Answer:
It is a basic postulate of the theory of relativity that the speed of light is constant. This can be broken down into two parts: The speed of light is independent of the motion of the observer. The speed of light does not vary with time or place.
Explanation:
A physics student sits in a chair. The chair pushes up on the student's body. Identify the other force of the interaction force pair.
Answer:
The other force is the weight of the student.
Explanation:
With respect to Newton's third law of motion, for the student to sit and balance on the chair, there must be two equal and opposite forces involved. The student applies his/ her weight on the chair which acts downwards, while the chair applies an equal but opposite force to the weight of the student.
The force applied by the chair on the student's body is counter balanced by the student's weight. Note that, if the weight of the student is greater than the opposing force from the chair, the chair would collapse.
Name the four layers of the atmosphere (in order starting at the bottom
Answer:
Troposphere, stratosphere, mesosphere and thermosphere. The next region is the exosphere, but that region is 500+ km from the Earth's surface.
[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]
Actually Welcome to the Concept of the Atmosphere.
The four layers of Atmosphere starting from bottom are as follows:
1.) Troposphere - The troposphere is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level.
2.) Stratosphere - The next layer up is called the stratosphere. The stratosphere extends from the top of the troposphere to about 50 km (31 miles) above the ground.
3.) Mesosphere - Above the stratosphere is the mesosphere. It extends upward to a height of about 85 km (53 miles) above our planet. Most meteors burn up in the mesosphere.
4.) Thermosphere - The layer of very rare air above the mesosphere is called the thermosphere. High-energy X-rays and UV radiation from the Sun are absorbed in the thermosphere, raising its temperature to hundreds or at times thousands of degrees.
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:
One molecule of dinitrogen tetroxide contains...
A.
four nitrogen atoms and four oxygen atoms.
B.
two nitrogen atoms and four oxygen atoms.
C.
four nitrogen atoms and two oxygen atoms.
D.
two nitrogen atoms and two oxygen atoms.
Answer:
C
Explanation:
C is correct
Ok!
151617
A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Answer:
3.9 × 10^7 J
Explanation:
Given that a tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 15 m and h = 5 m.) W = 1.59 J
Solution
Since the tank is half full, the height = 2.5m
Pressure = density × gravity × height
Pressure = 900 × 9.8 × 2.5
Pressure = 22050 Pascal
The cross sectional area of the pump will be area of a circle.
A = πr^2
A = π × 15^2
A = 706.858 m^2
Using the formula
Density = mass/volume
Mass = density × volume
Mass = 900 × 706.86 × 2.5
Mass = 1590.435
Energy = mgh
Energy = 1590.435 × 9.8 × 2.5
Energy = 38965657.8 J
Since the work done = energy
Therefore, the work done = 3.9 × 10^7 J
PLEASE CLICK ON THIS IMAGE I NEED HELP
Answer:
Second option
Explanation:
"Uniform" pretty much means the same thing happens.
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 wave with a frequency of 5Hz travels a distance of 40mm in 2 seconds.What is the speed of the wave
Answer:
20mm per second
Explanation:
4. When you are holding a book, energy is stored between the book and the Earth.
This type of energy is called
potential energy.
A. Elastic potential energy
B. Chemical potential energy
C. Gravitational potential energy
D. Kinetic energy
Answer:
gravitational potential energy
why type of volcano is built almost entirely from ejected lava fragments
Answer:
Shield volcanoes
Explanation:
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
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
Select the correct answer.
A pair of pliers is an example of a simple machine with two levers. Which part of the pliers is the fulcrum?
A. A
B. B
C. C
D. D
E. E
Answer:
There isnt enough in your question to answer the question bro, like we need a picture or something bro.
Explanation:
You don't have a image attached
a car moved 120km to the north. what is its displacement?
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
****PLEASE HELP**** WILL MARK BRAINLIEST
Assuming that voltage remains constant, what happens to the current in a
wire if the length of the wire increases?
O A. The current decreases.
OB. The current alternates between high and low values.
O C. The current increases.
O D. The current is not affected by a change in wire length.
Answer:
The Current decreases
Explanation:
HOPE THIS HELPS!
3. If a spring extends by 3 cm when a 4 N weight is suspended from it, find the extension
when the weight is changed to
(a) 8 N
(b) 10 N
(c) 14 N
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
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.
Who watching all star draft? Luka better get picked first ong
We intend to measure the open-loop gain (LaTeX: A_{open}A o p e n ) of an actual operational amplifier. The magnitude of LaTeX: A_{open}A o p e n is in the range of 106 V/V. However, the signal generator in measurement setup can supply minimal voltage of 1 mV, and the oscilloscope used at amplifier output can measure maximal voltage level of 10 V. Can you design a simple measurement setup using this signal generator and oscilloscope, and accurately measure the LaTeX: A_{open}A o p e n
Answer:
voltage divider, R₂ = 1000 R₁
measuring the output in the resistance R₁
Explanation:
Let's analyze the situation, in an op amp in open gain loop, the gain is maximum G = 10⁶ V / V
in this case the signal generator gives a minimum wave of 1 10⁻³ V, after passing through the amplified it becomes 10³ V which saturates the oscilloscope.
To solve this problem we must use a simple voltage divider, for this we use the fact that in a series circuit the voltage is the sum of the voltages of each element.
If we use two resistors whose relationship is
R₂ / R₁ = 10³
R₂ = 1000 R₁
When measuring the output in the resistance R₁ we have the desired divider, with a tolerance range, for the minimum output of the generator (1 10⁻³V) we have a reading of V = 1 V in the oscilloscope, for which we can use voltage up to 10V on the generator
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]
A 50 kg mass is sitting on a frictionless surface. An unknown constant force called force A pushes the mass for 2 seconds until the mass reaches a velocity of 3 m/s. If the 50 kg mass is now pushed by an unknown force B and reaches the velocity of 3 m/s in 4 seconds, compare the impulse delivered to the mass when acted upon by force A with the impulse delivered to the mass when acted on by force B? *
A) The impulse delivered to the mass when acted upon by force A is greater
B) The impulse delivered to the mass when acted upon by force B is greater
C) The impulse is the same in each case
D) We need to know the value of force A and force B in order to determine this
Answer:
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When individuals improve their aerobic endurance, which body systems are affected?
When individuals improve their aerobic endurance, which body systems are affected?
Anwser: your heart and lungs
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
An elevator filled with passengers has a mass of 1663 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 for 3.25 s. Calculate the tension in the cable (in N) supporting the elevator. 18317 Correct: Your answer is correct. N (b) The elevator continues upward at constant velocity for 8.92 s. What is the tension in the cable (in N) during this time
Answer:
(a) T = 18309.63 N = 18.31 KN
(b) T = 16314.03 N = 16.314 KN
Explanation:
(a)
The tension in an elevator while moving upward with some acceleration is given by the following formula:
[tex]T = m(g+a)\\[/tex]
where,
T = Tension = ?
m = mass = 1663 kg
g = acceleration due to gravity = 9.81 m/s²
a = acceleration of elevator = 1.2 m/s²
Therefore,
[tex]T = (1663\ kg)(9.81\ m/s^2 + 1.2\ m/s^2)\\[/tex]
T = 18309.63 N = 18.31 KN
(b)
Constant velocity means no acceleration. So, in that case, the tension will be equal to the weight of the elevator:
[tex]T = mg\\T = (1663\ kg)(9.81\ m/s^2)\\[/tex]
T = 16314.03 N = 16.314 KN
A 10 kg object moving to the left collides with and sticks to a 3 kg object moving to the right. Which of the following is true of the motion of the combined objects immediately after the collision?
Answer:
"Cannot be determined," if that's an answer choice. It depends on the velocities of both objects, since momentum=mass*velocity.
Explanation:
The earth rotates through one complete revolution every 24 hours. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 24 hours. Find the angular and linear velocity of a person standing on the equator. The radius of earth is approximately 4000 miles.
Answer:
ω = 7.27 x 10⁻⁵ rad/s
v = 467.99 m/s
Explanation:
First, we will find the angular velocity of the person:
[tex]Angular\ Velocity = \omega = \frac{Angular\ Distance}{Time}[/tex]
Angular distance covered = 1 rotation = 2π radians
Time = (24 h)(3600 s/ 1 h) = 86400 s
Therefore,
[tex]\omega = \frac{2\pi\ rad}{86400\ s}[/tex]
ω = 7.27 x 10⁻⁵ rad/s
Now, for the linear velocity:
[tex]v = r\omega[/tex]
where,
v = linear velocity = ?
r = radius of earth = (4000 miles)(1609.34 m/1 mile) = 6437360 m
Therefore,
[tex]v = (6437360\ m)(7.27\ x\ 10^{-5}\ rad/s)[/tex]
v = 467.99 m/s
1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?
Answer:
2km
Explanation:
Given data
We are told that the direction traveled are
North>>>East>>>South
Hence the displacement is defined as the distance away from the initial position is
Initial position =18km
FInal position = 16km
The displacement = 18-16= 2km
Hence the displacement is 2km
Which word best completes the sentence?
Select the word from the drop-down menu
He is quite
Choose...
despite never having left his smalL TOWEN
Answer:
it’s cosmopolitan
Explanation:
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