Answer:
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____ is the study of things getting faster as they move.
A. Anatomy
B. Force
C. Physics
D. Dynamics
Answer: b force
Explanation:
yes because the world comin g up with more technique
state the two motion equations.
Answer:
Equations of motion relate the displacement of an object with its velocity, acceleration and time. s=vt where s is the displacement, v the (constant) speed and t the time over which the motion occurred. ...
Displacement with negative acceleration: s, equals, v, t, minus, one half, a, t, square...
Displacement with positive acceleration: s, equals, u, t, plus, one half, a, t, squared,s...
Velocity squared: v, squared, equals, u, squared, plus, 2, a, s,v2=u2+2as
Velocity: v, equals, u, plus, a, t,v=u+at
the boiling point of F2 much lower than the boiling point of NH3
Answer:yeah it A
Explanation:
At 2:00, Alice was traveling in a car at 40 miles/hour. She then slows down, and at 4:00, she was traveling at 20 miles/hour.
What was Alice’s average acceleration between 2:00 and 4:00?
Show your complete calculation and state your answer with the correct units.
Answer:
-10miles/hr²
Explanation:
a = ∆v/∆t
Where:
a = acceleration (miles/hr²)
∆V = change in velocity (miles/hr)
t = time (hour)
The change in time is from 2:00 - 4:00 ∆t = 2 hours.
The distance covered is as follows: 20miles/hour - 40 miles/hr
∆v = -20miles/hr
Using a = ∆v/∆t
a = -20/2
a = -10miles/hr²
PLS HELP ME. A 0.0780 kg lemming runs off a 5.36m high cliff at 4.84 m/s what is it potential energy when it lands?
Answer:
p.e=0.078kg×1/2×5.36m
p.e=0.913j
A. What is the change in internal energy for each of the following situations? q = 7.9 J out of the system and w = 3.6 J done on the system q = 1.5 J into the system and w = 7.5 J done on the system
Answer: [tex]-4.3\ J,\ 9\ J[/tex]
Explanation:
Given
(a)
Heat transfer [tex]Q=-7.9\ J\quad \text{taken}[/tex]
Work done [tex]W=-3.6\ J\quad \text{on the system}[/tex]
Change in the internal kinetic energy is
[tex]\Delta U=Q-W\\\Rightarrow \Delta U=-7.9-(-3.6)\\\Rightarrow \Delta U=-4.3\ J[/tex]
(b)
Heat transfer [tex]Q=1.5\ J\quad \text{given}[/tex]
Work done [tex]W=-7.5\ J\quad \text{on the system}[/tex]
Change in the internal kinetic energy is
[tex]\Delta U=Q-W\\\Rightarrow \Delta U=1.5-(-7.5)\\\Rightarrow \Delta U=9\ J[/tex]
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.85 keV and a current of 5.40 mA produced by the generator. (a) What is the speed of the protons (in m/s)
Answer:
The speed of the proton is [tex]v =5.95\times 10^{5} m/s[/tex]
Explanation:
Energy, E = 1.85 keV
mass of proton, m = 1.67 x 10^-27 kg
Let the speed is v.
The kinetic energy is given by
[tex]K = 0.5m\times v^{2}\\\\1.85\times \1.6\times 10^{-16} = 0.5\times 1.67\times 10^{-27}\times v^{2}\\\\v =5.95\times 10^{5} m/s[/tex]
Which statements are true of noble gases?
Check all that apply.
A. They are metalloids.
B. Their valence shells are full of electrons.
C. They are not very reactive.
D. All of the noble gases have at least two electron shielding layers.
If a spider can travel 3.5 meters in 25 minutes, how fast can they go?
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon? A. Emission spectra of atoms B. Blackbody radiation C. Interference patterns of electrons D. Photoelectric effect
Answer:
In some situations, matter demonstrates wave behavior rather than particle behavior. This is best illustrated by which phenomenon is:
C. Interference patterns of electrons.
Answer:
C. Interference patterns of electrons
Iron is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron forms a crystal with an fcc unit cell and a lattice constant . Calculate the density of Iron.
The question is incomplete. The complete question is :
Iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant, a = 0.352 nm. Calculate the density of Iron β.
Solution :
The density is given by :
[tex]$\rho = \frac{ZM}{a^3N_0} \ \ g/cm^3$[/tex] ..................(i)
Here, Z = number of atoms in a unit cell
M = atomic mass
[tex]$N_0$[/tex] = Avogadro's number = [tex]$6.022 \times 10^{23}$[/tex]
a = edge length or the lattice constant
Now for FCC lattice, the number of atoms in a unit cell is 4.
So, Z = 4
Atomic mass of iron, M = 55.84 g/ mole
Given a = 0.352 nm = [tex]$3.52 \times 10^{-8}$[/tex] cm
From (i),
[tex]$\rho = \frac{ZM}{a^3N_0} $[/tex]
[tex]$\rho = \frac{4 \times 55.84}{(3.52 \times 10^{-8})^3 \times 6.022 \times 10^{23}} $[/tex]
[tex]$= 8.51 \ \ g \ cm ^{-3}$[/tex]
Therefore, the density of Iron β is [tex]$ 8.51 \ \ g \ cm ^{-3}$[/tex].
Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimumpossible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relationbetweenEandp. Give your answer to 2 significant figures. (The large value you will find is a strongargument against the presence of electrons inside nuclei, since no known mechanism could contain anelectron with this much energy.)
Answer:
39.40 MeV
Explanation:
Determine the minimum possible Kinetic energy
width of region = 5 fm
From Heisenberg's uncertainty relation below
ΔxΔp ≥ h/2 , where : 2Δx = 5fm , Δpc = hc/2Δx = 39.4 MeV
when we apply this values using the relativistic energy-momentum relation
E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,
Also in a nuclear confinement ( E, P >> mc )
while The large value will portray a Non-relativistic limit as calculated below
K = h^2 / 2ma^2 = 1.52 GeV
Consider a solid sphere and a solid disk with the same radius and the same mass. Explain why the solid disk has a greater moment of inertia than the solid sphere, even though it has the same overall mass and radius.
Answer:
Explanation:
In a Solid sphere; the moment of inertia around its geometrical axis can be expressed by using the formula:
[tex]\mathtt{I_s = \dfrac{2}{5} M_s R^2_s}[/tex]
For the solid disk; the moment of inertia around the central axis is:
[tex]\mathtt{I_D= \dfrac{1}{2}M_DR_D^2}[/tex]
Suppose [tex]M_D = M_S[/tex]; then we can say both to be equal to M
As well as [tex]R_D = R_S[/tex]; then that too can be equal to R
Now;
[tex]\mathtt{I_s = \dfrac{2}{5} M R^2} --- (1)[/tex]
[tex]\mathtt{I_D= \dfrac{1}{2}MR^2}---(2)[/tex]
Multiplying equation (1) by 2, followed by dividing it by 2; we have:
[tex]\mathtt{I_s= \dfrac{2}{5}MR^2} \times \dfrac{2}{2}[/tex]
[tex]I_s = \dfrac{4}{5} \times \dfrac{1}{2}MR^2 \\ \\ I_s = \dfrac{4}{5}\times I_D \\ \\ I_s > I_D[/tex]
Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 60 mi/h.How far must the faster car travel before it has a 15-min lead on the slower car
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}[/tex]
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
[tex]\frac{x}{55} - \frac{x}{60} = \frac{15}{60}\\\\multiple \ through \ by \ 660\\\\12x - 11x = 165\\\\x = 165 \ miles[/tex]
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Charge q is a test charge used to measure the electric field created by a source charge Q. What happens to the magnitude of the electric force on these charges if q is doubled?
Answer:
The electric force will also double.
When the test charge q is double the change in magnitude of the electric force is that it also gets doubled.
What is the electric force due to charge?Electric force is defined as force between the chages which is directly proportional to the product of charges and inversely proportional to the square of the distance between them. The electric force is a vector quantity, It have magnitude as well as direction.
Electric force = k*Q*q/r²
The electric field is defined as the electric force per unit charge. The electric field is a vector quantity.
Electric field = Electric force / test charge
Electric field = k*Q/r²
Given that in question there is the test charge, q for electric field and the source charge Q when the test charge q will become 2q then electric force is,
Electric force = kQ(2q)/r²
So, the magnitude of the electric force is increased by twice when the test charge will become 2q.
To learn more about the electric force refer to the link
brainly.com/question/2526815
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That light has a dual nature is referring to light: having energy and momentum. having high- or low-energy photons. acting as waves and particles. undergoing pair production.
Answer:
Option C, acting as waves and particles
Explanation:
Light has dual nature because it acts both as a wave and particle. It has high energy particle i.e photons and it also behave as an electromagnetic wave. This property of light is studied under the quantum mechanics. Einstein also proved that light is a stream of photons possessing both electrical and magnetic properties.
Hence, option C is correct
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 43 cm from the end of the plank with force F2. If the plank has a mass of 13.7 kg and its center of gravity is at the middle of the plank, what is the force F1
Answer: [tex]115.52\ N[/tex]
Explanation:
Given
Length of plank is 1.6 m
Force [tex]F_1[/tex] is applied on the left side of plank
Force [tex]F_2[/tex] is applied 43 cm from the left end O.
Mass of the plank is [tex]m=13.7\ kg[/tex]
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
[tex]\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N[/tex]
Net vertical force must be zero on the plank
[tex]\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N[/tex]
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity [tex]I[/tex] is proportional to 1/(distance)²
i.e
[tex]I[/tex] ∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2
Hence,
[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
g A ball is tossed straight up from the surface of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the asteroid's radius and then falls straight down toward the surface of the asteroid. What forces, if any, act on the ball while it is on the way up
Answer:
Only a decreasing gravitational force that acts downward
Explanation:
The gravitational force is the gravitational pull which attract a mass of smaller size by the mass of a bigger size. It is the force which attract two masses close to each other.
In the context, when a ball is tossed up from the surface of an asteroid that have no atmosphere, the ball rises up and then falls back to the surface of the asteroid. The ball falls back because the gravitational pull of the asteroid pulls back the ball to its surface. Thus a decreasing gravitational force acts on the ball in the downward direction while the ball is in its way up.
People who do very detailed work close up, such as jewellers, often can see objects clearly at much closer distance than the normal 25 cm. a. What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.5 cm
This question is incomplete, the complete question is;
People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.
a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.) __ mm
Answer:
1) the power in D of the eyes of a woman is 61.7647 D
2) the size in mm of an image of a 8.00 mm object is -1.882 mm
Explanation:
Given the data in the question;
a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,
so
u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m
v = 2.00 cm = ( 2 / 100 )m = 0.02 m
Now, we know that power of lens p = 1 / u + 1 / v
so we substitute
p = ( 1 / 0.085 ) + ( 1 / 0.02 )
p = 11.7647 + 50
p = 61.7647 D
Therefore, the power in D of the eyes of a woman is 61.7647 D
b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)
we know that;
m = -v / u
we substitute
m = -0.02 / 0.085
m = -0.2353
since H₀ = 8.0 mm
H[tex]_i[/tex] = m × H₀
H[tex]_i[/tex] = -0.2353 × 8.0
H[tex]_i[/tex] = -1.882 mm
the size in mm of an image of a 8.00 mm object is -1.882 mm
What is the relationship between the density of the equipotential lines, the density of the electric field lines and the strength of the electric field?
Answer:
I dont. understand the question, maybe insert the picture?
The human eye has two lenses.
O true
O False
Answer:
true its truedjjs sjsnsns
Which of the following is the best definition of a physical change?
A. A change in a substance where a new substance is formed
B. A change in a substance in which bonds are broken
C. A change in a substance with no new substances being formed
D. A change in a substance in which mass is conserved
Answer:
The answer is C. A change in a substance with no new substances being formed
Explanation:
I did the quiz.
The best definition of physical change is a change in a substance, with no new substances being formed. Hence, option C is correct.
What is a Physical Change?A chemical substance's form, not its chemical composition, can change due to physical changes. In most cases, compounds cannot be separated into chemical components or simpler compounds; instead, mixtures are separated into their constituent compounds through physical changes.
Whenever something changes physically but not chemically, we say that something has changed physically. This is in contrast to the idea of a chemical change, which occurs when a substance's composition changes or when one or more compounds join or fragment to generate new substances. In general, physical means can be employed to undo a physical alteration. For instance, by letting the water evaporate, salt that has been dissolved in it can be reclaimed.
Therefore, this concludes that option C is correct.
To know more about Physical change:
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Keesha is looking at a beetle with a magnifying glass. She wants to see an upright, enlarged image at a distance of 25 cm. The focal length of the magnifying glass is +5.0 cm. Assume that Keesha's eye is close to the magnifying glass.
(a) What should be the distance between the magnifying glass and the beetle?
(b) What is the angular magnification?
Answer:
a) p = 4.167 cm, b) m = + 6
Explanation:
a) For this exercise we must use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and the image, respectively
In this case the distance to the image q = 25 cm and the focal length is f = 5.0 cm
Since the object and its image are on the same side of the lens, the distance to the image by the sign convention must be negative.
[tex]\frac{1}{p } = \frac{1}{f} - \frac{1}{q}[/tex]
[tex]\frac{1}{p} = \frac{1}{5} - \frac{1}{-25}[/tex]
[tex]\frac{1}{ p}[/tex] = 024
p = 4.167 cm
b) angular magnification
m = h ’/ h = - q / p
m = - (-25) /4.167
m = + 6
the positive sign indicates that the image is straight and enlarged
Two identical particles each of mass M and charge Q are placed a certain distance apart. If they are in equilibrium
under mutual gravitational and electric force then calculate the order of Q/M in SI units.
Answer:
Q / m = 8.61 10⁻¹¹ C / kg
Explanation:
For this exercise we use the gravitational force of attraction
[tex]F_g = G \frac{m_1m_2}{r^2}[/tex]
the electric force
[tex]F_e = k \frac{q1q2}{r^2}[/tex]
indicate that the two forces are equal
G m₁ m₂ / r² = k q₁ q₂ / r²
they also say that the two masses are equal and the two charges are equal
G m² = k Q²
Q / m = [tex]\sqrt{\frac{G}{k} }[/tex]
we calculate
Q / m = [tex]\sqrt{\frac{6.67 \ 10^{-11} }{8.99 \ 10^9} }[/tex]
Q / m = [tex]\sqrt{ 0.7419 \ 10 ^{-20}}[/tex]
Q / m = 0.861 10⁻¹⁰
Q / m = 8.61 10⁻¹¹ C / kg
As you can see from the Group 1 stars, the cooler or hotter a star is, the brighter it will be. Group 2 and Group 3 stars do not follow this pattern. Hence, there must be something besides temperature that can affect how bright a star is. Describe your own hypothesis about these stars (Group 2 and Group 3). Why would their brightness not be strictly related to their temperature
Answer:
brightness that we observe from a star is related to the energy produced and the distance to the Earth
Explanation:
In stars, the color that we observe is directly related to the temperature of the star by the y of the Wien displacement.
λ_{max} T = 2,898 10³
the brightness that we observe from a star is related to the energy produced and the distance to the Earth
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same
Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]
Hence, this is the required solution.
An oscillating LC circuit consisting of a 1.0 nF capacitor and a 3.0 mH coil has a maximum voltage of 3.0 V. What are (a) the maximum charge on the capacitor, (b) the maximum current through the circuit, and (c) the maximum energy stored in the magnetic field of the coil
Answer:
[tex]E=4.5*10^-^9J[/tex]
Explanation:
From the question we are told that:
Capacitor [tex]C=1.0nf[/tex]
Induction [tex]I=3.0mH[/tex]
Voltage [tex]V=3.0[/tex]
Generally the equation for Max charge on Capacitor is mathematically given by
[tex]Q_{max}=C*V[/tex]
[tex]Q_{max}=1*10^{-9}*3[/tex]
[tex]Q_{max}=3*10^{-9}C[/tex]
Generally the equation for Energy in magnetic field of the coil is mathematically given by
Since
Energy stored in capacitor = Energy in magnetic field of the coil
Therefore
[tex]E = (1/2)* C * V^2[/tex]
[tex]E= 0.5 * 1*10^{-9} *3^2[/tex]
[tex]E=4.5*10^-^9J[/tex]
Last stage of a medium mass star
White Dwarf Stage
This ring is called a planetary nebula. when the last of the helium atoms in the core are fused into carbon atoms, the medium size star begins to die. Gravity causes the last of the star's matter to collapse inward and compact. This is the white dwarf stage.