Answer:depends on the masses of the objects and the distance between them
Explanation:
According to Newton's law of universal gravitation,the force of attraction between two objects depends on the masses of the objects and the distance between them
Question 7
Review
Which particles are not affected by the strong force?
A.
hadrons
B.
protons
C.
neutrons
D
electrons
Submit A
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Answer:
Electron
Explanation:
Because electron are not hadrons so electron are not affected by strong force
Particles that can not be affected by strong forces are electrons.
What are electrons?Electrons are the rotating material around the nucleus of an atomic element in orbit.
Atoms have electrostatic energy between their electrons. This force is not broken by a force as strong as nuclear power.
What are strong forces?Strong force is a fundamental interaction of nature that acts between subatomic particles of matter.
There are four basic forces in nature:
Gravity: the gravitational force used between any heavy objects. It has an infinite range.Electrical energy: energy used between electrically charged objects. It can be either attraction or repulsion.Nuclear power: is the magnetic field that responds to the binding of protons and neutrons within the nucleus of an atom. It only works for very short distances.Weak nuclear power: a force that causes nuclear decay. It only works for very short distances.Therefore, particles that are unaffected by strong force are electrons.
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To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
3. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?
4.If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?
Answer:
Question 3: 2.25 watts
Question 4: 405 joules
Explanation:
question 3:
Current =0.5 amps
Voltage =4.5 volts
Power= current x voltage
Power=0.5 x 4.5
power=2.25 watts
Question 4
Current =0.5 amps
Voltage =4.5v
Time=3 minutes
Time =3x60
Time =180 seconds
Energy=current x voltage x time
Energy =0.5 x 4.5 x 180
Energy =405 joules
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500.0 N/m (assume that the spring is massless). The block is held in position such that the spring is compressed 4.00 cm shorter than its undisturbed length. The block is suddenly released and allowed to slide away on the frictionless surface. Find the speed the block will be traveling when it leaves the spring.
Answer:
6 m/s
Explanation:
Given that :
mass of the block m = 200.0 g = 200 × 10⁻³ kg
the horizontal spring constant k = 4500.0 N/m
position of the block (distance x) = 4.00 cm = 0.04 m
To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.
i.e [tex]\frac{1}{2} kx^2 = \frac{1}{2} mv^2[/tex]
[tex]kx^2 = mv^2[/tex]
[tex]4500* 0.04^2 = 200*10^{-3} *v^2[/tex]
[tex]7.2 =200*10^{-3}*v^{2}[/tex]
[tex]v^{2} =\frac{7.2}{200*10^{-3}}[/tex]
[tex]v =\sqrt{\frac{7.2}{200*10^{-3}}}[/tex]
v = 6 m/s
Hence,the speed the block will be traveling when it leaves the spring is 6 m/s
Dual Nature of Light
Assignment
Active
Explaining the Nature of Light
Why do scientists believe that light is made of streams of
particles?
The main component of all computer memory is
Answer: R.A.M
Explanation:
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
Refracted light rays...
A- are bent as they pass into a different medium
B- are absorbed by an object
C- are reflected from an object at a variety of angles
D- bounce off a medium
Answer:
A.
Explanation:
Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.
A 46.0-kg box is being pushed a distance of 8.80 m across the floor by a force P whose magnitude is 171 N. The force P is parallel to the displacement of the box. The coefficient of kinetic friction is 0.250. Determine the work done on the box by (a) the applied force, (b) the friction force, (c) the normal force, and (d) by the force of gravity. Be sure to include the proper plus or minus sign for the work done by each force.
Answer:
a) 1504.8 J
b) 991.76 J
c) 0J
d) 0J
Explanation:
(a) The work done by the force P on the box is given by the following formula:
[tex]W_P=Px[/tex]
P: applied force = 171N
x: distance in which the for P is applied = 8.80m
you replace the values of P and x and obtain:
[tex]W_P=(171N)(8.80m)=1504.8J[/tex]
(b) The work don by the friction force is:
[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]
μ = coefficient of kinetic friction = 0.250
M: mass of the box = 46.0kg
g: gravitational constant = 9.8 m/s^2
[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]
(c) The Normal force is
[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]
but this force does not do work on the box because the direction is perpendicular to the direction of the force P.
[tex]W_N=0J[/tex]
(d) the same as before:
[tex]W_g=0J[/tex]
Part A - At what angle does it leave?
Part B - At what distance x does it exit the field?
Answer:
Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.
Explanation: Hope i helped!!!
A 0.009 kg bullet fired through a door enters at 803 m/s and leaves at 617 m/s. If the door material is known to exert an average resistive force of 5620 N on bullets of this type at usual speeds, find the thickness of the door.
Answer:
The thickness of the door is 0.4230 m
Explanation:
Given;
mass of bullet, m = 0.009 kg
initial velocity of the bullet, u = 803 m/s
final velocity of the bullet, v = 617 m/s
average resistive force of the door on the bullet, F = 5620 N
Apply Newton's second law of motion;
Force exerted by the door on the bullet = Force of the moving bullet
F = ma
where;
F is applied force
m is mass
a is acceleration
Also, Force exerted by the door on the bullet = Force of the moving bullet
[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]
where;
v is the final velocity of the bullet
u is initial velocity of the bullet
t is time
We need to calculate the time spent by the bullet before it passes through the door.
[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]
Distance traveled by the bullet within this time period = thickness of the door
This distance is equivalent to the product of average velocity and time
[tex]S = (\frac{u+v}{2}) t[/tex]
where;
s is the distance traveled
[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]
Therefore, the thickness of the door is 0.4230 m
Consider a circular vertical loop-the-loop on a roller coaster. A car coasts without power around the loop. Determine the difference between the normal force exerted by the car on a passenger with a mass of mm at the top of the loop and the normal force exerted by the car on her at the bottom of the loop. Express your answer in terms of mmm and the acceleration due to gravity ggg.
Answer:
Explanation:
Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .
Let N₁ and N₂ be the normal reaction force .
At the top position
centripetal force = N₂ + mg ; so
N₂ + mg = m v₂² / r
At the bottom position
centripetal force = N₁ - mg ; so
N₁ - mg = m v₁² / r
subtracting these two equations
N₁ - mg - N₂ - mg = m v₁² / r - m v₂² / r
N₁ - N₂ - 2mg = 1/r (m v₁² - m v₂² )
N₁ - N₂ - 2mg = 1/r x mg x 2r ( loss of potential energy = gain of kinetic energy )
N₁ - N₂ = 2mg + 2mg
= 4 mg .
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
A light bulb dissipates 100 Watts of power when it is supplied a voltage of 220 volts.
a) What is the current flowing through this light bulb?
b) What is the resistance of the light bulb?
Given Information:
Power = P = 100 Watts
Voltage = V = 220 Volts
Required Information:
a) Current = I = ?
b) Resistance = R = ?
Answer:
a) Current = I = 0.4545 A
b) Resistance = R = 484 Ω
Explanation:
According to the Ohm’s law, the power dissipated in the light bulb is given by
[tex]P = VI[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.
Re-arranging the above equation for current I yields,
[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]
Therefore, 0.4545 A current is flowing through the light bulb.
According to the Ohm’s law, the voltage across the light bulb is given by
[tex]V = IR[/tex]
Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.
Re-arranging the above equation for resistance R yields,
[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]
Therefore, the resistance of the bulb is 484 Ω
Answer:
bulb will burn out!
Explanation:
what do hydroelectric plants use to generate electrical energy?
Answer:
A. falling water
Explanation:
I got it right on Edgenuity. Good luck on your quiz.
In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.
What is hydroelectric power?Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.
Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.
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A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
A pipe branches symmetrically into two legs of length L, and the whole system rotates with angular speed ω around its axis of symmetry. Each branch is inclined at angle α to the axis of rotation. Liquid enters the pipe steadily, with zero angular momentum, at volume flow rate Q. The pipe diameter, D, is much smaller than L. Obtain an expression for the external torque required to turn the pipe. What additional torque would be required to impart angular acceleration ω_ ?
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
If Jim could drive a Jetson's flying car at a constant speed of 440 km/hr across oceans and space, approximately how long (in millions of years, in 106 years) would he take to drive to a nearby star that is 12.0 light-years away? Use 9.461 × 1012 km/light-year and 8766 hours per year (365.25 days).
Answer:
t = 2.94 x 10⁶ years
Explanation:
The equation used in the case of constant speed is:
s = vt
t = s/v
where,
s = distance = 12 light years
s = (12 light years)(9.461 x 10¹² km/light year) = 113.532 x 10¹² km
v = speed = 440 km/hr
t = time passed = ?
Therefore,
t = (113.532 x 10¹² km)/(440 km/hr)
t = 2.58 x 10¹¹ hr
Now, converting it to years:
t = (2.58 x 10¹¹ hr)(1 year/8766 hr)
t = 2.94 x 10⁶ years
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
What types of mediums are involved in the energy transfer
Answer:
In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.
Which term BEST describes the movement of air from the ocean toward the land in the daytime? (AKS 4b DOK 1) *
1 point
Sea breeze
Land Breeze
Valley Breeze
Current Breeze
Answer:
Option A, Sea Breeze
Explanation:
Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.
Hence, option A is correct
"It is impossible to devise a process which may convert heat, extracted from a single
reservoir, entirely into work without leaving any change in the working system”.
Use the Second Law of Thermodynamics to state our inability to utilize the heat contents
of oceans and atmosphere which contains a large amount of heat energy but cannot be
converted into useful mechanical work You may use the concept of heat engine to discuss
this
Answer:
According to the second law of thermodynamics, we are unable to use the heat of the ocean and the atmosphere because we do not have a reservoir that has a temperature lower than the ocean or the atmosphere.
Explanation:
As you already know, the ocean and atmosphere have a lot of thermal energy, however, we are unable to convert this energy into mechanical energy that would be useful for our activities. This can be explained by the second law of thermodynamics, since it states that the presence of two bodies with different temperatures is necessary for it to be possible to transform heat into work.
In this case, to transform the thermal energy of the ocean and the atmosphere into mechanical energy we would need the existence of a thermal motor, which is only possible to be established when there is a body with high thermal energy and a sink, a reservoir, with low thermal energy, which will be the place where the heat will be expelled, to be converted into work. We do not have a reservoir with less thermal energy than the ocean and the atmosphere, so we cannot use their energy.
PIUDICITIS CONSECulvely and Circle your aliswers. Lilyo
proper significant digits.
53. When you turn on your CD player, the turntable accelerates from zero to 41.8 rad/s in
3.0 s. What is the angular acceleration?
or
Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Electric fields are MOST associated with ________.
Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
I need some help!!!!!!!!!
Answer:
The Object will immediately begin moving toward the left
Explanation:
Because the force of thirteen is greater than ten and applied to the opposite side
A long solid conducting cylinder with radius a = 12 cm carries current I1 = 5 A going into the page. This current is distributed uniformly over the cross section of the cylinder. A cylindrical shell with radius b = 21 cm is concentric with the solid cylinder and carries a current I2 = 3 A coming out of the page. 1)Calculate the y component of the magnetic field By at point P, which lies on the x axis a distance r = 41 cm from the center of the cylinders.
Answer:
Explanation:
We shall use Ampere's circuital law to find magnetic field at required point.
The point is outside the circumference of two given wires so whole current will be accounted for .
Ampere's circuital law
B = ∫ Bdl = μ₀ I
line integral will be over circular path of radius r = 41 cm .
Total current I = 5A -3A = 2A .
∫ Bdl = μ₀ I
2π r B = μ₀ I
2π x .41 B = 4π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ x 2 / .41
= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.
Calculate potential energy of a 5 kg object sitting on 3 meter ledge
Answer:147 joules
Explanation:
Mass=m=5kg
Acceleration due to gravity=g=9.8m/s^2
Height=h=3 meter
Potential energy=m x g x h
Potential energy=5 x 9.8 x 3
Potential energy=147 joules
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
[tex]v(t)=ate^{-6t}[/tex] (1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex] (2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]
hence, the maximum speed is v_max = ((1/6)e^-1)a
