 (m).
A- What is the frequency of light used for this experiment?
B- What is the distance between second bright fringe and central fringe?
Answer:
A) f = 1.28x10¹⁵ Hz
B) y = 0.20 m
Explanation:
A) The frequency of light can be found as follows:
[tex] f = \frac{c}{\lambda} [/tex]
Where:
c: is the speed of light = 3.0x10⁸ m/s
λ: is the wavelength
The wavelength can be calculated using the following equation:
[tex] sin(\theta) = \frac{\lambda(m - 1/2)}{d} [/tex]
Where:
m = 3, d = 6 μm, θ = 5.6°
[tex] \lambda = \frac{d*sin(\theta)}{m - 1/2} = \frac{6 \cdot 10^{-6} m*sin(5.6)}{3 - 1/2} = 2.34 \cdot 10^{-7} m [/tex]
Now, the frequency is:
[tex] f = \frac{c}{\lambda} = \frac{3.0 \cdot 10^{8} m/s}{2.34 \cdot 10^{-7} m} = 1.28 \cdot 10^{15} Hz [/tex]
Hence, the frequency of light used for this experiment is 1.28x10¹⁵ Hz.
B) The distance between the second bright fringe and central fringe (y) is:
[tex] tan(\theta) = \frac{y}{D} [/tex]
[tex] y = D*tan(\theta) = 2 m*tan(5.6) = 0.20 m [/tex]
Therefore, the distance between the second bright fringe and central fringe is 0.20 m.
I hope it helps you!
A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on the same track at a speed of 1 m/s. What is the initial momentum of the orange train?.......... A. 200 kgm/s B. -200 kgm/s C. 100 kgm/s D. 80 kgm/s
Answer:The answer is option A.
Momentum=mass×velocity.
Orange train=mass is 200 and velocity is 1.
which is 200×1=200kgm/s.
The initial momentum of the orange train is 200 kgm/s.
What is momentum?Momentum is the product of mass and velocity of any object.Momentum is denoted by P.Momentum P = mv , where m = mass and v = velocity.Momentum is a vector quantity.How to calculate momentum of the orange train?
The orange train is having mass of 200 kg and it is travelling with speed of 1 m/s.
So, the orange trains momentum will be ,
P = mv
∴ P = 200 x 1
∴ P = 200 kgm/s
Therefore, the initial momentum of the orange train is 200kgm/s.
Learn more about momentum here -
https://brainly.com/question/22257327
#SPJ2
An ion has 109 neutrons, 87 protons and 84 electrons. What is the net charge in Coulombs on the ion?
Answer:
4.8E-19C
Explanation:
Charge is calculated by getting the net sum of protons and electrons
So
87protons - 84electrons= +3
So the charge of the ion is +3
But 1proton= 1.6*10^-19C
So charge will be 3 x 1.6E-19C
4.8x10^-19C
Why is it cooler to wear a white shirt than a black shirt on a hot, sunny day?
Answer:
because the colour black absorbs heat which will make you feel much hotter than a white shirt which blocks away heat.
How many significant figures does the number 0.0030 have?
A) 1
B) 2
C) 4
D) 3
Q10. Explain the Retention time, in GC. What will happen to retention time, if you increase the detector temperature
Answer:
The detector temperature doesn't affect retention time
Explanation:
Retention time is one of the chromatographic parameters. Is defined as the time of a compound spends from injection to detection.
A solute in GC is added to the injector where is volatilized. When volatilized, it pass through a column until the detector.
The detector temperature doesn't affect retention time. To change retention time you must change injector temperature or column temperature. An increase in column or injector temperature results in a decrease in retention time.
Kramer goes bowling and decides to employ the force of gravity to "pick up a spare." He rolls the 7.0-kg bowling ball very slowly so that it comes to rest a center-to-center distance of 0.25 m from the one remaining 1.5-kg bowling pin.
Required:
Determine the force of gravity between the ball and the pin and comment on the efficacy of the technique.
Answer:
[tex]1.12056\times 10^{-8}\ N[/tex].
Kramer could try blowing air on the bowling pin.
Explanation:
[tex]m_1[/tex] = Mass of bowling ball = 7 kg
[tex]m_2[/tex] = Mass of bowling pin = 1.5 kg
[tex]r[/tex] = Distance between them = 0.25 m
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
Force of gravity is given by
[tex]F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 7\times 1.5}{0.25^2}\\\Rightarrow F=1.12056\times 10^{-8}\ N[/tex]
The force of gravity between the ball and the pin is [tex]1.12056\times 10^{-8}\ N[/tex].
Kramer could try blowing air on the bowling pin.
A spring with force constant of 59 N/m is compressed by 1.3 cm in a hockey game machine. The compressed spring is used to accelerate a small metal "puck" with mass 39.2 grams which will then slide up an inclined surface inside the machine. Assume friction is so tiny it may be ignored in this problem. How high in cm above its initial vertical position will the puck rise before stopping (momentarily at least)? Note: Convert units as necessary.
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26.0 cm.(a) Determine the magnetic energy density in the field.uB = 1 J / m3(b) Determine the energy stored in the magnetic field within the solenoid.UB = 2 kJ
Answer:
(a) The magnetic energy density in the field is 6.366 J/m³
(b) The energy stored in the magnetic field within the solenoid is 5 kJ
Explanation:
magnitude of magnetic field inside solenoid, B = 4 T
inner diameter of solenoid, d = 6.2 cm
inner radius of the solenoid, r = 3.1 cm = 0.031 m
length of solenoid, L = 26 cm = 0.26 m
(a) The magnetic energy density in the field is given by;
[tex]u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3[/tex]
(b) The energy stored in the magnetic field within the solenoid
[tex]U_B = u_B V\\\\U_B = u_B AL[/tex]
[tex]U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\[/tex]
While in a stream 39 cm deep, they look down into the water and see a craw fish at the bottom. How deep does the stream appear to the student? (nwater = 1.33)
Answer:
The depth of stream to the student is [tex]d_1 = 0.2932 \ m[/tex]
Explanation:
From the question we are told that
The actual depth of the stream is [tex]d = 39 \ cm = 0.39 \ m[/tex]
The refractive index of the water is [tex]n = 1.33[/tex]
Generally the apparent depth of the stream is mathematically represented as
[tex]d_1 = \frac{d}{1.33}[/tex]
substituting values
[tex]d_1 = \frac{ 0.39}{1.33}[/tex]
[tex]d_1 = 0.2932 \ m[/tex]