Answer:
d.) form supernovas
Explanation:
Learned this in physical school last year
How does the principle of electromagnetism explain the interaction between Earth’s magnetic field and the solar wind? I get that the magnetosphere deflects solar wind, and that solar wind is electrically charged particles from the sun, so electricity is interacting with magnetism, but how does that let electromagnetism explain it?? and please dont just tell me that the magnetosphere just reflects solar particles. tell me how electromagnetism fits into it.
Answer:
Explained below.
Explanation:
Electromagnetism is defined as the study of the electromagnetic force that takes place between two or more electrically charged particles.
Now, in electromagnetic interaction, the charged particles tend to interact in an electromagnetic manner through the process of exchanging photons. Therefore, it was concluded from various experiments that the magnetic fields lines of forces were always closed, thereby eliminating the possibility of any magnetic monopole in such a manner that within a particular magnetic field, there will be two poles and in the solar system, the electromagnetic attraction between earth magnetic field and solar wind will be by way of a field.
Pls Help Me with this problem
Answer:
explanation
Explanation:
1 = C
2 = A
3 = D
4 = E
5 = B
An AC circuit has resistor, capacitor and inductor connected in series. It is
connected to a voltage supply at resonant frequency. If the frequency of the
voltage supply is reduced, which of the following quantities will be reduced? *
Answer:
Z = (R^2 + X^2)^1/2
At the resonant frequency, the quantity X equal zero and
w L - 1 / (w * C) equals zero where w is omega the angular frequency
4. Which point on this position vs. time graph has the fastest speed?
31
Position
Time
Point 1
Point 2
Point 3
Point 4
Answer:
Since v = (x(2) - x(1)) / t
point 2 obviously has the greatest displacement in a given time
Also, point 2 is the steepest line on this graph.
PLEASE HELP ME WITH THIS ONE QUESTION
A converging lens has a focal length of 40 cm. If an object is placed 50 cm in front of the image, where will the image be formed?
Answer:
the image will be real and inverted
examples of transfer of energy
Answer:
by radiation
by heating
electrically
mechanically
What happens to the body’s immune system when attacked by HIV ?
Answer:
HIV attacks a specific type of immune system cell in the body. It's known as the CD4 helper cell or T cell. When HIV destroys this cell, it becomes harder for the body to fight off other infections.
Explanation:
William B. Hartsfield was a man of humble origins who became one of the greatest mayors of Atlanta. He served as mayor for six terms (1937–41, 1942–61), longer than any other person in the city's history. Hartsfield held office during a critical period when the color line separating the races began to change and the city grew . . . to a metropolitan population of one million. He is credited with developing Atlanta into the aviation powerhouse that it is today . . . . . . . On August 30, 1961, the city peacefully integrated its public schools. As a result, Atlanta began to acquire its reputation as "A City Too Busy to Hate."
–New Georgia Encyclopedia
During Hartsfield’s time in office, Atlanta became known as "A City Too Busy to Hate.” What factors supported this nickname? Check all that apply.
The city integrated its schools.
Atlanta became a leader in aviation.
The economy declined.
The population increased to one million.
Many people served as the city mayor.
Answer:
It's A, B, and D
A. The city integrated its school
B. Atlanta became a leader in aviation
D. The population increased to one million
Explanation:
Calculate the force of an object that has a mass of 10kg and an acceleration of 4m/s²
A-40N
B-8N
C-2.5N
D-0.4N
A forward horizontal force of 12 N is used to pull a 240-N crate at constant velocity across a
horizontal floor. The coefficient of friction is:
Explanation:
[tex]let \: coefficient \: of \: friction \: be \: \gamma \\ from \: third \: law \: of \: solid \: friction \\friction \: force = \gamma \times normal \: reaction \\ F = \gamma R \\ but \: R = mg = 240N \\ \gamma = \frac{ F}{R} \\ \gamma = \frac{12}{240} \\ \gamma = \frac{1}{20} [/tex]
Which wave has the longest wavelength?
A. A
B. B
C. C
D.D
Answer:
In my opinion, ''A" would be the longest wavelength.
Why would citizens need to be wary of debris on the road?
A uniform electric field of magnitude 375N/C pointing in the positive x-direction acts on an electron, which is initially at rest.After the electron has moved 3.20cm,what is
(a)the work done by the field on the electron
(b)the charge in potential energy associated with the electron
(c)the velocity of the electron (mass of electron=9.11^-31 kg)
Answer:
a) W = - 1.92 10⁻¹⁸ J, b) U = 1.92 10⁻¹⁸ J, c) v = 2.05 10⁶ m / s
Explanation:
a) Work is defined by
W = F. x
the electric force is
F = q E
we substitute
W = q E x
where the displacement is parallel to the electric field
all quantities must be in the SI system
x = 3.20 cm = 0.0320 m
W = - 1.6 10-19 375 0.0320
W = - 1.92 10⁻¹⁸ J
b) potential energy
U = [tex]k \frac{q_1q_2}{r}[/tex]
the electrical power is
V = k q₁ / r
we can write this potential as a function of the electric field
V = -E x
we substitute
U = -q E x
U = - (-1.6 10⁻¹⁹) 375 0.0320
U = 1.92 10⁻¹⁸ J
Observe that the variation of the red potential is equal to the electrical work
c) let's use conservation of energy
starting point
Em₀ = U = e E x
final point
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
e E x = ½ m v²
v² = 2 e E x / m
v = [tex]\sqrt{ \frac{{2 \ 1.6 10^{-19} \ 375 \ 0.0320 } }{9.1 \ 10^{-31} } }[/tex]
v = [tex]\sqrt{4.2198 \ 10^{12}}[/tex]
v = 2.05 10⁶ m / s
Waldo needs to know how much force to apply in order to move a 4000-kg object at 2 m/S2. Which law should he refer to
A.
first law
B.
second law
C.
third law
D.
law of gravity
Answer:
B . Second law
Explanation:
According to second law:
Net Force acting on the body produce acceleration. The magnitude acceleration of the body is directly proportional to net force and inversly proportional to the mass.
Mathematically:
a = [tex]\frac{F}{m}[/tex]
and
F = ma
So according to the given condition Waldo should use Second law.
Fish can adjust their buoyancy with an organ called a swim bladder. The swim bladder is a flexible gas-filled sac; a fish can increase or decrease the amount of gas in its swim bladder so that it stays neutrally buoyant, neither sinking nor floating. Suppose a fish is neutrally buoyant at some depth and then goes deeper. What needs to happen to the volume of air in the swim bladder
As the fish goes deeper, the volume of air in the swim bladder increases.
What is Buoyant force?Buoyant force is the upward force a fluid exerts on an object.Buoyant force is calculated using the following formula;
F = ρVg
where;
ρ is the density of the fluid.V is the volume of the fluid displaced.g is acceleration due to gravity.The volume of the fluid displaced is calculated as follows;
V = Ah
where;
A is the area of the object.h is the depth of the fluid.Thus, we can conclude that as the fish goes deeper, the volume of air in the swim bladder increases.
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A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.55 m/s to the right relative to the plank.
Required:
What is the velocity of the plank relative to the surface of the ice?
Answer:
The speed of the plank relative to the ice is:
[tex]v_{p}=-0.33\: m/s[/tex]
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
[tex]m_{g}v_{g}+m_{p}v_{p}=0[/tex] (1)
Where:
m(g) is the mass of the girlm(p) is the mass of the plankv(g) is the speed of the girlv(p) is the speed of the plankNow, as we have relative velocities, we have:
[tex]v_{g/b}=v_{g}-v_{p}=1.55 \: m/s[/tex] (2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
[tex]45v_{g}+168v_{p}=0[/tex]
[tex]v_{g}-v_{p}=1.55[/tex]
[tex]v_{p}=-0.33\: m/s[/tex]
I hope it helps you!
A solid, uniform cylinder with mass 8.35kg and diameter 15.0cm is spinning with angular velocity 250rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction-brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.338. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.05 revolutions?
Answer:
354
Explanation:
What is another example of convection
Answer:
boiling water - When water heats, the heat from the burner is transferred to the container, heating the water at the bottom. This hot water rises, creating a circular motion, as colder water descends to replace it.
A 0.156-kg ball is hung vertically from a spring. The spring stretches by 4.18 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position
Answer:
The answer is "[tex]=2.72 \ seconds \\\\[/tex]"
Explanation:
Using formula:
[tex]kr=mg \\\\k=\frac{0.156 \times 9.8}{0.418}\\\\[/tex]
[tex]=3.65741627\ \frac{N}{M}\\\\=3.65\ \frac{N}{M}\\\\[/tex]
Calculating the time period:
[tex]T=2\pi \sqrt{\frac{m}{k}}\\\\[/tex]
[tex]= 2 \pi \sqrt{\frac{0.156}{3.65}}\\\\= 2 \times 3.14 \sqrt{\frac{0.156}{3.65}}\\\\= 6.28 \sqrt{\frac{0.156}{3.65}}\\\\=6.28 \times 0.108210508\\\\=0.67956199\\\\=0.68\\\\[/tex]
The time to calculating the 4 oscillation:
[tex]\to 4T=4\times 0.68 =2.72 \ seconds \\\\[/tex]
The time taken by the spring to complete 4 complete oscillations will be t=2.72 sec
What are oscillations?Oscillation is the process of repeating variation of the quantity or the recitative or periodic variations with respect to the time.
Here it is given that
mass =0.156 kg
stretch x= 4.18 cm
Now from the formula
[tex]kx=mg[/tex]
[tex]k= \dfrac{0.156\times 9.81}{0.418}[/tex]
[tex]k=3.65\ \frac{N}{m}[/tex]
Now the time period will be calculated as:
[tex]T=2\pi\sqrt{\dfrac{m}{k}[/tex]
[tex]T=2\pi\sqrt{\dfrac{0.156}{3.65}[/tex]
[tex]T=6.28\times 0.108210508[/tex]
[tex]T=0.68\ sec[/tex]
Now for the 4 oscillations the total time will be
[tex]T=4\times 0.68=2.72\ sec[/tex]
Hence the time taken by the spring to complete 4 complete oscillations will be t=2.72 sec
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A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 m along the surface before stopping.
Required:
What was the initial speed of the bullet?
Answer:
The initial speed of bullet is "164 m/s".
Explanation:
The given values are:
mass of bullet,
[tex]m'=9.00 \ g[/tex]
or,
[tex]=0.009 \ kg[/tex]
mass of wooden block,
[tex]m=1.20 \ kg[/tex]
speed,
[tex]s=0.390 \ m[/tex]
Coefficient of kinetic friction,
[tex]\mu=0.20[/tex]
As we know,
The Kinematic equation is:
⇒ [tex]v^2=u^2+2as[/tex]
then,
Initial velocity will be:
⇒ [tex]u=v^2-2as[/tex]
[tex]=v^2-2 \mu gs[/tex]
On substituting the given values, we get
⇒ [tex]u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}[/tex]
[tex]=\sqrt{-1.5288}[/tex]
[tex]=1.23 \ m/s[/tex]
As we know,
The conservation of momentum is:
⇒ [tex]mu=m'u'[/tex]
or,
⇒ Initial speed, [tex]u'=\frac{mu}{m'}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{1.20\times 1.23}{0.009}[/tex]
⇒ [tex]=\frac{1.476}{0.009}[/tex]
⇒ [tex]=164 \ m/s[/tex]
Black holes result from
Answer:
supernova explosion or death of massive star
Explanation:
"Most black holes form from the remnants of a large star that dies in a supernova explosion."
The sun is made up mostly of
gas material.
liquid material.
solid material.
plasma material.
The sun is made up mostly of plasma material. Option D
Plasma is considered the fourth state of matter, distinct from gases, liquids, and solids. It is a highly ionized gas consisting of charged particles, including electrons and ions. The sun is primarily composed of plasma due to the extremely high temperatures and intense energy present in its core.
The sun's core has a temperature of about 15 million degrees Celsius (27 million degrees Fahrenheit), which is hot enough to strip electrons from atoms, creating a plasma state. The high temperature and pressure cause hydrogen atoms to undergo nuclear fusion, resulting in the release of vast amounts of energy in the form of light and heat.
This fusion process involves the conversion of hydrogen nuclei (protons) into helium nuclei, releasing enormous energy in the process. The intense heat and pressure within the sun's core sustain this fusion reaction, powering the sun and providing the energy that is radiated out into space as sunlight.
Plasma is an electrically conductive state of matter, meaning it can carry electric currents. The sun's plasma exhibits complex dynamics, including the generation of magnetic fields, solar flares, and coronal mass ejections.
In summary, the sun is primarily composed of plasma material due to the extreme temperatures and intense energy within its core. The plasma state allows for nuclear fusion to occur, releasing the immense energy that sustains the sun's radiative output.
Option D
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The Hydrogen Spectrum
When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms will emit electromagnetic radiation and visible light can be observed. If this light passes through a diffraction grating, the resulting spectrum appears as a pattern of four isolated, sharp parallel lines, called spectral lines. Each spectral line corresponds to one specific wavelength that is present in the light emitted by the source. Such a discrete spectrum is referred to as a line spectrum
By the early 19th century, it was found that discrete spectra were produced by every chemical element in its gaseous state. Even though these spectra were found to share the common feature of appearing as a set of isolated lines, it was observed that each element produces its own unique pattern of lines. This indicated that the light emitted by each element contains a specific set of wavelengths that is characteristic of that element.
The first quantitative description of the hydrogen spectrum was given by Johann Balmer, a Swiss school teacher, in 1885, By trial and error, he found that the correct wavelength λ of each line observed in the hydrogen spectrum was given by
1/λ = R ( 1/2^2-1/n^2)
where R is a constant, later called the Rydberg constant, and n may have the integer values 3, 4, 5, If λ is in meters, the numerical value of the Rydberg constant (determined from measurements of wavelengths) s R= 1.097x10^7 m-1
Balmer knew only the four lines in the visible spectrum of hydrogen. Thus, the original formula was written for a limited set of values of T. However, as more techniques to detect other regions of the spectrum were developed, it became clear that Balmer's formula was valid for all values of n. The entire series of spectral lines predicted by Balmers formula is now referred to as the Balmer series
Part A) What is the wavelength of the line corresponding to n =4 in the Balmer series? Express your answer in nanometers to three significant figures .
Part B) What is the wavelength of the line corresponding to n =5 in the Balmer series?
Answer:
a) λ = 4.862 10⁻⁷ m, b) λ = 4.341 10⁻⁷ m
Explanation:
The spectrum of hydrogen can be described by the expression
[tex]\frac{1}{\lambda} = R_H ( \frac{1}{n_o^2} - \frac{1}{n^2} ) \ \ \ \ n>n_o[/tex]
in the case of the initial state n = 2 this series is the Balmer series
a) Find the wavelength for n = 4
let's calculate
[tex]\frac{1}{ \lambda}[/tex] = 1,097 10⁷ ([tex]\frac{1}{2^2} - \frac{1}{4^2}[/tex])
\frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷
λ = 4.862 10⁻⁷ m
b) n = 5
\frac{1}{ \lambda} = 1,097 10⁷ ([tex]\frac{1}{2^2} - \frac{1}{5^2}[/tex])
\frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷
λ = 4.341 10⁻⁷ m
Helppp pls yes or no question
Answer:
yes, should be
Explanation:
This is a hard yes or no question becuase the amplitudes are the same height but in different beating orders.
Answer:
They appear to be about the same height, so the answer is true.
Explanation:
The electric potential at a point in
space is 235 V. If a 0.0485 C
charge is placed there, what will its
potential energy U be?
(Unit = J)
Answer:
11.3975
Explanation:
Accellus said so
Answer:
11.3975
Explanation:
its for Acellus
What are the two main types of defense that teams employ?
O Front and Back
O Person and Zone
O Goal and Midline.
Thing
Answer:
they employ front and Back defenses
Explanation:
there are the most positions in these types
A painter can paint 2 small rooms and a balcony in 50 hours. If a small room takes twice as long to paint as a balcony, how long will it take him to paint 1 small room plus 3 balconies
Explanation:
ddhjdkdjjjkjkjkkkkkkkkk
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a coordinate system, unable to move at all times. The second object is initially placed 3.00 cm along the positive x-axis and is free to move. The moment the second object is released at x = 3.00 cm, what is the acceleration of this second object? This experiment is done far away from other massive objects, in outer space.
Answer:
a = 640 m/s²
Explanation:
From work-kinetic energy principles,
The net force acting on the second object is the gravitational force and the electric force due to the first object.
So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²
Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²
The net Force F = ma
So, -F₁ + F₂ = F (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)
-Gm²/r² + kq²/r² = ma
ma = -Gm²/r² + kq²/r²
a = (-Gm²/r² + kq²/r²)/m
a = (-G + kq²/m²)m/r²
Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.
Substituting the variables into the equation, we have
a = (m/r²)(-G + k(q/m)²)]
a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]
a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg² ]
a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)
a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²
a = 0.064 × 10⁴N/kg
a = 64 × 10 N/kg)
a = 640 m/s²
Answer it with 1 reason.1.Mirrors having a curved reflecting surface are called as:
a.Plane mirror
b.Spherical mirror
c.Simple mirror
d.None of above
2.Type of spherical mirror are:
aConcave
b.Convex
c.both A and B
d.none of the above
3.Spherical mirror with reflecting surface curved inwards is called ……………
a. convex mirror.
b. concave mirror
c. curved mirror
d. none of the above
4.Spherical mirror with reflecting surface curved ………………… is called concave mirror.
a.Outwards
b.Inwards
c.Backwards
d.none of the above
5.The center of reflecting surface of a spherical mirror is known as……………
a.Pole
b.Aperture
c.Axis
d.Curvature
1 a. Spherical Mirror
2. Both A and B
3. Concave mirror
4.Inwards
5. Pole.
These are your answer. You can mark me Brainliest. Thanks.
hope it helps you have a good day
Explanation:
spherical mirror
both A and B
inwards
pole
As you rise upwards in the atmosphere air pressure___
Answer:
Gases - Water vapor, Nitrogen, Oxygen etc. Describe what happens to air pressure as you rise upwards in the atmosphere. What causes this change in air pressure? As altitude increases, air pressure will decrease As altitude increases the gas molecules that make up the air spread further apart
Answer: As you rise upwards in the atmosphere, air pressure decreases
Explanation: Because there are fewer molecules in the air, the air has a lower density, which results in a drop in air pressure. At different altitudes, the air is not the same. The area close to the earth is denser. As we rise, it continues to thin out. As a result of more air nearby, there is also increased air pressure. The weight and pressure of the air also continue to reduce as the volume of air does.