Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.
Answers
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
[tex]F_{pull}=m_{T}.a[/tex]
[tex]m_{T}=\frac{F_{pull}}{a}[/tex]
[tex]m_{T}=\frac{3615}{1.516}[/tex]
[tex]m_{T}=2384.5[/tex]
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
For [tex]m_{1}[/tex]:
The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
[tex]m_{1} = \frac{F_{12}}{a}[/tex]
[tex]m_{1} = \frac{1387}{1.516}[/tex]
[tex]m_{1}[/tex] = 915kg
For [tex]m_{2}[/tex]:
There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]
[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]
[tex]m_{2}[/tex] = 605kg
For [tex]m_{3}[/tex]:
[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]
[tex]m_{3} = 2384.5-1520.0[/tex]
[tex]m_{3}[/tex] = 864.5kg
Related Questions
An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens has a focal length of 14 mm , the eyepiece a focal length of 21 mm .
A) Where is the image formed by the objective lens? Give your answer as the distance from the image to the lens. Express your answer using two significant figures.
B) How tall is the image mentioned in part A? Express your answer using two significant figures.
C) If you want to place the eyepiece so that the image it produces is at infinity, how far should this lens be from the image produced by the objective lens? Express your answer using two significant figures.
D) Under the conditions of part C, find the overall magnification of the microscope. Express your answer using two significant figures.
Answers
Answer:
Explanation:
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
lens formula
[tex]\frac{1}{v} -\frac{1}{u} =\frac{1}{f}[/tex]
Putting the values
[tex]\frac{1}{v} +\frac{1}{15} =\frac{1}{14}[/tex]
v = 210 mm .
B )
magnification = v / u
= 210 / 15
= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\frac{210}{15} \times \frac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
= 14 x 250 / 21
= 166.67
= 170 ( in two significant figures )
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
What is objective lens?The objective lens of a microscope is the one at the bottom near the sample. At its simplest, it is a very high-powered magnifying glass, with very short focal length. This is brought very close to the specimen being examined so that the light from the specimen comes to a focus inside the microscope tube
For image formation in objective lens
object distance u = 14 +1 = 15 mm
focal length f = 14 mm .
image distance v = ?
By using lens formula
[tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
Putting the values
[tex]\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{14}[/tex]
v = 210 mm .
B ) Magnification is the ratio of the size of the image to the size of the an object.
[tex]\rm magnification = \dfrac{v} { u}[/tex]
[tex]M= \dfrac{210} { 15}[/tex]
M= 14
size of image = 14 x 1.1 mm
= 15.4 mm
= 15 mm approx
C )
For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .
21 mm is the answer .
D )
overall magnification =
[tex]\dfrac{210}{15}\times \dfrac{D}{f_e}[/tex]
D = 25 cm , f_e = focal length of eye piece
[tex]= 14 \times \dfrac{ 250} { 21}[/tex]
= 166.67
= 170 ( in two significant figures )
Hence all the answers are:
(a) The distance of the image v=220mm
(b) SIze of the image 15 mm
(c) Distance of lens be from the image produced by the objective lens 21 mm
(d) overall magnification of the microscope 170
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An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answers
Answer:
The value is [tex]N = 3.708*10^{7} \ \ atoms[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]
The temperature is [tex]T = 26 + 273 = 299 \ K[/tex]
The volume is 1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]
Generally according to the ideal gas law we have that
[tex]PV = NkT[/tex]
here k is the Boltzmann constant with a value [tex]k = 1.38 *10^{-23} \ J/K[/tex]
=> [tex]N = \frac{PV}{ k T}[/tex]
=> [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]
=> [tex]N = 3.708*10^{7} \ \ atoms[/tex]
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
Answers
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.
Answers
Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answers
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
In which type of indicating valve is the valve stem housed in a hollow metal post that contains a movable plate with a small glass window
Answers
Answer:
Post indicator valve
Explanation:
Post Indicator Valves are commonly used to control the water flow of sprinkler systems used in public and private buildings, warehouses, and factories for fire suppression. PIVs control water flow from the public system into the building's fire suppression system.
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.
Answers
Answer:
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal: speed of light(c)
Explanation:
Generally the ratio of the E(electric field ) and the B(magnetic field ) is equal to the speed of the electromagnetic wave i.e the speed of light (c) the value is
[tex]c = 3.0 *10^{8} \ m/s[/tex]
The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both
Answers
Answer:
a. Moon around Earth.
Explanation:
Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.
The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.
The object that should exhibit the longest orbital period is option a. Moon around Earth.
What is Charon's orbit?Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.
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Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.
Answers
Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer
Explanation:
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.
What is friction?Friction is the force that prevents one hard material from scooting or rolling over the other.
Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.
We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.
Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.
In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.
Thus, the correct option is B.
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Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answers
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b
Answers
Answer:
Will be equal to alpha x r; less than UsN
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answers
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is
Answers
Answer:
N2 = ¼N1
Explanation:
First of all, let's define the terms;
N1 = number of electric field lines going through the sphere of radius R
N2 = number of electric field lines going through the sphere of radius 2R
Q = the charge enclosed at the centre of concentric spheres
ε_o = a constant known as "permittivity of the free space"
E1 = Electric field in the sphere of radius R.
E2 = Electric field in the sphere of radius 2R.
A1 = Area of sphere of radius R.
A2 = Area of sphere of radius 2R
Now, from Gauss's law, the electric flux through the sphere of radius R is given by;
Φ = Q/ε_o
We also know that;
Φ = EA
Thus;
E1 × A1 = Q/ε_o
E1 = Q/(ε_o × A1)
Where A1 = 4πR²
E1 = Q/(ε_o × 4πR²)
Similarly, for the sphere of radius 2R,we have;
E2 = Q/(ε_o × 4π(2R)²)
Factorizing out to get;
E2 = ¼Q/(ε_o × 4πR²)
Comparing E2 with E1, we arrive at;
E2 = ¼E1
Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;
N2 = ¼N1
A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answers
Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s
Answers
(a) The car is undergoing an acceleration of
[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]
so that in 9.02 s, it will have covered a distance of
[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]
The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:
[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]
(b) The wheels have average angular velocity
[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]
where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.
The wheels start at rest, so
[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]
A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?
Answers
Answer:
The period is [tex]T = 0.00255 \ s[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 392 \ Hz[/tex]
Generally the period is mathematically represented as
[tex]T = \frac{1}{f}[/tex]
=> [tex]T = \frac{1}{ 392}[/tex]
=> [tex]T = 0.00255 \ s[/tex]
A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Answers
Answer:
The amplitude [tex]A(5) = 1 \ m[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 100 \ N/m[/tex]
The damping constant is [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]
The mass is [tex]m = 0.050 \ kg[/tex]
The maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]
The time considered is t = 5.0 s
Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as
[tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]
substituting values
[tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]
[tex]A(5) = 1 \ m[/tex]
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answers
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the first-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm
Answers
Answer:
[tex]\theta_1 = 0.400^o[/tex]
[tex]\theta_2 =0.378^o[/tex]
Explanation:
From the question we are told that
The number of slits per cm is k = [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]
The order of the maxima is n = 1
The wavelength are [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]
The spacing between the slit is mathematically represented as
[tex]d = \frac{ 0.01}{k}[/tex]
=> [tex]d = \frac{ 0.01}{161}[/tex]
=> [tex]d = 6.211 *10^{-5} \ m[/tex]
Generally the condition for constructive interference is
[tex]n\lambda = d \ sin \theta[/tex]
At [tex]\lambda_1[/tex]
[tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_1 = 0.400^o[/tex]
At [tex]\lambda_2[/tex]
[tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]
[tex]\theta_2 =0.378^o[/tex]
Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact
Answers
Answer:
Explanation:
Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.
While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.
Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''
The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.
A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.
Answers
Answer:
24.34 m/s
Explanation:
recall that one of the equations of motions takes the form:
v = u + at
where,
v = final velocity
u = initial velocity (given as 22.2 m/s)
a = acceleration (given as 2.68m/s²)
t = time elapsed during acceleration (given as 0.80s)
since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:
v = u + at
v = 22.2 + (2.68) (0.8)
v = 24.34 m/s
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant when it has turned through 0.40 radian, what is the magnitude of the total linear acceleration of a point on the rim (radius = 13 cm)?
a. 0.31 m/s^2
b. 0.27 m/s^2
c. 0.35 m/s^2
d. 0.39 m/s^2
e. 0.45 m/s^2
Answers
Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;
[tex]a_t = ar[/tex]
where;
a is the angular acceleration and
r is the radius of the circular path
[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]
Determine time of the rotation;
[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]
Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration
[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]
The magnitude of total linear acceleration is given by;
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]
Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?
Answers
Answer:
counterclockwise
Explanation:
given data
area = 25 cm²
solution
We know that a changing magnetic field induces the current and induced emf is express as
[tex]\epsilon = -N \frac{d \phi }{dt}[/tex] ..................................1
and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field
so when magnetic field decrease and point coming out of the paper.
so induced current in the loop will be counterclockwise
Convert 7,348 grams to kilograms
Answers
A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which a wave travels on this string.
Answers
Answer:
Velocity of wave (V) = 5.2 × 10² m/s
Explanation:
Given:
Per unit length mass (U) = 4.80 × 10⁻³ kg/m
Tension (T)= 1,300 N
Find:
Velocity of wave (V)
Computation:
Velocity of wave (V) = √T / U
Velocity of wave (V) = √1300 / 4.80 × 10⁻³
Velocity of wave (V) = √ 270.84 × 10³
Velocity of wave (V) = 5.2 × 10² m/s
16. If one body is positively charged and another body is negatively charged, free electrons tend to
O A. move from the negatively charged body to the positively charged body
O B. remain in the positively charged body
OC. move from the positively charged body to the negatively charged body
O D. remain in the negatively charged body
Answers
Answer:
Hey there!
The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body
Let me know if this helps :)
what is defect of vision
Answers
Answer:
The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric field of the wave is given by = (0.082 V/m) . What is the magnetic vector of the wave at the point P at that instant? (c = 3.0 × 108 m/s)
Answers
Answer:
[tex]B=2.74\times 10^{-10}\ T[/tex]
Explanation:
It is given that,
A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m
We need to find the magnetic vector of the wave at the point P at that instant.
The relation between electric field and magnetic field is given by :
[tex]c=\dfrac{E}{B}[/tex]
c is speed of light
B is magnetic field
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T[/tex]
So, the magnetic vector at point P at that instant is [tex]2.74\times 10^{-10}\ T[/tex].
The magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]
The formula relating electric field and the magnetic field is given as;
[tex]c=\frac{E}{B}[/tex]
E is the electric field strengthB is the magnetic vector of the wavec is the speed of lightFrom the formula shown:
[tex]B=\frac{E}{c}\\B=\frac{0.082}{3.0\times 10^8}\\B=2.73 \times 10 ^{-10}T[/tex]
Hence the magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]
Learn more on magnetic field here: https://brainly.com/question/21040756
A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less than the actual distance, the same as the actual distance, or more than the actual distance? Explain.
Answers
Answer:
p = -q
he distance is equal to the current distance, so the distance does not change
Explanation:
For this exercise we can solve it using the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies
1 / p = - 1 / q
p = -q
Therefore the distance is equal to the current distance, so the distance does not change
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answers
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
