Answer:
[tex]1863.9\ \text{kg}[/tex]
Explanation:
[tex]m_1[/tex] = Mass of car = 1571 kg
[tex]u_1[/tex] = Initial velocity of car = 129 km/h
[tex]u_2[/tex] = Initial velocity of the animal = 0
[tex]v[/tex] = Velocity of combined mass = 59 km/h
[tex]m_2[/tex] = Mass of the animal
As the momentum of the system is conserved we have
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow \dfrac{m_1u_1+m_2u_2}{v}-m_1=m_2\\\Rightarrow m_2=\dfrac{1571\times 129+0}{59}-1571\\\Rightarrow m_2=1863.9\ \text{kg}[/tex]
The mass of the animal is [tex]1863.9\ \text{kg}[/tex].
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.
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 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
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:
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Using Newton's 2nd Law of Motion Formula (F=MA) answer the following.
the net force on a vehicle that is accelerating at a rate of 1.5m/s^2 is 1,800 newtons. what is the mass of the vehicle to the nearest kilogram?
Explanation:
1200 is your answer for this question
A vertical spring scale can measure weights up to 225 N . The scale extends by an amount of 12.5 cm from its equilibrium position at 0 N to the 225 N mark. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.65 Hz. Ignoring the mass of the spring, what is the mass m of the fish?
Answer:
The mass of the fish is 6.493 kilograms.
Explanation:
The spring-mass system experiments a Simple Harmonic Motion, then the angular frequency ([tex]\omega[/tex]), in radians per second, is determined by the following equation:
[tex]\omega = 2\pi\cdot f[/tex] (1)
Where [tex]f[/tex] is the frequency, in hertz.
In addition, the mass of the fish ([tex]m[/tex]), in kilograms, is:
[tex]m = \frac{k}{\omega^{2}}[/tex] (2)
Where [tex]k[/tex] is the spring constant, in newtons per meter.
And the spring constant is determined by Hooke's Law:
[tex]k = \frac{F}{\Delta x}[/tex] (3)
Where:
[tex]F[/tex] - Elastic force, in newtons.
[tex]\Delta x[/tex] - Spring elongation, in meters.
If we know that [tex]F = 225\,N[/tex], [tex]\Delta x = 0.125\,m[/tex] and [tex]f = 2.65\,hz[/tex], then the mass of the fish is:
[tex]\omega = 2\pi\cdot f[/tex]
[tex]\omega \approx 16.650\,\frac{rad}{s}[/tex]
[tex]k = \frac{F}{\Delta x}[/tex]
[tex]k = 1800\,N[/tex]
[tex]m = \frac{k}{\omega^{2}}[/tex]
[tex]m = 6.493\,kg[/tex]
The mass of the fish is 6.493 kilograms.
This question involves the concepts of Hooke's Law, and the frequency of spring-mass system.
The mass of the fish is "6.5 kg".
First, we will calculate the angular speed of the system:
[tex]\omega=2\pi f[/tex]
where,
ω = angular speed = ?
f = frequency = 2.65 Hz
Therefore,
[tex]\omega = 2\pi(2.65\ Hz)\\\omega = 16.65\ rad/s\\[/tex]
Now, we will use the Hooke's Law to find out the spring constant of the spring:
[tex]K =\frac{F}{\Delta x}[/tex]
where,
K = spring constant = ?
F = force applied = maximum weight = 225 N
Δx = extension = 12.5 cm = 0.125 m
Therefore,
[tex]k=\frac{225\ N}{0.125\ m}\\\\k = 1800\ N/m\\[/tex]
Now, we will use the formula for the angular speed of a spring-mass system to find out the mass of the fish:
[tex]\omega = \sqrt{\frac{k}{m}}\\\\m = \frac{k}{\omega^2}\\\\m=\frac{1800\ N/m}{(16.65\ rad/s)^2}[/tex]
m = 6.5 kg
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The attached picture illustrates Hooke's Law.
Question below in photo!! Please answer! Will mark BRAINLIEST! ⬇⬇⬇⬇⬇⬇⬇
its wave length
its wave lenght because how its measure
An astronaut brings a cube from the Earth to the Moon.
What is true about the inertial mass and weight of the cube?
Note that the gravitational acceleration on the Moon is about 1.6m/s^2
A) The inertial mass increases but the weight decreases.
B) Both the inertial mass and weight decrease.
C) The inertial mass decreases but the weight increases.
D) The inertial mass remains constant but the weight decreases.
Answer:
Answer D. Inertial mass constant, weight decreases.
Explanation:
Khan Academy
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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The half-life of argon-44 is 12 minutes. Suppose you start with 20 atoms of
argon-44 and wait 12 minutes. How many atoms of argon-44 will be left?
A. 20 atoms
B. 40 atoms
C. 10 atoms
D. 5 atoms
SUBMIT
Answer:
C. 10
Explanation:
The half-life of argon-44 is 12 minutes, 10 atoms of argon-44 will be left. The correct option is C.
A radioactive substance's half-life is the amount of time it takes for half of the atoms in a sample to decay.
The half-life of argon-44 in this situation is 12 minutes. Starting with 20 argon-44 atoms, half of them will have disintegrated within 12 minutes, leaving 10 atoms.
This happens because radioactive decay is an exponential process with a constant rate of decay. Every half-life cuts the number of atoms in half.
So, if we waited another 12 minutes, half of the remaining 10 atoms would decay, leaving us with 5 atoms. This process is repeated, with half of the remaining atoms dying with each half-life.
Thus, the correct option is C.
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The solar glare of sunlight bouncing off water or snow can be a real problem for drivers. The reflecting sunlight is horizontally polarized, meaning that the light waves oscillate at an angle of 90 degrees with respect to a vertical line. At what angle relative to this vertical line should transmission axis of polarized sunglasses be oriented, if they are to be effective against solar glare
Answer:
Explanation:
The light waves in the reflected sunlight are horizontally polarized, which illustrates that they oscillate at a [tex]90^o[/tex] angle related to a vertical line.
Depending on the condition of the height of the light, the glare can be almost entirely horizontally polarized. Furthermore, all reflections from over-water surfaces are partially polarized. The water becomes more translucent when using polarized sunglasses.
If polarized sunglasses are to be efficient against solar glare, the transmission axis should be positioned at an angle of [tex]\theta = 45^{o}[/tex]
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
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:
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
Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acceleration?
Answer:
her acceleration is 1 m/sec
Explanation:
The following information is given in the question
The initial velocity is 5 m/s
After 10 seconds, she would be moved at 15 m/s
We need to find the acceleration
As we know that
Acceleration = Change in speed ÷ time
Acceleration = (15 - 5) ÷ (10)
= 1 m/sec
Hence, her acceleration is 1 m/sec
The same would be considered
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.
Electricity involves transfer of charge. The charge(s) involved in electrical forces are composed of which?
- Protons
- Neutrons
- Electrons
- Quarks
- Protons & Electrons
-2 m
2m-2m
Determine the reactions on the beam as shown.
270 kN
67.5 KN.m
0.3 m
760
B
-2.1 m
3 m-
-1.2 m
Water from a river or reservoir enters a hydroelectric power plant through a.....????
The most common type of hydroelectric power plant is an impoundment facility. An impoundment facility, typically a large hydropower system, uses a dam to store river water in a reservoir. Water released from the reservoir flows through a turbine, spinning it, which in turn activates a generator to produce electricity.
What is turbine ?
"A turbine is a rotary mechanical device that extracts energy from a fluid flow and converts it into useful work." The work produced by a turbine can be used for generating electrical power when combined with a generator.
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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
supergiant stars
a.) form from red giants
b.) fuse hydrogen into carbon
c.) form planetary nebulae
d.) form supernovas
Answer:
d.) form supernovas
Explanation:
Learned this in physical school last year
The gauge pressure in your car tires is 3.00 ✕ 10^5 N/m2 at a temperature of 35.0°C when you drive it onto a ship in Los Angeles to be sent to Alaska. What is their gauge pressure (in atm) later, when their temperature has dropped to
−42.0°C?
Assume the tires have not gained or lost any air.
Explanation:
Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.
So, we have:
P1/T1=P2/T2
Solving for P2, we have:
P2=(P1*T2)/T1
NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!
Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:
P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]
P2=228930.7156 N/m^2
Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2
Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!
Hope this helps!
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In the diagrams, the solid green line represents Earth's rotational axis and the dashed red line represents the magnetic field axis. Which diagram accurately shows Earth's magnetic field?
Answer: Picture 2 or B!!!!!
Answer:
B
Explanation:
the answer as said is picture 2
what is inside a black hole
[tex]\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}[/tex]
A black hole is a tremendous amount of matter crammed into a very small — in fact, zero — amount of space. The result is a powerful gravitational pull, from which not even light can escape — and, therefore, we have no information or insight as to what life is like inside. A black hole is not empty, It's actually a lot of matter condensed into a single point. This point is known as a singularity.
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
You decide to fly over to Fred Meyer’s using your jet pack but during your vertical ascent you drop your wallet. If your wallet is dropped when you are ascending at a constant speed of 6 m/s and the wallet is released with the same upward velocity of 6 m/s when t = 0. Determine the speed of your wallet when it hits the ground at t = 8s. Also determine the altitude of your jet pack.
Answer:
Here we only need to look at the vertical problem, so first, let's look at the forces acting vertically on the wallet.
When the wallet starts to fall, the only force acting on it will be the gravitational force (where we are ignoring the effects of air friction).
Then the acceleration of the wallet will be equal to the gravitational acceleration, g = 9.8m/s^2
Then we can write:
a(t) = (-9.8m/s^2)
Where the negative sign is because this acceleration is downwards.
To get the vertical velocity equation of the wallet we need to integrate over time to get:
v(t) = (-9.8m/s^2)*t + v0
Where v0 is the constant of integration, and in this case is the initial velocity of the wallet, which we know is equal to 6m/s, then the velocity equation is:
v(t) = (-9.8 m/s^2)*t + 6m/s
To get the position equation we need to integrate over time again, we get:
p(t) = (1/2)*(-9.8 m/s^2)*t^2 + (6m/s)*t + p0
Where p0 is the initial vertical position, in this case, is the height at which the wallet is dropped, which is also the altitude of your jet pack when the wallet falls.
Now we want to know two things:
Determine the speed of your wallet when it hits the ground at t = 8s
Here we just need to evaluate the velocity equation in t = 8s.
v(8s) = (-9.8 m/s^2)*8s + 6m/s = -72.4 m/s
We also want to determine the altitude of the jet pack (when the wallet drops).
To find this, we can use the fact that the wall hits the ground at t = 8s.
The wallet hits the ground when it's vertical position is equal to zero, then:
p(8s) = 0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0
Now we can solve this for p0.
0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0
(1/2)*(9.8 m/s^2)*(8s)^2 - (6m/s)*8s = p0
265.6m = p0
This means that the altitude of the jet pack when the wallet drops is 265.6m
A hammer with mass m is dropped from rest from a height h above the earth's surface. This height is not necessarily small compared with the radius RE of the earth. If you ignore air resistance, derive an expression for the speed v of the hammer when it reaches the surface of the earth. Your expression should involve h, RE, and mE, the mass of the earth.
Answer:
v = √{2(GmE(1/RE - 1/(RE + h)]}
Explanation:
From the law of conservation of energy, the total kinetic + potential energy at h = total kinetic + potential energy at the surface of the earth(h = 0)
So, K + U = K' + U'
where K = kinetic energy of hammer at h = 0
U = gravitational potential energy of hammer at h = -GmEm/(RE + h)
K' = kinetic energy of hammer at the earth's surface (h = 0) = 1/2mv²
U = gravitational potential energy of hammer at earth's surface (h = 0) = -GmEm/RE
So, K + U = K' + U'
0 + [-GmEm/(RE + h)] = 1/2mv² + [-GmEm/RE]
0 - GmEm/(RE + h) = 1/2mv² - GmEm/RE
collecting like terms,we have
1/2mv² = GmEm/RE - GmEm/(RE + h)
Factorizing GmE and multiplying both sides by 2, we have
v² = 2(GmE(1/RE - 1/(RE + h)]
taking square-root of both sides, we have
v = √{2(GmE(1/RE - 1/(RE + h)]}
The expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
What will be the speed of the hammer?The law of conservation says that the total energy at the height h and the total energy at the surface of the earth will remain constant.
From the law of conservation of energy
[tex]KE+PE=KE'+PE'[/tex]
where
KE = kinetic energy of hammer at h = 0
PE = gravitational potential energy of hammer at h height
[tex]PE=\dfrac{-Gm_em}{R_e+h}[/tex]
KE' = kinetic energy of hammer at the earth's surface (h = 0)
[tex]KE'=\dfrac{1}{2} mv^2[/tex]
PE' = gravitational potential energy of hammer at earth's surface (h = 0)
[tex]PE'=\dfrac{Gm_em}{R_e}[/tex]
Now by putting the values
[tex]KE+PE=KE'+PE'[/tex]
[tex]0+ \dfrac{-Gm_em}{R_e+h}[/tex] [tex]=\dfrac{1}{2} mv^2[/tex] [tex]-\dfrac{Gm_em}{R_e}[/tex]
[tex]\dfrac{1}{2} mv^2=\dfrac{Gm_em}{R_e} -\dfrac{Gm_em}{R_e+h}[/tex]
[tex]v^2=2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )[/tex]
[tex]v=\sqrt{2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )}[/tex]
Thus the expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
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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
To know more about oscillations, follow
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someone please help!! tysm
Hi there!
[tex]\large\boxed{5N}[/tex]
From the given diagram, 20 N is going left, and 15 N is going right.
To find the net force, we must subtract the two since the two forces go in opposite directions:
20 N - 15 N = 5N.
How much energy (in kWh) is produced in one day by a solar panel of surface area A =15
m? in a region where the average solar power density if 4.33 kWh/m²/day. Assume the
efficiency of the panel to be 18 %. Round off your result to 2 decimal places, and do not
write the unit
Question 2
20 pts
The average electricity consumption of a house in Gainesville is known to be 907 kWh in a
month (One month - 30 days). They would like to install solar panels of 12 % efficiency to
generate this electricity. Given that the average solar power density in Gainesville is 5,47
kWh/m2/day, how much surface area must the panels occupy? Calculate the result in m²
but do not write the unit. Round off you answer to a whole number (zero decimal place.)
Answer:
I am calculating the total area of a solar panel for a particular load demand by ... designing according to energy demand for a day then how will it affect total solar area? ... Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency ... would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.
Explanation:
