Answer:
Dark matter:
- Doesn't interact with baryonic matter.
- It has not been observed directly
- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.
Explanation:
Dark matter:
- Doesn't interact with baryonic matter.
- It has not been observed directly
- dark matter does not absorb, reflect or even emit light, thereby making it to be extremely hard to spot. Therefore, it does not interact with electromagnetic radiation.
Baryonic matter:
- Has been observed directly because it includes nearly all the matter that we see in the world daily.
- It interacts with baryonic matter
- interacts with electromagnetic radiation
Dark Matter:
It has not been observed directly.It does not interact with baryonic matter.It does not interact with electromagnetic radiation.Baryonic Matter:
It has been observed directly.It interacts with baryonic matter.It interacts with electromagnetic radiation.Dark matter can be defined as often invisible substances that are difficult to spot because they don't absorb, emit or reflect light.
Hence, dark matter do not affect human view because they do not interact or interfere with electromagnetic radiation (force).
Although, humans can see right through the (weakly interacting) dark matter but it has not been observed directly.
Baryonic matter can be defined as a dark matter that is made up of baryons such as neutrons, and protons. Also, they are ordinary matter (both fermions and hadrons), as distinct from exotic forms.
In conclusion. baryonic matter has been observed directly and it can interact with electromagnetic radiation.
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Two copper spheres are currently 1.2 meters apart. One sphere has a charge of +2.2 x 10-4 C and the other has a charge of -8.9 x 10-4C.
What is the force between the charged spheres?
Is the force attractive or repulsive?
Answer:
The force between the charged spheres is 1,223.75 N
The force is attractive because opposite charges mutually attract.
Explanation:
Coulomb's Law
The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.
Expressing the above as a formula:
[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]
Where:
[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]
q1, q2 = the objects' charge
d= The distance between the objects
The copper spheres have charges of:
[tex]q_1=+2.2\cdot 10^{-4}\ c[/tex]
[tex]q_2=-8.9\cdot 10^{-4}\ c[/tex]
Note the charges have opposite signs.
They are separated by d=1.2 m
Applying Coulomb's formula:
[tex]\displaystyle F=9\cdot 10^9\frac{2.2\cdot 10^{-4}\cdot 8.9\cdot 10^{-4}}{1.2^2}[/tex]
[tex]\displaystyle F=9\cdot 10^9\frac{1.958\cdot 10^{-7}}{1.44}[/tex]
F = 1,223.75 N
The force between the charged spheres is 1,223.75 N
The force is attractive because opposite charges mutually attract.
a 12-kg block is pulled by a string at a 30 degree angle. if the string exerts a 50 n force, the block accelerates at 4.13 m/s2. what is the frictional force and normal force exerted on the block?
Let, frictional force is f and normal force is N.
Balancing horizontal forces :
[tex]50cos \ 30^o - f = 12 \times 4.13\\\\f = 50cos \ 30^o - 49.56\\\\f = -6.26\ N[/tex]
Also, balancing vertical forces, we get :
[tex]N = mg + 50sin \ 30^o\\\\N =( 12\times 9.8 ) + (50\times sin \ 30^o)\\\\N = 142.6\ N[/tex]
Therefore, the frictional force and normal force is -6.26 N and 142.6 N.
The gravity on Earth is 9.8m/s2, the mass of a bowling ball is about 5kg. What is the weight of the bowling ball on the Earth?
Answer:
yo can someone answer i need this
Explanation:
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.3 m/s. How long does it take for the glove to reach its maximum height
Hello!!
For the maximum height the final velocity is zero, because can't up more.
Then, use the formula:
V = Vi + gt
Replacing, we have:
0 m/s = 5,3 m/s + (-9,8 m/s² * t)
0 m/s - 5,3 m/s = -9,8 m/s² * t
(-5,3 m/s) / -9,8 m/s² = t
t = 0,54 s
The time it will take to reach the maximum height is 0,54 seconds.
A runaway train car that has a mass of 15,000 kg travels at a speed of 5.4 m/s down a track. Compute the time required for a force of 1500 N to bring the car to rest.
The 1500 N force would need to act in the direction opposite the train car's movement, so if we take the current direction it's moving in to be positive, we have by Newton's second law
-1500 N = (15,000 kg) a → a = -0.10 m/s²
If t is the time it takes for the train to come to rest, then
-0.10 m/s² = (0 - 5.4 m/s) / t
t = (5.4 m/s) / (0.10 m/s²) → t = 54 s
You did 550 J of work lifting a 220 N backpack. How high did you lift the backpack?
Answer:
0.255 meters
Explanation:
cause formula to calculate work done against gravity is mgh where m refers mass g refers acceleration due to gravity and h refers height
A typical donation of whole blood is equal to a pint (473 ml). If the average density of blood rhoBLOOD = 1060 kg/m^3, what is the mass of donated blood?
a. 5.01 kg
b. 0.501 lb
c. 0.501 kg
d. 0.501 g
e. 5.01 g
Answer:
c. 0.501 kg
Explanation:
Given;
volume of the denoted blood, V = 473 ml = 0.000473 m³
density of the denoted blood, ρ = 1060 kg/m³
The mass of donated blood is given as;
mass of the donated blood = density of the denoted blood x volume of the denoted blood
m = 1060 x 0.000473
m = 0.501 kg
Therefore, the mass of donated blood is 0.501 kg
Consider a box with the same mass and velocity instead sliding without friction toward the incline. How does this object's kinetic energy before the incline compare to the sphere
Answer:
Object's kinetic energy is high on frictionless medium.
Explanation:
There is a great difference in object's kinetic energy which is present between sliding with friction and sliding without friction because friction decreases the speed of an object so if an object moves on the frictionless medium, it moves and covers more distance and we can say that the object has more kinetic energy while on the other hand, if a body moves on the medium and experience friction so it moves with little speed so we can say that the body has low kinetic energy.
A 1500 kg car sits on a 3.5° inclined hill. Find the force of friction required to keep it from
sliding down the hill. The coefficient of static friction is μ=0.45
Answer:
The force of friction required to keep the car from sliding down the hill is 898.054 newtons and since [tex]f<f_{max}[/tex], the car will not slide down the hill.
Explanation:
The free body diagram is included below and now we prepare each equation of equilibrium for respective orthogonal axis:
[tex]\Sigma F_{x'} = f-m\cdot g \cdot \sin \theta = 0[/tex] (1)
[tex]\Sigma F_{y'} = N-m\cdot g \cdot \cos \theta = 0[/tex] (2)
Where:
[tex]f[/tex] - Static friction force, measured in newtons.
[tex]N[/tex] - Normal force from the inclined hill to the car, measured in newtons.
[tex]m[/tex] - Mass of the car, measured in newtons.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]\theta[/tex] - Angle of inclination of the hill, measured in sexagesimal degrees.
If we know that [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the friction force required to keep the car from sliding down the hill is:
[tex]f = m\cdot g \cdot \sin \theta[/tex]
[tex]f = (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \sin 3.5^{\circ}[/tex]
[tex]f = 898.054\,N[/tex]
The force of friction required to keep the car from sliding down the hill is 898.054 newtons.
If the car does not slide down the hill, then the following condition must be observed:
[tex]f\le \mu_{s}\cdot m\cdot g \cdot \cos \theta[/tex] (2)
Where [tex]\mu_{s}[/tex] is the static coefficient of friction, dimensionless.
If we know that [tex]\mu_{s} = 0.45[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 3.5^{\circ}[/tex], then the maximum allowable static friction force is:
[tex]f_{max} = \mu_{s}\cdot m \cdot g \cdot \cos \theta[/tex]
[tex]f_{max} = (0.45)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 3.5^{\circ}[/tex]
[tex]f_{max} = 6607.378\,N[/tex]
Since [tex]f<f_{max}[/tex], the car will not slide down the hill.
which law of thermodynamics would be violated if heat were to spontaneously flow between two abject which are in thermal equilibrium?
Answer:
The law zero of thermodynamics.
Explanation:
The law zero of thermodynamics, which tells us that heat flows from a body at a higher temperature to another body with lower temperature, when the heat transfer is zero, it is said that the two bodies are in thermal equilibrium, their temperatures are equal
please help!!!
In an electric motor, how do the permanent magnets and the electromagnet
interact?
PLEASE ANSWER FAST
A. The electromagnet causes the permanent magnets to reverse
poles.
B. The permanent magnets reverse the current in the electromagnet.
C. The electromagnet reverses the current in the permanent
magnets.
D. The permanent magnets cause the metal loop in the
electromagnet to turn.
if you don’t know don’t answer
The correct asnwer is D. The permanent magnets cause the metal loop in the
electromagnet to turn.
In an electric motor, the permanent magnets cause the metal loop in the
electromagnet to turn. So, option (D) is correct.
What is electric motor?An electrical device that transforms electrical energy into mechanical energy is an electric motor. The majority of electric motors work by creating force in the form of torque imparted to the motor shaft through the interplay of the magnetic field of the motor and electric current in a wire winding.
In an electric motor, their remains a permanent magnet to produce constant magnetic field. When electric current is passed through the conducting loop, it acts ling electromagnet and begins to rotate. Thus an electric motor works. Hence, option(D) is correct.
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A ball is thrown with a speed of 100 ft/s in a direction of 30 degrees above the horizontal. determine the horizontal distance in m, the time of flight, and the height to which it rises..Note: you have to label your answer and final answer must be two decimal place
Answer:
Horizontal distance covered is 270.63ft
Time of flight is 3.125secs
Height to which it rises is 39.06feet
Explanation:
The horizontal distance covered is the range;
Range = U²sin2theta/g
Range = 100²sin2(30)/32
Range = 10000sin60/32
Range = 10000(0.8660)/32
Range = 8660/32
Range = 270.625ft
Hence the horizontal distance covered is 270.63ft
The time of flight is expressed using the formula;
T = 2Usin theta/g
g is the acceleration due to gravity
T = 2(100)sin30/32
T = 200(0.5)/32
T = 100/32
T = 3.125secs
Hence the time of flight is 3.125secs
Maximum height H = u²sin²theta/2g
H = 100²(sin30)²/2(32)
H = 10000(0.5²)/64
H = 10000(0.25)/64
H = 2500/64
H = 39.0625
Hence the height to which it rises is 39.06feet
How has the magnet affected the block of material next to it?
A. The domains in the end of the block farthest from the magnet are aligned with its magnetic field.
B. The domains in the end of the block closest to the magnet are aligned with its magnetic field.
C. The domains in the center of the block are aligned with the magnetis magnetic field
D. The domains in the entire block are aligned with the magnets magnetic field
Answer: (APE X) B
Explanation: I am taking it and I just got it right
The magnetic characteristics of a substance are determined by the magnetic moments of its atoms, the domains which are closest to the magnet are aligned with its magnetic field, hence option b is correct.
What is the effect of a magnet?Long-range alignments of magnetic moments in ferromagnetic materials, known as domains, include magnetic moments that all point in the same direction.
This indicates that the atoms' individual magnetic moments are lined up with one another and pointing in the same general direction.
Therefore, a magnetic substance will get magnetized if it is placed close to a magnet. Magnetism is the name for this kind of electricity, hence option b is correct.
Learn more about the magnet, here:
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An 5-kg animal accelerates from rest to 10 m/s in 10 m and encounters an average drag force of 12 N. What is the average power output in Watts if this takes 2.0 seconds
Answer:
60 watts
Explanation:
Given that
Mass of the animal, m = 5 kg
Velocity of the animal, v = 10 m/s
Distance moved by the animal, d = 10 m
Drag force, F(d) = 12 N
Time taken, t = 2 seconds
To start with, we need to find the power output in itself before proceeding to find the average power output. And as such, we have
Power = Force * Distance/Time
But Distance/Time is velocity, so
Power = Force * Velocity
Power = 12 * 10
Power = 120 W.
We then use this power gotten to find the average power output.
Power(avg) = P/t
Power(avg) = 120 / 2
Power(avg) = 60 Watts.
Therefore, the average power output is found to be 60 Watts
The signal from the oscillating electrode is fed into an amplifier, which reports the measured voltage as an rms value, 1.0 nV . What is the potential difference between the two extremes
Answer:
The value is [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]
Explanation:
From the question we are told that
The measure voltage is [tex]E_{rms} = 1.0 \ n V = 1.0 *10^{-9} \ V[/tex]
Generally the peak voltage is mathematically represented as
[tex]E_{max} = \sqrt{2} * E_{rms}[/tex]
=> [tex]E_{max} = \sqrt{2} * 1.0 *10^{-9}[/tex]
=> [tex]E_{max} = 1.4142 *10^{-9} \ volts[/tex]
Generally the potential difference between the two extremes is mathematically represented as
[tex]V = 2 * E_{max}[/tex]
=> [tex]V = 2 * 1.4142 *10^{-9}[/tex]
=> [tex]V = 2.8284 *10^{-9 } \ Volts[/tex]
The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value of the magnetic field in the wave is closest to:
Answer:
5.096*10^-8
Explanation:
Given that
The average value of the electromagnetic wave is 310 mW/m²
To find the maximum value of the magnetic field the wave is closest to, we say
Emax = √Erms
Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]
Emax = √233.7648
Emax = 15.289
Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer
15.289 / (3*10^8) = 5.096*10^-8 T
Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8
Which of these is considered a contact force?
a the force between two magnets.
b the gravitational pull of a planet.
c the friction between an object and air.
d The force between two charged particles.
Answer:
C -) The friction between an object and air.
Explanation:
The frictional force is the force that exists between two surfaces in contact, which is opposed to the movement.
That is, this type of force exists as long as there is physical contact between surfaces.
While the other types of force always act at a separation distance between bodies.
Car 1 drives 20 mph to the south, and car 2 drives 30 mph to the north. From
the frame of reference of car 1, what is the velocity of car 2?
Answer:
10 mph faster than car 1 is going
Explanation:
Define each of the three heat transfer methods:
Conduction
Convection
Radiation
7. an electron with initial kinetic energy of 6.0 eV encounters a barrier with height 11.0 eV. What is the probability of tunneling if he width of he barrier is a. 0.80 nm b. 0.40 nm.
Explanation:
In quantum tunneling through a brier, the energy of the tunneled particle is same but the probability amplitude is decreased.
The expression for the probability of tunneling is expressed as follows:
[tex]T=16 \frac{E}{U_{0}}\left(1-\frac{E}{U_{0}}\right) e^{-2 \sqrt{\frac{2 m\left(U_{0}-E\right)}{h}} L}[/tex]
Here, T is the transmission probability; E is the kinetic energy, U_{0} is the barrier height, $m$ is the mass of the electron, and L is the width of the barrier.
a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power
Work done against gravity becomes potential energy of the object-Earth system. Its given by:
W = Ep
W = mgh but weigh(w) = mg
W = wh
W = 350 N × 5 m
W = 1750 JPower is the amount of energy transferred or converted per unit time. This is:
P = W/t
P = 1750 J / 20 s
P = 87.5 WQuestion 4(Multiple Choice Worth 2 points)
Satellite technology has allowed us to monitor areas that were once so remote that we had no way of monitoring them. Review the statements and choose the one that best describes how we can use this technology to benefit the environment.
Large ocean liners use satellites to aid in navigation. This type of navigation aid is also used on cruise ships. Using satellites can help these large ships navigate waters during bad storms.
Satellite surveillance can be used to aid in port security. Satellite data can be monitored and evaluated by complex computer systems. These systems can alert officials when something unusual occurs.
Satellite technology allows us to monitor and track the changes in the polar ice sheets. This can help us plan and direct conservation efforts for the animals that depend on polar ice in order to survive.
Satellite surveillance has allowed us to map areas that were previously too remote to observe. This helps developers to locate and choose sites for large resorts.
Answer:
satellite technology helped large ocean liners guide you to our space mission.
Explanation:
Large ocean liners use satellites to aid in navigation. This type of navigation aid is also used on cruise ships. Using satellites can help these large ships navigate waters during bad storms.
Satellite surveillance can be used to aid in port security. Satellite data can be monitored and evaluated by complex computer systems. These systems can alert officials when something unusual occurs.
Satellite technology allows us to monitor and track the changes in the polar ice sheets. This can help us plan and direct conservation efforts for the animals that depend on polar ice in order to survive.
Satellite surveillance has allowed us to map areas that were previously too remote to observe. This helps developers to locate and choose sites for large resorts.
hello what is a major theme of this speech?
Pitch refers to A. how high or low a note is. B. the loudness of a note. C. the length of the note. D. the range of notes in the composition.
This is music by the way.
Answer:
A
Explanation:
OKAY REAL QUICK A NOTE REPRESENTS THE 12 TONES ON THE CHROMATIC SCALE WHICH IN TURN REPRESENTS THE WHOLE OF SOUND IN ALL MUSIC- EACH NOTE HAS A PITCH THAT REPEATS IN EVERY OCTAVE.
1. Answer the question below. Use the rubric in the materials for help if needed.
Devon drove 50 miles west and then drove 15 miles north. Determine Devon's displacement. Show your work.
I need help
Answer:
d = 52.20 miles
Explanation:
Given that,
Devon drove 50 miles west and then drove 15 miles north.
We need to find Devon's displacement. Let it is d. It can be calculated as follows :
[tex]d=\sqrt{50^2+15^2} \\\\=52.20\ \text{miles}[/tex]
Hence, Devon's displacement is 52.20 miles.
HELP NOW PLS !!
If you exert 950 N*s of impulse on a 12 kg frictionless cart over the course of 5 seconds, how far will it travel during those seconds?
Answer:
The value is [tex]s =197.88 \ m[/tex]
Explanation:
From the question we are told that
The impulse is [tex]I = 950 \ N\cdot s[/tex]
The mass is [tex]m = 12 \ kg[/tex]
The time t = 5 s
Generally impulse is mathematically represented as
[tex]I = F * t[/tex]
=> [tex]F = \frac{I}{t}[/tex]
=> [tex]F = \frac{950 }{ 5 }[/tex]
=> [tex]F = 190 \ N[/tex]
Generally force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]a = \frac{F }{ m }[/tex]
=> [tex]a = \frac{190 }{ 12 }[/tex]
=> [tex]a = 15.83 \ m/s^2[/tex]
Generally from kinematic equation , the distance covered is
[tex]s = ut + \frac{1}{2} * at^2[/tex]
Here u is the initial velocity of the cart and the value is [tex]u = m/s[/tex]
=> [tex]s = 0 * 5 + \frac{1}{2} * 15.83 * 5^2[/tex]
=> [tex]s =197.88 \ m[/tex]
how much work is needed to move an object 20 meters with 40 newtons of force?
Answer: c I think is the answer
Explanation:
Two spherical balls are placed so that their centers are 3.61 m apart. The
force between them is 1.65 x 10-7 N. If the mass of the smaller ball is 81 kg,
what is the mass in kilograms of the other ball?
Answer:
The mass of the other ball is 397.775 kg.
Explanation:
Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.
The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.
The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.
In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:
[tex]F=G\frac{M1*M2}{d^{2} }[/tex]
where:
F = It is the module of the force exerted between both bodies, and its direction is found on the axis that joins both bodies. G = It is the constant of Universal Gravitation, whose value is 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]In this case:
F= 1.65*10⁻⁷ NG= 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]M1= 81 kgM2= ?d= 3.61 mReplacing:
[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]
Solving for M2:
[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]
M2= 397.775 kg
The mass of the other ball is 397.775 kg.
A football quarterback throws a football for a long pass. While in the motion of throwing, the quarterback moves the ball , starting from rest, and completes the motion in . Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
football or basketball which one is easier to get you offers
Answer:
basketball
Explanation: cause it is
Answer:
It honestly depends on which one you're better at.
Explanation: I mean that's what I think.
