Answer:
i need help with the same question
Explanation:
Astronauts living on the International Space Station have zero gravitational acceleration.
True
False
I'm asking for a quick favor. I'm trying to understand an equation that has to do with Projectile motion, Bernoulli's principle, and Magnus Effect. Basically focused on understanding air resistance on a projectile. I would like to discuss this privately rather than have it on this public forum. I'll give you 100 of my points if you help.
Explanation:
Projectile motion, Bernoulli's principle, and Magnus Effect.
Sure I would be happy to discuss projectile motion!
I'll do it if you mark brainliest :) I need the points thanks
Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to
A. Paul Costa
B. Hans Eysenck
C. Gordon Allport
D. Robert McCrae
The answer is b I just took the test
Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.
What is Hans Eysenck's theory?Eysenck's theory of personality is based on three logical attributes namely:
a) Introversion vs. extroversion – Extroversion leads to sociable life , while introversion causes need of being alone and limited interactions
b) Neuroticism vs. stability – Neuroticism leads to anxiousness and an overactive sympathetic nervous system while stability leads to emotional stability.
c) Nsychoticism vs. socialization. - psychoticism leads to independent thinking, and hostility. While socialization leads to co-operative and conventional behaviour.
Therefore Neuroticism, psychoticism, and extraversion are the three dominant personality traits according to Hans Eysenck.
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During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of 112 N/m. If the hose is stretched by 4.70 m and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length
Answer:
1237 J
Explanation:
The work done by the hose on the balloon is the work done by a spring which is
W = 1/2k(x₀² - x₁²) where k = spring constant = 112 N/m, x₀ = 4.70 m and x₁ = 0 m.
Substituting the values of the variables into the equation, we have
W = 1/2k(x₀² - x₁²)
W = 1/2 × 112 N/m((4.70 m)² - (0 m)²)
W = 56 N/m(22.09 m² - 0 m²)
W = 56 N/m(22.09 m²)
W = 1237.04 J
W ≅ 1237 J
In 1976, a 53 kg helicopter was built in Denmark. Suppose this helicopter flew east with a speed of 60.0 m/s and the total momentum of the helicopter and pilot was 7,200 kg∙m/s to the east. What was the mass of the pilot?
Answer:
67 kg.
Explanation:
From the question,
Total momentum = (mass of the pilot+ mass of the helicopter)×velocity.
I = (m+M)v............................... Equation 1
Where I = Total momentum of the pilot and the helicopter, m = mass of the pilot, M = mass of the helicopter, v = velocity.
make m the subject of the equation
m = (I/v)-M.............................. Equation 2
Given: I = 7200 kg.m/s, v = 60 m/s, M = 53 kg.
Substitute these values into equation 2
m = (7200/60)-53
m = 120-53
m = 67 kg.
Hence the mass of the pilot is 67 kg
If the hoop has speed vo at the bottom of
the pipe, what is its speed when it has rolled
halfway up the side of the pipe?
The speed of the hoop when it has rolled halfway up the side of the pipe is √(v₀² - gR).
Conservation of energyThe speed of the hoop when it has rolled halfway up the side of the pipe is calculated as follows;
K.E = P.E
- ¹/₂mv₀² + ¹/₂Iω² = (mgh₀ - mghf)
- ¹/₂mv₀² + ¹/₂Iω² = (0 - 0.5mgh) (hf = 0.5h) (half way up)
¹/₂Iω² = ¹/₂mv₀² - 0.5mgh
where;
I is moment of inertia of the hoop = mr²ω is angular speed = v/r¹/₂(mr²)(vf/r)² = ¹/₂mv₀² - 0.5mgh
¹/₂vf² = ¹/₂v₀² - ¹/₂gh
vf² = v₀² - gh
vf = √(v₀² - gh)
where;
h is the distance traveled half-way up the pipe = Rvf = √(v₀² - gR)
[tex]v_f = \sqrt{v_0^2 - gR}[/tex]
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Ultraviolet waves are transmitted by ....... ?
Answer:
Sunlight is the greatest source of UV radiation. Man-made ultraviolet sources include several types of UV lamps, arc welding, and mercury vapour lamps.
Explanation:
A 20g bullet moving at 200m/s hits a bag of sand and comes to rest in 0.011s, calculate the momentum of the bullet just before hitting the bag
Answer:
momentum = mass * velocity ... kg•m/s
acceleration = v / t = 200 m/s / 11 ms ... m/s^2
force = mass * acceleration = .02 kg * (200 m/s / 11 ms) N
Explanation:
Pleaseee I need help!!
Answer:
R1=3.333 Ohms
R2=10 Ohms
R3=16.666 Ohms
Explanation:
30 total
30=R1+R2R3
30=1(x):3(x):5(x)
x=3.3333
R1=3.333 Ohms
R2=10 Ohms
R3=16.666 Ohms
A 7.80 g bullet has a speed of 620 m/s when it hits a target, causing the target to move 6.30 cm in the direction of the bullet's velocity before stopping. (a) Use work and energy considerations to find the average force (in N) that stops the bullet. (Enter the magnitude.) N (b) Assuming the force is constant, determine how much time elapses (in s) between the moment the bullet strikes the target and the moment it stops moving. s
Answer:
[tex]23796.19\ \text{N}[/tex]
[tex]0.0002032\ \text{s}[/tex]
Explanation:
F = Force
s = Displacement = 6.3 cm
m = Mass of bullet = 7.8 g
v = Velocity of bullet = 620 m/s
t = Time taken
Work done is given by
[tex]W=Fs[/tex]
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2[/tex]
Using work energy considerations we get
[tex]Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}[/tex]
The average force that stops the bullet is [tex]23796.19\ \text{N}[/tex].
Force is given by
[tex]F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}[/tex]
The time taken to stop the bullet is [tex]0.0002032\ \text{s}[/tex].
A vertical piston cylinder assembly contains 10.0kg of a saturated liquid-vapor water mixture with initial quality of 0.85 The water receives energy by heat transfer until the temperature reaches 320*C. The piston has a mass of 204kg and area of 0.005m2. Atmospheric pressure of 100kPa acts on the top side of the piston. Local gravitational acceleration is 9.81m/s2 Calculate the amount of heat transfer between the water and the surroundings in kJ. Enter a numeric value only. 6735.66
Answer:
Explanation:
From the given information:
At state 1:
Initial Quality [tex]= x_1 = 0.85[/tex]
mass = 10.0 kg
At state 2:
Temperature [tex]T_2 = 320^0[/tex]
mass of the piston [tex]m_p = 204 \ kg[/tex]
area of the piston [tex]A_p = 0.00 5 \ m^2[/tex]
Atmospheric pressure [tex]P_{atm}= 100 \ kPa = 100 \times 10^3 \ Pa[/tex]
Gravitational acceleration = 9.81 m/s²
[tex]\mathbf{P= P_1=P_2}[/tex], This is because there exists no restriction to the movement of the piston and provided the process is frictionless. So, the process 1-2 is regarded as constant.
To calculate the applying force balance over the piston by using force balance in the vertical direction:
[tex]\mathbf{P_{AP} = P_{atmA_p} + m_pg}[/tex]
∴
(100 × 10³)×0.005 + 204 × 9.31 = P × 0.05
P = 500248 Pa
P = 500.25 kPa
At state 1:
[tex]\mathbf{P_1 = P = 500.25 \ kPa}[/tex]
[tex]x_1 = 0.85[/tex]
Hence, this is a saturated mixture of liquid and vapor
Using the steam tables at 500.25 kPa
[tex]V_f = 1.093 \times 10^{-3} \ m^3/kg \\ \\ V_g = 0.375 \ m^3/kg \\ \\ U_f = 639.72 \ kJ/kg \\ \\ U_g = 2560.72 \ kJ/kg[/tex]
∴
Specific volume at state 1 is given as:
[tex]V_1 = [ V_f +x_1(v_g -v_f) ] \ at \ 500.25 \ kPa \\ \\ V_1 = 0.319 \ m^3/kg[/tex]
volume at state 1 is given by:
[tex]V_1 = mV_1 = 10 \times 0.319 \\ \\ V_1 = 3.19 \ m^3[/tex]
Similarly, the specific internal energy is:
[tex]U_1 = [U_f +x_1 (U_o-Uf)] \ at \ 500.25 \ kPa[/tex]
[tex]U_1 = 639.72 +0.82 (2560.72 -639.72)[/tex]
[tex]U_1 = 2272.57 \ kJ/kg[/tex]
At state 2:
[tex]P = P_1 = P_2 = 500.25 \ kPa \\ \\ T_2 = 320^0 \ C[/tex]
Using steam tables at P = 500.25 kPa and T = 320° C
[tex]V_2 = 0.541 \ m^3/kg \\ \\ U_2 = 2835.08 \ kJ/kg[/tex]
∴
[tex]V_2 = mV-2 = 10 \times V_2 = 5.41 \ m^3[/tex]
[tex]\text{Now; Applying the 1st law of thermodynamics to the system}[/tex]
[tex]_1Q_2 -_1W_2 = \Delta V =m(u_2-u_1) \\ \\ where;\ _1W_2 = P(V_2-V_1) \\ \\ _1Q_2 -P(V_2-V_1) = m(u_2-u_1) \\ \\ _1Q_2 - 500.25(5.91 -3.19) = 10( 2835.08 -2272.57) \\ \\ \mathbf{ _1Q_2 = 6735.66 \ kJ}[/tex]
What is the magnitude of the electric field at the point (8.70 - 9.10 7.20 ) m if the electric potential is given by V
This question is incomplete, the complete question is;
What is the magnitude of the electric field at the point (8.70i - 9.10j + 7.20k ) m if the electric potential is given by V = 4.30xyz², where V is in volts and x, y and z are in meters.
Answer:
the magnitude of the electric field is 5648.67 N/C
Explanation:
Given the data in the question;
Suppose electric field at this point is;
E = Exi + Eyj + Ezk
Now, electric field is given by;
E = -dV/dr
so,
Ex = -d( 4.30xyz² )/dx = -4.30yz²
Ey = -d( 4.30xyz² )/dy = -4.30xz²
Ez = -d( 4.30xyz² )/dz = -8.60xyz
so
E = -4.30yz² i - 4.30xz² j - 8.60xyz k
now, at the point (8.70i - 9.10j + 7.20k )
E = (-4.30(-9.10)(7.20)²) i + (-4.30(8.70)(7.20)²) j + (-8.60(8.70)(-9.10)(7.20) k
E = 2028.499 i - 1939.334 j + 4902.206 k
so, Magnitude of electric field will be;
|E| = √( Ex² + Ey² + Ez² )
we substitute
|E| = √( (2028.499)² + (-1939.334)² + (4902.206)² )
|E| = √( 31907448.222993 )
|E| = 5648.67 N/C
Therefore, the magnitude of the electric field is 5648.67 N/C
What is the name of the theory describing how the lithosphere is broken into segments, or plates, which "float" on the asthenosphere, and associates the interactions between these plates with earthquakes and volcanic activity and the formation of mid-ocean ridges, trenches, mountains, and chains of volcanic islands?
A. Conservation of Matter
B. Big Bang Theory
C. Plate Tectonics
D. Theory of Evolution
Answer:
C. Plate Tectonics
Explanation:
The theory of plate tectonics is when the lithosphere is separated into plates. These plates move over or float over the asthenosphere. The movement of these plates cause earthquakes and can interact with the volcanic activity.
what is tangential velocity
Answer:
Tangential velocity is the linear speed of any object moving along a circular path
Explanation:
Answer:
Tangential velocity is the linear speed of any object moving along a circular path. A point on the outside edge of a turntable moves a greater distance in one complete rotation than a point near to the center.
Explanation:
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an egg is Free Falling Down from a nest in a tree (neglect air resistance)
We are to calculate the acceleration.
Answer:
-9.8 m/s²
Explanation:
Since the egg is in a free fall, it means that the force due to gravity will be equal to the normal force.
Now,
Force due to gravity: F_g = mg
Normal force; F_n = ma
Thus;
mg = ma
m will cancel out to get
g = a
Since it is a free fall motion, then gravity is negative;
-g = a
g has a constant value of 9.8 m/s². Thus;
a = -9.8 m/s²
The amplitude of a wave
determines the volume of a
sound.
True
O False
A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.
Question:
How far north and east is B from A?
Find the displacement of the turtle from the origin at the point.
Answer:
5cm east& 1cm west from A
Explanation:
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The total distance travelled by bug is 19 cm while the displacement of the bug is 5.1 cm.
Given here,
A bug stat to crawl from point A
8 cm to east
5 cm to south
3.0 to west
4.0 cm north reach the point B,
Displacement:
It is the shortest distance between two point or the length of strait line between two points.
Therefore, the total distance travelled by bug is 19 cm while the displacement of the bug is 5.1 cm.
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At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 55 kg, 90 kg, and 42.5 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vC = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming car A hits car B before car C does. Assume positive sign denoting forward motion and negative sign denoting backward motion.
Answer:
Vb = 0.334 m/s
Va = -1.265 m/s
Vc = 1.424 m/s
Explanation:
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Initial momentum = 255(2) – 242.5(1.5) = 146.25
Final momentum = 255Va + 290Vb + 242.5 Vc = 146.25
Vb - Va = 0.8(2) = 1.6
Vc - Vb = 0.8(1.5) = 1.2
Va = Vb -1.6
Vc = Vb + 1.2
255(Vb -1.6) + 290Vb + 242.5(Vb + 1.2) = 146.25
255 Vb – 408 + 290 Vb + 242.5 Vb + 291 = 146.25
787.5 Vb = 263.25
Vb = 0.334 m/s
Va = Vb -1.6 = 0.334 – 1.6 = -1.265 m/s
Vc = Vb + 1.2 = 0.224 + 1.2 = 1.424 m/s
In a mass spectrometer, a specific velocity can be selected from a distribution by injecting charged particles between a set of plates with a constant electric field between them and a magnetic field across them (perpendicular to the direction of particle travel). If the fields are tuned exactly right, only particles of a specific velocity will pass through this region undeflected. Consider such a velocity selector in a mass spectrometer with a 0.105 T magnetic field.
Part (a) What electric field strength, in volts per meter, is needed to select a speed of 3.8 × 106 m/s?
Part (b) What is the voltage, in kilovolts, between the plates if they are separated by 0.75 cm?
Answer:
a).[tex]$3.99 \times 10^5 \ v/m$[/tex]
b). 2.9925 kV
Explanation:
Given :
For mass spectrometer
The magnetic field = B
B = 0.105 T
a). Given speed, v = [tex]$3.8 \times 10^6 \ m/s$[/tex]
We known
[tex]$\frac{E}{B}=v$[/tex]
∴ [tex]$E= 3.8 \times 10^6 \times 0.105$[/tex]
[tex]$=3.99 \times 10^5 \ v/m$[/tex]
b). Now spectrometer, d = 0.75 cm
[tex]$d=0.75 \times 10^{-2} \ m$[/tex]
We known
[tex]$E=\frac{V}{d}$[/tex]
[tex]$V = E\times d$[/tex]
[tex]$V = 3.99 \times 10^5 \times 0.75 \times 10^{-2}$[/tex]
[tex]$V = 2.9925 \times 10^3 \ V$[/tex]
= 2.9925 kV
what kind of weather would u assosiate with higher perssure system? How does the density, humidity and air motion compare to that of low pressure system?
Answer:
Low pressure systems typically arrive with storms and clouds. Air motion is usually upwards, as heated are is less dense and more buoyant than cooler air. A high pressure system is typically cooler than its counter-part, and skies are usually clear. Low pressure systems carry more water vapor due to rising hot air cooling and condensing.
An object which is dropped from a certain height has zero (0) initial velocity.
Answer:
0 m/s
Explanation:
if an object is dropped we know the initial velocity is zero when in free fall
Your teacher then challenges you to use the items in the envelope to create a model of a carbon-14 atom. Before you begin, you and your
group brainstorm ideas for creating your model. Your group suggests the following steps:
Step
Procedure
1
use 6 red circles for the nucleus of the atom
2
use 14 red circles for the nucleus of the atom
3
add 6 blue circles to the nucleus of the atom
4
add 8 blue circles to the nucleus of the atom
5
place 14 yellow circles around the nucleus of the atom
6
place 6 yellow circles around the nucleus of the atom
Exhibits
Which of the following steps should you follow to create a correct model of a Carbon-14 atom?
O A. steps 1,4,6
O B. steps 1, 3,5
o c. steps 2, 3, 6
O D. steps 2,4,5
O
08:11
Answer:
Look it up
Explanation:
We need to use 6 red circles as well as 8 blue circles to represent the 6 protons and 8 neutrons in the nucleus, after which we need to place 6 yellow circles around the nucleus to represent the 6 electrons in the atom. is required. Thus, the correct option is A.
How to create a Carbon-14 atom?To make an accurate model of a carbon-14 atom, we need to follow the steps that represent the correct number of protons, neutrons, and electrons in a carbon-14 atom.
Carbon-14 has 6 protons and 8 neutrons, so we will use a total of 6 red and 8 blue circles around the nucleus which also has 6 electrons, so we need to put 6 yellow circles around the nucleus.
The correct steps to follow to create a model of a Carbon-14 atom are:
Step 1: Use 6 red circles for the nucleus of the atom
Step 4: Add 8 blue circles to the nucleus of the atom
Step 6: Place 6 yellow circles around the nucleus of the atom
Thus, the correct option is A.
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The chemical equation for the decomposition of potassium chlorate into potassium chloride and oxygen gas is
KCIO: |_ KCI + ___ 02
Which coefficients correctly balances the equation?
A) 4.4,3
B 3,3,2
C) 2,2,3
D
2, 1,3
E The equation is already balanced.
Answer:
Option C. 2, 2, 3
Explanation:
__KClO₃ —> __ KCl + __O₂
The above equation can be balance as illustrated below:
KClO₃ —> KCl + O₂
There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by writing 2 before KClO₃ and 3 before O₂ as shown below:
2KClO₃ —> KCl + 3O₂
There are 2 atoms of K on the left side and 1 atom on the right side. It can be balance by writing 2 before KCl as shown below:
2KClO₃ —> 2KCl + 3O₂
Thus, the equation is balanced. The coefficients are: 2, 2, 3
You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a constant speed of 1.6 m/s. The rope makes an angle of 35 degrees with the horizontal. What is the net force on the block
Answer:
Fnet = 0
Explanation:
Since the block slides across the floor at constant speed, this means that it's not accelerated.According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:[tex]F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)[/tex]
In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:[tex]F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)[/tex]
⇒ 169 N + Fn = Fg = 216 N (3)
This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:Fn = 216 N - 169 N = 47 N (4)A 125 kg mail bag hangs by a vertical rope 3.3 m long. A postal worker then displaces the bag to a position 2.2 m sideways from its original position, always keeping the rope taut.
1) What horizontal force is necessary to hold the bag in the new position?
2) As the bag is moved to this position, how much work is done by the rope?
3) As the bag is moved to this position, how much work is done by the worker?
Answer:
1) the required horizontal force F is 1095.6 N
2) W = 0 J { work done by rope will be 0 since tension perpendicular }
3) work is done by the worker is 1029.4 J
Explanation:
Given that;
mass of bag m = 125 kg
length of rope [tex]l[/tex] = 3.3 m
displacement of bag d = 2.2 m
1) What horizontal force is necessary to hold the bag in the new position?
from the figure below; ( triangle )
SOH CAH TOA
sin = opp / hyp
sin[tex]\theta[/tex] = d / [tex]l[/tex]
sin[tex]\theta[/tex] = 2.2/ 3.3
sin[tex]\theta[/tex] = 0.6666
[tex]\theta[/tex] = sin⁻¹ ( 0.6666 )
[tex]\theta[/tex] = 41.81°
Now, tension in the string is resolved into components as illustrated in the image below;
Tsin[tex]\theta[/tex] = F
Tcos[tex]\theta[/tex] = mg
so
Tsin[tex]\theta[/tex] / Tcos[tex]\theta[/tex] = F / mg
sin[tex]\theta[/tex] / cos[tex]\theta[/tex] = F / mg
we know that; tangent = sine/cosine
so
tan[tex]\theta[/tex] = F / mg
F = mg tan[tex]\theta[/tex]
we substitute
Horizontal force F = (125kg)( 9.8 m/s²) tan( 41.81° )
F = 1225 × 0.8944
F = 1095.6 N
Therefore, the required horizontal force F is 1095.6 N
2) As the bag is moved to this position, how much work is done by the rope?
Tension in the rope and displacement of mass are perpendicular,
so, work done will be;
W = Tdcos90°
W = Td × 0
W = 0 J { work done by rope will be 0 since tension perpendicular }
3) As the bag is moved to this position, how much work is done by the worker
from the diagram in the image below;
SOH CAH TOA
cos = adj / hyp
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex]
we substitute
cos[tex]\theta[/tex] = ([tex]l[/tex] - h) / [tex]l[/tex] = 1 - h/[tex]l[/tex]
cos[tex]\theta[/tex] = 1 - h/[tex]l[/tex]
h/[tex]l[/tex] = 1 - cos[tex]\theta[/tex]
h = [tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
now, work done by the worker against gravity will be;
W = mgh = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
W = mf[tex]l[/tex]( 1 - cos[tex]\theta[/tex] )
we substitute
W = (125 kg)((9.8 m/s²)(3.3 m)( 1 - cos41.81° )
W = 4042.5 × ( 1 - 0.745359 )
W = 4042.5 × 0.254641
W = 1029.4 J
Therefore, work is done by the worker is 1029.4 J
A cyclist exerts a 15.0 N force while riding 251 m in 30.0 s. What power does the cyclist develop?
Answer:
P=126W
Explanation:
Sorry if im wrong!
Answer:
125.5 watts
Explanation:
P=work/time
work=F*d
P=(F*d)/t
P=(15*251)/30
P=125.5 watts
Why might video games increase creativity while the use of cell phones, the internet, or computers do not?
Explanation:
Video games are developed around a structure that is unique from your usual media, where there is a sense purpose in mind when playing a game, as decided by the player, and it allows them to explore creative options in order to solve scenarios, depending on the genre. In the use of phones, internet, or computers, this structure is more rare and diverse from video games, which does not lean more toward a creative purpose to build from. That idea gives people the inspiration and discover skills they never knew they even had.
HELP - What are some good topics to add to a Pipeline presentation? So yes, a presentation about pipelines, like for example in the US. Have any ideas??
In a thunderstorm at 32.0°C, Reginald sees a bolt of lightning and hears the thunderclap 2.00s later. How far from Reginald did the lightning strike? The speed of sound through air at 32.0°C is 350.2 m/s. Show your work.
PLEASE HELP, THANKS!
Answer:
d = 700.4 m
Explanation:
Given that,
The speed of sound through air at 32.0°C is 350.2 m/s.
Reginald sees a bolt of lightning and hears the thunderclap 2.00s later.
We need to find how far from Reginald did the lightning strike. Let the distance be d. So,
Speed = distance/time
d = vt
So,
d = 350.2× 2
d = 700.4 m
So, the required distance is 700.4 m.
A 40-pF capacitor is charged to a potential difference of 500 V. Its terminals are then connected to those of an uncharged 10-pF capacitor. Calculate: (a) the original charge on the 40-pF capacitor; (b) the charge on each capacitor after the connection is made; and (c) the potential difference across the plates of each capacitor after the connection.
(a) The original charge on the 40-pF capacitor is [tex]2 .0 \ \times \ 10^{-8} \ C[/tex].
(b) The charge on each capacitor after the connection is made is [tex]4 .0 \ \times \ 10^{-9} \ C[/tex].
(c) The potential difference across the plates of each capacitor after the connection is 100 V and 400 V.
Original charge of the capacitorThe original charge on the 40-pF capacitor is calculated as follows;
[tex]Q = CV\\\\Q = 40 \times 10^{-12} \times 500\\\\Q = 2 .0 \ \times \ 10^{-8} \ C[/tex]
Charge on each capacitor[tex]C = \frac{C_1C_2}{C_1 + C_2} \\\\C = \frac{10 \times 10^{-12} \times 40 \times 10^{-12}}{10\times 10^{-12} \ + \ 40 \times 10^{-12}} \\\\C = 8 \times 10^{-12} \ F[/tex]
[tex]Q = Q_1 = Q_2\\\\Q = 8 \times 10^{-12} \ \times \ 500\\\\Q = 4 \times 10^{-9} \ C[/tex]
Potential differenceThe potential difference across the plates of each capacitor after the connection is calculated as follows;
[tex]V = \frac{Q}{C} \\\\V_1 = \frac{Q}{C_1} \\\\V_1 = \frac{4 \times 10^{-9}}{40 \times 10^{-12}} \\\\V_1 = 100 \ V\\\\V_2 = \frac{Q}{C_2} \\\\V_2 = \frac{4 \times 10^{-9}}{10 \times 10^{-12} } \\\\V_2 = 400 \ V[/tex]
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