Answer:
B) Inertia is the resistance of any physical object
Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion with a negative velocity and a negative acceleration.
Answer:
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
Explanation:
In a vertical movement the acceleration is always downwards, therefore negative since it is created by the attraction of the Earth on the body.
An example of positive velocity is throwing a ball upwards
An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height
How do a parachutes work??4-5 sentences plsss help rn
Answer:
Explanation:
A parachute works by forcing air into the front of it and creating a structured 'wing' under which the canopy pilot can fly. Parachutes are controlled by pulling down on steering lines that change the shape of the wing, cause it to turn. The main forces acting on a parachute are gravity and drag. When you first release the parachute, the force of gravity pulls it downward, and the parachute speeds toward the ground. The faster the parachute falls, though, the more drag it creates.
A parallel plate vacuum capacitor has 8.40 J of energy stored. The separation between plates is 2.30 mm. If the separation is decreased to 1.15 mm what is the energy stored if (a) the charge Q on the plates is held constant, and (b) the voltage V across the plates is held constant
(a) 4.20 J
(b) 16.74 J
Explanation:For a parallel plate vacuum capacitor with area A and whose plates are separated by by a distance of d, its capacitance C is given by;
C = A∈₀ / d --------------------(i)
Where;
∈₀ = constant called permittivity of vacuum.
The energy U stored in such capacitor is given by;
U = [tex]\frac{1}{2}[/tex]CV² ----------------------(ii)
or
U = [tex]\frac{1}{2}[/tex](Q²/C) -------------------(**)
Where;
V = potential difference or voltage across the plates.
Q = charge on the plates.
(a) If the charge is held constant
Combine equations (i) and (**) to give;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d) -----------------------(iii)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iii)
8.40 = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / 0.023)
8.40 = [tex]\frac{1}{2}[/tex]Q² x (0.023 / A∈₀)
Multiply through by 2
2 x 8.40 = Q² x (0.023 / A∈₀)
16.80 = Q² x (0.023 / A∈₀)
Divide through by 0.023
16.80 / 0.023 = Q² x (0.023 / A∈₀) / 0.023
730.4 = Q² / (A∈₀)
Make Q² subject of the formula
Q² = 730.4(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
Q = constant [this means that Q² still remains 730.4(A∈₀) ]
The energy stored is found by substituting these values of d and Q² into equation (iii) as follows;
U = [tex]\frac{1}{2}[/tex]Q² / (A∈₀ / d)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀)) / (A∈₀ / 0.0115)
U = [tex]\frac{1}{2}[/tex](730.4(A∈₀))(0.0115 / A∈₀)
U = [tex]\frac{1}{2}[/tex](730.4)(0.0115)
U = 4.20J
Therefore, the energy stored if the charge Q on the plates is held constant is 4.20 J
(b) If the voltage is held constant
Combine equations (i) and (ii) to give;
U = [tex]\frac{1}{2}[/tex](A∈₀ / d)V² -----------------------(iv)
From the question;
The parallel plate capacitor has 8.40J energy stored and distance between plates is 2.30mm i.e
U = 8.40J
d = 2.30mm = 0.023m
Substitute these values into equation (iv)
8.40 = [tex]\frac{1}{2}[/tex](A∈₀ / 0.023)V²
Multiply through by 2 x 0.023
2 x 0.023 x 8.40 = (A∈₀)V²
2 x 0.023 x 8.40 = (A∈₀)V²
0.385 = (A∈₀)V²
Make V² subject of the formula
V² = 0.385/(A∈₀)
Now, if the separation distance is decreased to 1.15mm and the voltage is held constant i.e
d = 1.15mm = 0.0115m
V = constant [this means that V² still remains 0.385/(A∈₀) ]
The energy stored is found by substituting these values of d and V² into equation (iv) as follows;
U = [tex]\frac{1}{2}[/tex](A∈₀ / 0.0115)[0.385/(A∈₀)]
U = [tex]\frac{1}{2}[/tex](0.385/0.0115)
U = 16.74
Therefore, the energy stored if the voltage V across the plates is held constant is 16.74 J
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
Answer:
Option 2
Explanation:
The complete question is
A 50 kg child sits on the left side of the bathtub. A small toy boat of 0.5 kg is on the right side of the bathtub. Which part of the bathtub has the greatest pressure
TopBottomLeftRightSolution
It is the bottom of the bucket that will high pressure because of the additional weight of 50 Kg boy along with the weight of the water and the tub itself.
Pressure acts in the down ward direction and is equal to the force/weight divided by the area.
Hence, option 2 is correct
what is the meaning of friend ?
Answer:
person that you know and like (not a member of your family), and who likes you
how can you prove that acceleration is a derived unit
a = (dx / dt)²
Explanation: Unit of distance is m (metres) and unit of time is s (seconds) speed v is first derivative of distance x versus time:
v = dx / dt, unit is m/s. Acceleration is second derivative of
speed versus time a = (dx / dt)² = (dv/dt) , unit is m/s²
Answer:
Explanation:
Acceleration is derived unit because it has two fundamental units involved i.e. meter and second square.
You drop a ball from a height of 10 meters. Each time the ball bounces, it
reaches a lower height. Why does the ball lose height after each time it hits
the ground?
Two trains are moving at 50 m/s in opposite directions on the same track. The engineers see simultaneously that they are on a collision course and apply the brakes when they are 1,100 m apart. Assuming both trains have the same acceleration, what must this acceleration be (in m/s2) if the trains are to stop just short of colliding
Answer:
Acceleration is 2.5 m/s^2
Explanation:
Let train 1 travel x meters and train 2 travels 1000-x meters
As per the las of acceleration
[tex]v^2 = u^2 + 2ax[/tex]
Substituting the given values, we get -
[tex]0 = 50^2 + 2ax\\ax = -1250\\[/tex]---Eq (1)
Similarly
[tex]0 = 50^2 + 2a(1000-x)\\a(1000-x) = 1250[/tex]-- Eq (2)
Substituting the value of ax from eq (1) into Eq(2), we get -
1000*a - 1250 = 1250
a = 2500/1000 = 2.5 m/s^2
allung d Uall, Wily
In which condition is mirage seen ? Why is light dispersed ?
Answer:
Mirage is a phenomenon which can be seen when the surface air gets heated up and it becomes lighter. Lighter air moves up in the atmoshphere.
Explanation:
When the lighter air from cooler areas to warmer areas are refracted and they bent upwards.and it dispers
,
A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance
Answer:
[tex]R=2.78\ \Omega[/tex]
Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :
[tex]P=I^2R[/tex]
Put all the values to find R.
[tex]R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega[/tex]
So, the resistance is equal to [tex]2.78\ \Omega[/tex].
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming that the only forces affecting the speed are the weight and the drag, calculate the drag coefficient. The frontal area is 0.9m2 .
Answer:
0.9378
Explanation:
Weight (W) of the rider = 100 kg;
since 1 kg = 9.8067 N
100 kg will be = 980.67 N
W = 980.67 N
At the slope of 12%, the angle θ is calculated as:
[tex]tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0[/tex]
The drag force D = Wsinθ
[tex]\dfrac{1}{2}C_v \rho AV^2 = W sin \theta[/tex]
where;
[tex]\rho = 1.23 \ kg/m^3[/tex]
A = 0.9 m²
V = 15 m/s
∴
Drag coefficient [tex]C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}[/tex]
[tex]C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}[/tex]
[tex]C_D =0.9378[/tex]
If a car's velocity is 30 m/s and it drives at this velocity for 4 seconds, how far did it go?
Answer:
120 m
General Formulas and Concepts:
Kinematics
VelocityDisplacementDistanceTimeExplanation:
Step 1: Define
Identify
[Given] v = 30 m/s
[Given] t = 4 s
Step 2: Solve
Multiply [Cancel out units]: 30 m/s · 4 s = 120 mAnswer:
[tex]\boxed {\boxed {\sf 120 \ meters}}[/tex]
Explanation:
Distance, or how far an object travels, is the product of velocity and time.
[tex]d= v*t[/tex]
The velocity is 30 meters per second and the time is 4 seconds.
v= 30 m/s t=4 sSubstitute the values into the formula.
[tex]d= 30 \ m/s * 4 \ s[/tex]
Multiply. The units of seconds (s) cancel.
[tex]d= 30 \ m * 4[/tex]
[tex]d=120 \ m[/tex]
The car travels a distance of 120 meters in 4 seconds at a velocity of 30 meters per second.
A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
ocean floor and is received 15s later. What is the distance between the submarine and the ocean floor?
Answer:
the distance between the submarine and the ocean floor is 11,250 m
Explanation:
Given;
speed of the wave, v = 1500 m/s
time of motion of the wave, t = 15 s
The time taken to receive the echo is calculated as;
[tex]time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m[/tex]
Therefore, the distance between the submarine and the ocean floor is 11,250 m
the boiling point of F2 much lower than the boiling point of NH3
Answer:yeah it A
Explanation:
John is going to use a rope to pull his sister Laura across the ground in a sled through the snow. The rope makes an angle of 25 with the ground He is pulling horizontally with a constant force of 400 N. John and manages to get the sled going from 0 to 4 m/s in 5 s. The force due to friction on the sled is 310 N. What is the mass of Laura and the sled combined
Fill in the blanks. Power station produce electricity at __________.
Answer:
heat
Explanation:
heat is the answer hdhhdhdhdhdhdhdhdhdhd
A closely wound, circular coil with radius 2.80 cm has 790 turns. Part A What must the current in the coil be if the magnetic field at the center of the coil is 0.0760 T
Answer:
[tex]I=11.1A[/tex]
Explanation:
From the question we are told that:
Radius [tex]R=2.80[/tex]
Turns [tex]N=790[/tex]
Magnetic field B=0.0760
Generally the equation for Magnetic field at the center of the coil is mathematically given by
[tex]B=\frac{\mu NI}{2r}[/tex]
[tex]0.076=\frac{4\p*10^-^7*790*I}{2*0.028}[/tex]
[tex]I=\frac{0.076*2*0.028}{4\p*10^{-7}*790}[/tex]
[tex]I=11.1A[/tex]
PLEASE HELP!!!!
A person pushes attempts to push a couch with a 25 N force. The couch, however, doesn't move. What is the static friction force acting on the couch? *
A. 25 N
B. 0 N
C. 50 N
D. There is no static friction the ONLY force acting on the couch is the push
Static friction cancels out the force of the push, so it also has a magnitude of 25 N.
what happened in my room
Answer:
A GHOST CAME! Booooo!!!!!!Hah lol
C.
A palm fruit dropped to the ground from the top of
a tree 45m tall. How long does it take to reach the
ground? A. 9s B. 4.5s C. 6 D. 7.5s E. 35
(g = 10ms2).
Answer:
b 4.5
Explanation:
time=distance/speed
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A small tack with charge Q is lowered on a silk thread through the hole into the interior of the shell.
Required:
a. What is the charge on the inner surface of the shell?
b. What is the charge on the outer surface of the shell?
Answer:
(a) Negative Q
(b) Positive Q
Explanation:
Charge is the inherent property of matter due to the transference of electrons.
There are three methods of charging a body.
(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.
(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.
(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.
(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.
(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.
The charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Reasons for change of charge on a body
Due to the process of induction the inner surface of the shell creates negative charge because when a uncharged body bring near to the charged body, the uncharged body gets the same amount of charge but opposite in sign.
While on the other hand, there is no charge interaction with the outer surface so it remains positively charge so we can conclude that the charge on the inner surface of the shell is negative whereas the charge on the outer surface of the shell is positive.
Learn more about charge here: https://brainly.com/question/18102056
What did you enjoyed about the webinar?
Explanation:
hjjdjdjjddjjdndnbbhhhydgdhgdgdvgbbb! bbbhhhhhhhhBefore the 1970's, energy prices were A. relatively low B. extremely burdensome C. incredibly high D. very problematic
Answer:
C
Explanation:
correct me if I'm wrong.
Answer:
A
Explanation:
i got it right on acellus
The Heat Force
이
18
1 point
-
If two objects are the same temperature and are physically touching which of the following would be true?
The objects would be in thermodynamic equilibrium and would transfer energy through conduction.
ОООО
1
The objects would not be in thermodynamic equilibrium and heat would transfer through conduction.
The objects would not be in thermodynamic equilibrium and as a result there would be no heat transfer
The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.
2
If two objects are the SAME temperature and are physically touching,
then
. . .
. . .
. . .
The objects would be in thermodynamic equilibrium and as a result there would be no heat transfer.
Visible matter belonging to the Milky Way Galaxy can be traced out to about 50,000 light years from the center.
a. True
b. False
Answer:
b. False
Explanation:
The visible matter that belongs to the Milky way Galaxy are traced out to be about 50 kpc distance from the center.
Kpc stands for kiloparsec. It is the unit of measurement of distance.
A parsec is[tex]$\text{ used to measure large distances}$[/tex] of the astronomical objects that lies [tex]$\text{outside the solar system}$[/tex], mainly where galaxies are involved.
1 kiloparsec is 1000 parsec and is equal to 3260 light years.
So the visible matter is about 163,078 light years away.
Hence the answer is FALSE.
a. Green light shines through a 100mm-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.
b. Green light shines through a 100-μ m-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.
Answer:
a) size of the bright spot is proportional to the hollow size
b) as the size of the hole increases, the circular point decreases.
Explanation:
a) In this case the diameter of the hole is much greater than the wavelength, as the size of the hole is many orders greater than the wavelength we are in the part of geometric optics,
Consequently the size of the bright spot is proportional to the hollow size.
Consequently the size increases
b) in this case the hole diameter d = 100 10⁻⁶m and the wavelength that for the green color is lam = 500 nm = 5 10⁻⁷ m
We see that angles are very small so the wavelength of the office is greater than the wavelength, but you can observe the effects of diffraction
d sin θ = 1.22 m λ
the numerical constant appears by solving the equation in polar coorθdinates, because the hole is circular
the first zero occurs for m = 1
sin θ = 1.22 λ / d
In these experiments the angles are small
sin θ = θ
we substitute
θ = 1.22 λ/ d
θ = 1.22 500 10⁻⁹ / 100 10⁻⁶
θ = 6.1 10⁻³
without the hole diameter increases by 20%
d’ = 1.2 d
we substitute
θ'= 1.22 λ / d'
θ’ = 1.22 λ /1.2 d
θ‘= 1.22 λ /d [tex]\frac{1}{1.22}[/tex]
θ ’= θ 0.83
θ ’= 6.1 10⁻³ 0.83
θ' = 5 10⁻³ rad
Therefore, the answer is that as the size of the hole increases, the circular point decreases.
A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact
Answer:
The speed of the bullet just before the impact is 701 m/s
Explanation:
Given;
mass of the bullet, m₁ = 20 g = 0.02 kg
mass of the block, m₂ = 2.8 kg
displacement of the block, d = 9 mm = 9 x 10⁻³ m
duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s
Apply the principle of conservation of energy;
The final kinetic energy of the bullet = maximum potential energy of the block
[tex]\frac{1}{2} m_1v^2 = m_2gh\\\\v^2 = \frac{2m_2gh}{m_1} \\\\v= \sqrt{\frac{2m_2gh}{m_1} } \\\\v = \sqrt{\frac{2 \times 2.8 \times 9.8 \times (9\times 10^{-3})}{0.02} } \\\\v = 4.97 \ m/s[/tex]
Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₁ is the initial velocity of the bullet
u₂ is the initial velocity of the block = 0
m₁u₁ + 0 = v(m₁ + m₂)
m₁u₁ = v(m₁ + m₂)
0.02u₁ = 4.97(2.8 + 0.02)
0.02u₁ = 14.02
u₁ = 14.02 / 0.02
u₁ = 701 m/s
Therefore, the speed of the bullet just before the impact is 701 m/s
Who is a socio-economically disadvantaged child? Explain any four ways forhelping such a child
Explanation:
A socio-economically disadvantaged child is the one who is disadvantaged in terms of social position and economic position. Such children have limited resources in terms of education, money and future options. Four ways of helping such children are as follows:
1) Help them in education - You can help such children by giving them free tuition. If you belong to a well off family, you can get them admitted in schools as well. Provide them with books and uniform.
2) Encourage them to do well in school and pursue their passions.
3) Provide them with meals, if they do not have access to regular meals.
4) encourage them to go school regularly.
Answer:
sorry i dont know the answer bit mark me as BRAINLISTConvert 15000kg/m cube
into gm/cm cube
please write the process also
Answer:
15000 Kilograms/Cubic Meters (kg/m3) = 15 Grams/Cubic Centimeters (g/cm3)
Explanation:
1 g/cm3 is equal to 1000 kilogram/cubic meter. To convert 100 gram into kg then divide it by 1000 i.e. 100/1000 = 0.1 kg. To convert any value of gm/cm3 into kg/m3 then multiply it by 1000.
15000 kg / m^3 =
15000 × 10^3 g / m^3 =
15000 × 10^3 × 10^3 mg / m^3 =
15 × 10^9 mg / m^3 =
15 × 10^9 × 10^(-3) mg / dm^3 =
15 × 10^9 × 10^(-3) × 10^(-3) mg / cm^3 =
15 × 10^9 × 10^(-6) mg / cm^3 =
15 × 10^( 9 - 6 ) mg / cm^3 =
15 × 10^3 mg / cm^3 =
15000 mg / cm^3 =
Look : We found the exact thing we had ...
WoW ...
We got a point ;
Remember from now on :
kg / m^3 = mg / cm^3
A single-story retail store wishes to supply all its lighting requirement with batteries charged by photovoltaic cells. The PV cells will be mounted on the horizontal rooftop. The time-averaged lighting requirement is 10 W/m2 , the annual average solar irradiance is 150 W/m2 , the PV efficiency is 10%, and the battery charging/discharging efficiency is 80%. What percentage of the roof area will the PV cells occupy
Answer:
83.33% of the roof area will be occupied by the PV cells
Explanation:
Given the data in the question;
time-averaged lighting requirement [tex]P_{lighting[/tex] = 10 W/m²
the annual average solar irradiance [tex]q_{solar[/tex] = 150 W/m²
the PV efficiency η[tex]_{pv[/tex] = 10% = 0.1
battery charging/discharging efficiency η[tex]_{battery[/tex] = 80% = 0.8
we know that; Annual average power to the light = [tex]P_{lighting[/tex] × A[tex]_{roof[/tex]
Now, the electrical power delivered by the solar cell battery system will be;
⇒ [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
[tex]P_{lighting[/tex]A[tex]_{roof[/tex] = [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
Such that;
A[tex]_{pv[/tex] = [tex]P_{lighting[/tex]A[tex]_{roof[/tex] / [tex]q_{solar[/tex] × A[tex]_{pv[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = [tex]P_{lighting[/tex] / [tex]q_{solar[/tex] × η[tex]_{pv[/tex] × η[tex]_{battery[/tex]
so we substitute
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / [ 150 W/m² × 0.1 × 0.8 ]
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 10 W/m² / 12 W/m²
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 0.8333
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = (0.8333 × 100)%
A[tex]_{pv[/tex] / A[tex]_{roof[/tex] = 83.33%
Therefore, 83.33% of the roof area will be occupied by the PV cells.