Answer:
An isothermal process (change)
Explanation:
The initial volume of the gas, V₁ = 15 L
The initial volume of the gas, V₂ = 39 L
The amount of work it performs, W = 1.8 kJ
For the above change, the initial and final pressure of the gas can be calculated, using Boyle's law which is based on an isothermal change process as follows;
P₁·V₁ = P₂·V₂
P ∝ 1/T
What are examples of water on Earth that are part of the water cycle
The coefficients of static and kinetic friction between a 50.0-kg box and a horizontal surface are 0.500 and 0.400 respectively. (a) What is the acceleration of the object if a 250-N horizontal force is applied to the box
Answer:
1.08 m/s^2
Explanation:
The maximum value of static friction F_s = μ_s mg
where μ_s = coefficient of static friction
m= mass of box= 50kg
g= 9.81 m/s^2 , acceleration due to gravity
F_s = 0.5×50×9.81 =245 N
Since, the applied force is greater than the maximum value of static friction F_s, the object must be in motion. Hence, kinetic friction must be taken into account and not static friction.
By equilibrium condition
ma = F - μ_k m g
where F = 250 N, μ_k = coefficient of kinetic friction = 0.4, m= 50 Kg
a = acceleration of the box
50a = 250-0.4×50×9.81
Solving we get a= 1.08 m/s^2
This is one popular brand of exercise machine for a crossword puzzle
Answer:
Aerobics I think.
Explanation:
After supper, your mother runs the warm pan under cold water. The pan cools off quickly. This is an example of -
conduction
convection
radiation
Answer:
conduction (the heat is transferring to the air)
In LEDs, the longer leg (wire) is always
connected to the ____ terminal
a) negative
b) neutral
c) positive
d) Any terminal
Positive
In an LED, the longer lead is always connected to the positive terminal
3. Tabulate the difference between a Static friction and Kinetic friction.
Answer:
Static friction is what keeps the box from moving without being pushed, and it must be overcome with a sufficient opposing force before the box will move. Kinetic friction is the force that resists the relative movement of the surfaces once they're in motion.
a toy of mass 600 is whirled by a child in a horizontal circle using a string of length 2m with a linear speed of 5 m/s determine the angular velocity of the toy?
Explanation:
angular velocity = velocity/radius
= 5/2
= 2.5 rad/s
A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1) Calculate the final velocity of the 2.00-kg ball. (Express your answer to three significant figures.)
Answer:
The final velocity of the 2kg ball is 1.270 m/s
Explanation:
According to Newton's second and third laws of motion
Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".
Newton's third law state that "for every action, there must be an equal and opposite reaction".
The combinations of these two laws resulted in an elastic collision
Given that:
m1 = 2kg
u1 = 2.20m/s
m2 = 4.00kg
u2 = 0m/s
An Elastic collision is when kinetic energy before = kinetic energy after
E.K before = [tex]1/2mv^{2}[/tex]
E.K before = 1/2 * 2 * (2.20)^2
E.K = 1/2 * 2 * 4.84
E.K before = 4.84j
E.K after = 1/2 x (4 + 2)v^2
E.K after = 1/2(6v^2)
E.K after = 3v^2
Since E.K before = E.K after
4.84 = 3v^2
Divide through by 3
4.84/3 = 3v^2/3
1.6133 = v^2
[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]
What impulse occurs when a cart that is originally at rest experiences an average force of
N for 2.5 s? *
(10 Points)
25 N
25 Nm
25 Ns
25 kg m/s
Explanation:
What impulse occurs when a cart that is originally at rest experiences an average force of
N for 2.5 s? *
(10 Points)
25 N
25 Nm
25 Ns
25 kg m/s
As a part of a project to construct a Rube Goldberg machine, a student wishes to construct a spring-loaded cube launcher. For examples of Rube Goldberg machines, see the beautiful Cog video or the Mythbusters Happy Holidays Rube Machine. Our student plans to trigger the release of the spring with a falling ball. The small cube to be launched has a mass of 119.0 grams. The available spring is a very stiff but light spring that has a spring constant of 461.0 N/m (the spring is so light that it may be considered massless). The spring will be mounted at the bottom of a wooden ramp which makes an angle of 50.0 degrees with the horizontal; the length of the spring is parallel to the ramp with the lower end of the spring fixed in place. For simplicity, assume that the part of the wooden ramp which is underneath the spring is highly polished and very slick; you may assume no friction on the cube by the ramp when the cube is moving on this portion of the ramp. For the rest of the wooden ramp, the coefficients of friction between the ramp surface and the cube surface are 0.590 for static friction and 0.470 for kinetic friction. Measured from the equilibrium position of the free end of the mounted spring, the distance to the top of the ramp is 17.0 cm (this is measured along the ramp) The plan is to compress the spring and maintain the compression with a simple trigger, which will be released by a falling ball. The cube will be at rest on the free end of the compressed spring. When the trigger is released, the spring will expand, shooting the cube up the ramp. The desired speed for the cube when it reaches the top of the ramp (where the cube will leave the ramp as a projectile to continue the Rube Goldberg sequence) is 45.0 cm/s. Your ultimate task will be to calculate the necessary amount of compression for the spring. The Spring is Compressed by 5.90 cm Before attempting the ultimate task, answer the following questions (Parts A-B) for the case in which the spring is compressed by 5.90 cm and then released, shooting the cube up the ramp and into the air.
A) Find the work done on the cube by the spring during the time that the cube is in contact with the spring.
B) What is the speed of the cube at the the instant just before the sliding cube leaves the ramp?
Answer:
1. The work done on the cube during the time the cube is in contact with the spring is 0.8023705 J
2. The speed of the cube at the instant just before the sliding cube leaves the ramp is approximately 31.5 cm/s
Explanation:
The given parameters of the Rube Goldberg machine are;
The distance from the free end of the spring to the top of the ramp, d = 17.0 cm = 0.17 m
The mass of the small cube to be launched, m = 119.0 g = 0.119 kg
The spring constant of the spring, k = 461.0 N/m
The angle of elevation of the ramp to the horizontal, θ = 50.0°
The coefficient of static friction of the wood, [tex]\mu_s[/tex] = 0.590
The coefficient of dynamic friction of the wood, [tex]\mu_k[/tex] = 0.470
The velocity of the cube at the top of the ramp, v = 45.0 cm/s = 0.45 m/s
The amount by which the cube is compressed, x = 5.90 cm = 0.059 m
The work done on the cube during the time the cube is in contact with the spring = The energy of the spring, E = (1/2)·k·x²
∴ E = (1/2) × 461.0 N/m × (0.059 m)² = 0.8023705 J
The work done on the cube during the time the cube is in contact with the spring= E = 0.8023705 J
2. The frictional force, [tex]F_f[/tex] = [tex]\mu_k[/tex]·m·g·cos(θ)
∴ [tex]F_f[/tex] = 0.470 × 0.119 × 9.8 × cos(50) ≈ 0.35232 N
The work loss to friction, W = [tex]F_f[/tex] × d
∴ W = 0.35232 N × 0.17 m ≈ 0.05989 J
The work lost to friction, W ≈ 0.05989 J
The potential energy of the cube at the top of the ramp, P.E. = m·g·h
∴ P.E. = 0.119 kg × 9.8 m/s² × 0.17 m × sin(50°) ≈ 0.151871375 J
By conservation of energy principle, the Kinetic Energy of the cube at the top of the ramp, K.E. = E - W - P.E.
∴ K.E. = 0.8023705 J - 0.05989 J - 0.151871375 J ≈ 0.590609125 J
K.E. = (1/2)·m·v²
Where;
v = The speed of the cube at the instant just before the sliding cube leaves the ramp
∴ K.E. = (1/2) × 0.119 kg × v² ≈ 0.590609125 J
v² ≈ 0.590609125 J/((1/2) × 0.119 kg) ≈ 9.92620378 m²/s²
v = √(9.92620378 m²/s²) ≈ 3.15058785 m/s ≈ 31.5 cm/s
The speed of the cube at the instant just before the sliding cube leaves the ramp, v ≈ 31.5 cm/s.
Question 9 of 15
Locate the polyatomic ion in the compound MgSO4-
A. Mgs
B. o
C. Mg
D. SO4
Answer:
d
Explanation:
What is a overly-simplified definition of Einstein's theory of general relativity?
Answer:
the laws of physics are the same for all non-accelerating observers
Explanation:
The only way that heat can travel through outer space is ______
convection
radiation
conduction
none of the above
plssssssssssss answer correctly
Which of the following explains a projectile's parabolic motion? Choose all that apply
The law of inertia
acting on the x axis
The acceleration on
the x axis
The applied force
keeping the
projectile moving
The downward
force of gravity
HELP PLS!!!!!!! 20 POINTS
Answer:
Cayla ? please whats going on ?
Explanation:
Which of the following electromagnets is the strongest? Why?
Answer:
Bitter Magnet inside a superconducting magnet
Explanation:
Since there are no options available, generally, the electromagnet that is considered the strongest is the Bitter Magnet inside a superconducting magnet.
This electromagnet produces 45 Tesla units which is a result of bitter magnet producing 33.5 Tesla and the superconducting coil produces the additional 11.5 Tesla.
Hence, justifying that the greater the current in the coil the stronger the electromagnet.
Sound wave A is moving through a medium that is decreasing temperature.
Sound wave B is moving from a liquid to a gas medium
Which statement can be made about both waves?
1. They will speed up
2. They will increase the
density of their medium
3. They will slow down
4. They will decrease the density of their medium
The satement that can be made about sound wave A and sound wave B is, they will slow down.
Relationship between sound wave and temperatureThe relationship between sound waves and temperature is given by the following formula;
[tex]v = \sqrt{\frac{\gamma RT}{M} }[/tex]
The speed of sound wave increases with increase in temperature, and vice versa.
Speed of sound wave in liquid and gaseous mediumSound wave is mechanical wave, because it requires material medium for its propagation. Sound will travel faster in liquid medium than gaseous medium because of number of molecules per unit volume.
Thus, the satement that can be made about sound wave A and sound wave B is, they will slow down.
Learn more about speed of sound waves here: https://brainly.com/question/2142871
see bellow
fffffffffffffff
Answer:
third, The water has potential energy at the top of the waterfall and increasing kinetic energy as it falls.
Explanation:
Water at the top is not falling, but will fall soon, so has potential energy. It has increasing kinetic energy as it falls because the speed of water falling increases.
PLESE HELP !!!!!!!!!
What is the dependent variable of this testable question? How does the temperature of a tennis ball affect the height of its bounce?
Question 2 options:
brand of tennis balls
the age of the tennis ball
temperature of a tennis ball
height of its bounce
ILL GIVE BRANLIEST TO THE CORRECT ONE
Answer:
Height of its bounce
Explanation:
The dependent variable is always what is being measured or the data collected.
tank contains 335 kg of water at a uniform temperature of 60oC. The tank is insulated and not heated; it neither loses nor gains heat through the walls of the tank. A valve is opened and water exits the tank at a rate of 0.5 kg/sec and a temperature of 60oC. After 10 seconds the valve is closed again . Using the assumption that water at zero degrees centigrade contains zero energy and considering only internal, how much energy left the tank through the valve during this 10 second period; report as kJ.
Answer:
Explanation:
Thermal energy or internal energy gain or loss = mass x specific heat x temperature
specific heat of water = 4.2 kJ / kg degree Celsius
mass of water lost in 10 second = rate of loss x time = .5 x 10 = 5 kg .
heat energy associated with lost water = 5 x 4.2 x ( 60 - 0 ) = 1260 kJ .
Heat energy lost = 1260 kJ .
Which formula is used to find an object's acceleration?
a= Δt – Δν
a= Δv + Δt
a= Δv/ Δt
a= Δt/Δv
Answer:
its the third one
Directions: Follow the instructions to go through the simulation. Respond to the questions and prompts in the orange boxes. Vocabulary: air track, approach velocity, conservation of energy, conservation of momentum, elasticity, kinetic energy, momentum, separation velocity, velocity Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Imagine going to a bowling alley with a bowling ball and a ping pong ball. Why is a bowling ball better for knocking down pins than a ping pong ball?
Answer and Explanation: Momentum (Q) is defined as mass in motion, which means it relates mass of an object with its velocity:
Q = m.v
So, momentum only depends on mass of the object and the velocity it is moving.
Comparing bowling ball and ping pong ball, the first one has more mass than the second one. Therefore, a bowling pin will better knock down the pins than the ping pong ball.
The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the elevator, if upward is the positive direction. Write the numerical value only with 2 decimal places. Do not write the unit. If the acceleration is negative (downward) include the negative sign in your answer.
Answer: 1
Explanation:
Given
Tension is the string [tex]T=44\ N[/tex]
mass of object [tex]m=4\ kg[/tex]
Tension is greater than the weight of the object i.e. elevator is moving upward
we can write
[tex]\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2[/tex]
A wave has a wavelength of 1.5 meters and frequency of 125 Hz. What is the wave speed?
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light when it was hit from behind by car A, of mass 1500 kg. The cars locked bumpers during the collision and slid to a stop. Measurements of the skid marks left by the tires showed them to be 7.30 m long, and inspection of the tire tread revealed that the coefficient of kinetic friction between the tires and the road was 0.65.
(a) What was the speed of car A just before the collision?
(b) If the speed limit was 35 mph, was car A speeding, and if so, by how many miles per hour was it exceeding the speed limit?
Answer:
Explanation:
Force of friction at car B ( break was applied by car B ) =μ mg = .65 x 2100 X 9.8 = 13377 N .
work done by friction = 13377 x 7.30 = 97652.1 J
If v be the common velocity of both the cars after collision
kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²
= 1800 v²
so , applying work - energy theory ,
1800 v² = 97652.1
v² = 54.25
v = 7.365 m /s
This is the common velocity of both the cars .
To know the speed of car A , we shall apply law of conservation of momentum .Let the speed of car A before collision be v₁ .
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 7.365
v₁ = 17.676 m /s
= 63.63 mph .
( b )
yes Car A was crossing speed limit by a difference of
63.63 - 35 = 28.63 mph.
(a) The speed of car A just before the collision is 51.58 mph.
(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.
What is collision?
The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.
Given data -
The mass of car A is, mA = 1500 kg.
The mass of car B is, mB = 2100 kg.
The length of the skid mark is, d = 7.30 m.
The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].
(a)
The combined kinetic energy of both cars is,
[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]
Applying the work-energy principle as,
Work done due to kinetic friction = Combined kinetic energy of cars
[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]
Converting into mph as,
[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]
To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.
So , momentum before collision = momentum after collision of both the cars
1500 x v₁ = ( 1500 + 2100 ) x 21.49
v₁ = 51.58 mph
Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.
(b)
With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,
= 51.58 - 35 = 28.63 mph.
= 16.58 mph
Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.
Learn more about the average speed here:
https://brainly.com/question/12322912
What is the displacement for a driver who travels 10 km to get to a point that is 4 km from his starting point?
4 km
10 km
6 km
14 km
Answer:
6km
Explanation:
An experiment is designed to compare the differences in learning outcomes
between learning math from a video game and learning it from traditional
classroom activities. The experiment finds no difference between the two.
Which experiment method is important for controlling bias?
A. Developing a video game with subjects' input
B. Randomly assigning students to the two different groups
C. Asking the subjects which group they preferred to be in
D. Using classroom activities with researchers' input
Answer:
B. Randomly assigning students to the two different groups
Explanation:
The bias can be controlled by making use of the randomized block design where we divide the subject into subgroups also called blocks and randomly assign treatment conditions to each block.
In this case, the blocks or sub groups are learning math from a video game and learning it from traditional classroom activities.
So the best way to control the bias is by randomly assigning students to the different groups.
Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.250 m. What is the net force on 92?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
-5.00 x 10-6
-5.00 x 10-6 C
91
92
93
0.500 m
0.250 m
q1 = -5.00 x 10-6 C
q2 = -5.00 x 10-6 C
q3 = -5.00 x 10-6 C
E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left
E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right
E net = 720000 - 180000 = 540000 N/C to the right
F = qE
F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N
The force on q2 is 2.7 N to the left.
The net electrostatic force on the q2 is 2.7N owards left
The equation for electrostatic force is
[tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]
where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.
the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1= -5.0 ×[tex]10^{-6}[/tex] C and q3= -5.0 ×[tex]10^{-6}[/tex] C
distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m
distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m
force due to charge q1
[tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N = 0.9N towards right
[tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left
hence net force F = [tex]F_{1}+F_{2}[/tex]
= 0.9N - 3.6N = -2.7N
F = 2.7 N towards left
Learn more about electrostatic force:
https://brainly.com/question/11375177
A garden hose attached with a nozzle is used to fill a 20-gallon bucket. The inner diameter of the hose is 1 inch and it reduces to 0.5 inch at the nozzle exit. If the average velocity in the hose is 8ft/s, determine (a) the volume and mass flow rates of water through the hose (4-point), (b) how long it will take to fill the bucket with water (3-point), and (c) the average velocity of water at the nozzle exit (3-point).
Answer:
a. i. 0.044 ft³/s ii. 2.75 lb/s
b. 62.3 s
c. 32 ft/s
Explanation:
(a) the volume and mass flow rates of water through the hose
i. The volume flow rate
The volume flow rate, Q = Av where A = cross-sectional area of nozzle at inlet = πd²/4 where r = diameter of nozzle at inlet = 1 inch = 1/12 ft and v = average velocity in the nozzle = 8 ft/s.
So, Q = Av
= πd²v/4
= π(1/12 ft)² × 8 ft/s/4
= 0.175/4 ft³/s
= 0.044 ft³/s
ii. The mass flow rate
The mass flow rate, M = ρQ where ρ = density of water = 62.4 lb/ft³ and Q = volume flow rate = 0.044 ft³/s
So, M = ρQ
= 62.4 lb/ft³ × 0.044 ft³/s
= 2.745 lb/s
≅ 2.75 lb/s
(b) how long it will take to fill the bucket with water
Since we have 20 gallons = 20 × 1 gallon, we convert if to cubic feet. So, 20 gallons = 20 × 1 gallon = 20 × 0.137 ft³ = 2.74 ft³.
Since V = Qt where V = volume, Q = volume flow rate and t = time, making t subject of the formula, we have
t = V/Q
= 2.74 ft³/0.044 ft³/s
= 62.27 s
≅ 62.3 s
(c) the average velocity of water at the nozzle exit
Using the continuity equation,
A₁v₁ = A₂v₂ where A₁ = cross-sectional area of nozzle at inlet = πd₁²/4 where d₁ = diameter of nozzle at inlet = 1 inch = 1/12 ft, v₁ = average velocity in the nozzle = 8 ft/s, A₂ = cross-sectional area of nozzle at inlet = πd₂²/4 where d₂ = diameter of nozzle at inlet = 0.5 inch = 0.5/12 ft, v₂ = average velocity at exit nozzle.
So, A₁v₁ = A₂v₂
(πd₁²/4)v₁ = (πd₂²/4)v₂
πd₁²v₁/4 = πd₂²v₂/4
Dividing through by π/4. we have
So, d₁²v₁ = d₂²v₂
making v₂ subject of the formula, we have
v₂ = d₁²v₁/d₂²
v₂ = (d₁²/d₂²)v₁
v₂ = (d₁/d₂)²v₁
substituting the values of the variables into the equation, we have
v₂ = (1/0.5)² × 8 ft/s
v₂ = (2)² × 8 ft/s
v₂ = 4 × 8 ft/s
v₂ = 32 ft/s
PLEASE HELP ASAP!!!
Geese fly 34 mph north when they return from their winter migration.
What information were we given about the geese if speed and direction are given?
А
Speed
B
Velocity
С
Acceleration
Answer:
B - Velocity
Explanation:
Velocity definition: “The speed of something in a given direction.”