Answer:
Homogenous Mixture
Explanation:
The composition in paint is uniform so every part of it looks the same throughout the paint.
4 What type of circuit is described by each of the following statements?
Answer series or parallel
a All components are connected end-to-end.
b. The current in the circuit divides so that some flows through one component
and the rest through another component
Two lamps are connected side by side so that each lights brightly
d The current has the same valur everywhere in the circuit
C
A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?
The final speed of the combined masses is 0.186m/s
According to the law of collision, the sum of the momentum of the bodies before the collision is equal to the momentum after the collision.
Mathematically;
m1u1 - m2u2 = (m1+m1)v
v is the final speed of the combined masses.Substituting the given parameters;
0.24(0.6) - 0.26(0.2) = (0.24+0.26)v
0.144 - 0.052 = 0.5v
0.092 = 0.5v
v = 0.092/0.5
v = 0.184m/s
Hence the final speed of the combined masses is 0.186m/s
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What do you feel when you receive your homework?
The feeling can either be positive or negative.
How one feels depends on number of factorsA positive feeling can occur when one performs very well in the given home. Also, a positive feeling can come from a high level of satisfaction in the homework. When you complete your homework on time using the recommended steps, you will be sure to do well on the homework.
In other hand, a negative feeling may result from poor performance in the homework. In ability to complete the homework or missing some steps in the homework can increase your level of trepidation even before seeing your score.
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Please answer the following question!
What is the momentum of a 750-kg Volkswagen Beetle when at rest?
Please give the value and momentum!
Answer:
0
Explanation:
the momentum will always be 0 when it is at rest because the object isnt moving!
Hope this helped!
Here's a question from ~ [ AIEEE 2002 ]
The minimum velocity ( in m/s ) with which a car driver must traverse a flat curve of radius 150 m and Coefficient of friction 0.6 to avoid skidding is ~
[ I'm looking for Proper Information, and please don't get it from any Website ]
Thanks for Answering !
As car is avoid skidding
[tex]\\ \sf\hookrightarrow \dfrac{mv^2}{r}=\mu mg[/tex]
Cancel m[tex]\\ \sf\hookrightarrow \dfrac{v^2}{r}=\mu g[/tex]
[tex]\\ \sf\hookrightarrow v^2=\mu rg[/tex]
[tex]\\ \sf\hookrightarrow v^2=0.6(10)(150)[/tex]
[tex]\\ \sf\hookrightarrow v^2=60(150)[/tex]
[tex]\\ \sf\hookrightarrow v^2=900[/tex]
[tex]\\ \sf\hookrightarrow v=30ms^{-1}[/tex]
Done
The minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is 29.7 m/s.
The given parameters:
Radius of the curve, r = 150 mCoefficient of friction, μ = 0.6The minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is calculated as follows;
[tex]\frac{mv^2}{r} = \mu mg\\\\v^2 = \frac{\mu mgr}{m} \\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr} \\\\v = \sqrt{0.6 \times 9.8 \times 150} \\\\v = 29.7 \ m/s[/tex]
Thus, the minimum velocity of the with which the car driver must traverse the flat curve to avoid skidding is 29.7 m/s.
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For a uniformly accelerated motion the graph of displacement versus time would be
Explanation:
hope this may help you
have a nice day
anything that has mass and occupies space is called
Answer:
Matter
Hope you could get an idea from here.
Doubt clarification - use comment section.
A 0.24 kg mass with a speed of 0.60 m/s has a head-on collision with a 0.26 kg mass that is traveling in the opposite direction at a speed of 0.20 m/s. Assuming that the collision is perfectly inelastic, what is the final speed of the combined masses?
The final speed of the combined masses is 0.21 m/s
Applying the law of conservation of momentum:
Total momentum before collision = Total momentum after collision.
⇒ Formula:
MU+mu = V(M+m).................. Equation 1
⇒ Where:
M = mass of the first bodym = mass of the second bodyU = Initial speed of the first bodyu = Initial speed of the second bodyV = common final speed.From the question,
⇒ Given:
M = 0.24 kgU = 0.60 m/sm = 0.26 kgu = -0.20 m/s (traveling in opposite direction)⇒ Substitute these values into equation 1
0.24(0.6)+0.26(-0.20) = V(0.24+0.2)⇒ Solve for V
0.144-0.052 = 0.44V0.44V = 0.092V = 0.092/0.44V = 0.209V ≈ 0.21 m/sHence the final speed of the combined masses is 0.21 m/s
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The instant before a batter hits a 0.14-kilogram baseball, the velocity of the ball is 45 meters per second west. The instant after the batter hits the ball, the ball's velocity is 35 meters per second east. The bat and ball are in contact for 1.0 x 10-2 second.
1. find the magnitude and direction of the average acceleration of the baseball while it is in contact with the bat.
2. calculate the magnitude of the average force the Bat exerts on the ball while they are in contact.
(1) The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is 8,000 m/s² east.
(2) The magnitude of the average force the Bat exerts on the ball while they are in contact is 1,120 N.
The given parameters:
Mass of the baseball, m = 0.14 kgInitial velocity of the baseball, u = 45 m/s west (negative direction)Final velocity of the baseball, v = 35 m/s east (positive direction)Time of contact, t = 0.01 sThe magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is calculated as follows:
[tex]a = \frac{v- u}{t} \\\\a = \frac{35 - (-45)}{0.01} \\\\a = \frac{80}{0.01} \\\\a = 8,000 \ m/s^2[/tex]
The magnitude of the average force the Bat exerts on the ball while they are in contact is calculated as follows;
[tex]F = ma\\\\F = 0.14 \times 8,000\\\\F = 1,120 \ N[/tex]
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A roller coaster goes from 2.00 m/s [forward] to 10.0 m/s [forward) in 4.50 s. What is its acceleration?
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
A 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. If the ball is in contact with
the racket for 0.04 s, what is the average force on the ball by the racket?
0.57 N
32 N
98 N
0.00005 N
From Newton's second law of motion, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
Given that a 0.07 kg tennis ball, initially at rest, leaves a racket with a speed of 56 m/s. And the time for contact with the racket is 0.04 s, that is,
mass m = 0.07 kg
velocity v = 56 m/s
time t = 0.04 s
force f = ?
To calculate the average force on the ball by the racket, let us apply Newton's second law of motion.
Impulse = change in momentum
ft = mv
Substitute all the parameters into the equation above
0.04f = 0.07 x 56
make f the subject of the formula
f = 3.92 / 0.04
f = 98 N
Therefore, the average force on the ball by the racket is 98 Newtons. The correct answer is option C
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The average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
From the question, we are to determine the average force on the ball by the racket.
From the formula,
[tex]F = \frac{mv}{t}[/tex]
Where F is the force
m is the mass
v is the velocity
and t is the time
From the given information
m = 0.07 kg
v = 56 m/s
t = 0.04 s
Putting the parameters into the formula,
we get
[tex]F = \frac{0.07 \times 56}{0.04}[/tex]
[tex]F = \frac{3.92}{0.04}[/tex]
F = 98 N
Hence, the average force on the ball by the racket is 98 N. The correct option is the third option - 98 N
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The Earth currently has an orbital period of 1 year and a semi-major
axis length of 1 AU. If the Earth's orbit changed and the semi-major
axis length increased to 1.1 AU, how would the Earth's orbital period change?
A: It would increase by 0.15 years
B: It would increase by 0.032 years
C: It would decrease by 0.15 years
D: It would decrease by 0.032 years
The Earth's orbital period change is A: It would increase by 0.15 years
Using Kepler's third law which states that the square of the orbital period of the planet is directly proportional to the cube of its distance from the sun.
So, T² ∝ R³
T'²/T² = R'³/R³ where
T = orbital period at R = 1 yearR = initial axis length = 1 AUT' = orbital period at R'R' = final axis length = 1.1 AU.So, making T' subject of the formula, we have
T' = [√(R'/R)³]T
T' = [√(1.1 AU/1 AU)³] × 1 year
T' = [√(1.1)³] × 1 year
T' = √1.331 × 1 year
T' = 1.15 × 1 year
T' = 1.15 years.
So, the change in the Earth's orbital period ΔT = T' - T
= 1.15 years - 1 year
= 0.15 years
Since this is positive, the orbital period increases by 0.15 years.
So, the Earth's orbital period change is A: It would increase by 0.15 years
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1. Which of the following types of energy is
potential energy? More than one answer is
possible,
A kinetic energy
B.thermal energy
C.sound energy
D.gravitational energy
Answer:
Potential energy is stored energy and the energy of position––gravitational energy. There are several forms of potential energy. Electrical Energy is the movement of electrical charges. Everything is made of tiny particles called atoms.
A racing car has a mass of 750 kg. It undergoes an acceleration of 4.00 m/s2. What is the net force acting on the car?
Answer:
3000 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 750 × 4 = 3000
We have the final answer as
3000 NHope this helps you
What is number 10 for this?
Answer:
The wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.
Given data in the question;
Speed of wave;
Frequency of wave;
wavelength;
To determine the wavelength of the radio wave, we use the expression for the relations between wavelength, frequency and speed.
Where is wavelength, f is frequency and c is the speed.
We substitute our given values into the equation
Therefore, the wavelength of WFNX’s radio waves with the given speed and frequency is 2.95m.
Explanation:
1)A chocolate bar measures 10 cm long by 2 cm wide and is 2 cm thick
a)Calculate the volume of one bar.
b)How many bars each 2 cm long, 2 cm wide and 2cm thick have the same total volume?
c)A pendulum makes 10 complete oscillations in 8 seconds. Calculate the time period of the pendulum
Answer:
40cm³
Explanation:
I only have an answer to 1a
Volume = Length X breadth X height
where;
length = 10cm
breadth = 2cm and;
width = 2cm
*Find*
V = ?
V = L X B X H
V = 10 X 2 X 2
= 40cm³✓
if A and B are non zero vectors, is it possible for vector A×vector B and vector A.vector B both to be zero? Justify your answer
Answer:
not able to understand the question
Stewart (70 Kg) is attracted to Ms. Little (60 Kg) who sits 2 m away. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Happy Holidays!
We can use the following equation to solve for the gravitational force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = Gravitational constant
m1,m2 = masses of the objects (kg)
r = distance between the objects (m)
Plug in the given values into the equation:
[tex]F_g = (6.67*10^{-11})\frac{(70)(60)}{(2)^2}} = \boxed{7.0 * 10^{-8}N}[/tex]
economic importance of plant resources
Answer:
People depend upon plants to satisfy such basic human needs as food, clothing, shelter, and health care. These needs are growing rapidly because of a growing world population, increasing incomes, and urbanization . Plants provide food directly, of course, and also feed livestock that is then consumed itself.
Which is a possible food chain in the ocean from start to finish?
a.phytoplankton, zooplankton, small fish, large fish, large shark
b.phytoplankton, large shark, zooplankton, large fish, small fish
c.large shark, large fish, small fish, zooplankton, phytoplankton
d.zooplankton, phytoplankton, large fish, small fish, large shark
Answer:
a.phytoplankton, zooplankton, small fish, large fish, large shark
Explanation:
The bottom level of the ocean's food chain is made up of one-celled organisms called phytoplankton. These tiny organisms are microscopic. They are so small they cannot be seen without a microscope. Billions of phytoplankton live in the upper part of the ocean. They take in the sun's light.
Phytoplankton, or plant plankton, is the first link in the food chain in the ocean. They require sunlight for photosynthesis, which they must perform. They therefore reside higher up in the shallow sunlight. Furthermore, they transform themselves into an ocean animal that can eat the “food” from the sun. Thus, option A is correct.
What food chain in the ocean from start to finish?One-celled organisms known as phytoplankton make up the base of the ocean's food chain. These microscopic organisms are quite small. Without a microscope, they are invisible due to their small size. In the upper ocean, there are billions of phytoplanktons. They absorb the light from the sun.
Aquatic food webs are built on phytoplankton and algae. They are consumed by zooplankton, tiny fish, and crustaceans, which are the main consumers. Fish, tiny sharks, corals, and baleen whales all devour the primary consumers after them.
Therefore, phytoplankton, zooplankton, small fish, large fish, large shark is a possible food chain in the ocean from start to finish.
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A wave has wavelength of 10 m and a speed of 340 m/s. What is the frequency of the wave?
I need the Formula,Known,Substitute & Solve Answer with Units
Answer:
This is the answer that I got.
Explanation:
Hope it is right.
can you fit all the planets between earth and moon
Answer: No, planets of our solar system, with or without Pluto, cannot fit within the mean lunar distance. An additional 3,500 km is needed to squeeze in Neptune
how to tell a girl you like her?
A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to the cup or surroundings. What is the final temperature of the mixture? answer in celsius
Answer:
≈19.144°C.
Explanation:
all the details are in the attachment.
Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are the parameters of the sample of water.
P.S. change the provided design according Your requirements.
Three forces acting on an object are given by F1 = (−2.00 i + 2.00 j) N,F2 = (5.00 i − 3.00 j) N, and F3 = (−45.0 i) N. The object experiences an acceleration of magnitude 3.75 m/s2.
i. What is the direction of the acceleration?
ii. What is the mass of the object?
iii. If the object is initially at rest, what is its speed after 10.0 s?
iv. What are the velocity components of the object after 10.0 s?
From summation and resultant of forces, the four answers are:
i. ∅ = 34 degrees
ii. M = 0.48 kg
iii. V = 37.5 m/s
iv. V = 37.5i + 0j + 0k
Given that Three forces acting on an object are
F1 = (−2.00 i + 2.00 j) N
F2 = (5.00 i − 3.00 j) N
F3 = (−45.0 i) N.
If the object experiences an acceleration of magnitude 3.75 m/s2.
i. The direction of the force will be the direction of the acceleration.
Sum of the forces = (-2 + 5 - 4.5)i + (2 - 3 + 0)j
Sum of the forces = -1.5i -j
The acceleration direction will be
tan∅ = [tex]\frac{y}{x}[/tex]
tan∅ = [tex]\frac{1}{1.5}[/tex]
∅ = [tex]tan^{-1}[/tex] (0.67)
∅ = 34 degrees (approximately)
ii. The mass of the object can be calculated from Newton's law.
Resultant force = mass x acceleration.
Resultant force = [tex]\sqrt{1.5^{2} + 1^{2} }[/tex]
Resultant force = [tex]\sqrt{3.25}[/tex]
Resultant force = 1.8N
Then,
1.8 = 3.75M
M = 1.8 / 3.75
M = 0.48 kg
iii. If the object is initially at rest, the speed of the object after 10.0 will be calculated by using first equation of motion.
V = U + at
Where U = 0
V = 0 + 3.75 x 10
V = 37.5 m/s
Therefore, the speed of the object after 10.0 is 37.5 m/s
iv. The velocity components of the object after 10.0 are
V = 37.5i + 0j + 0k
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I change every second what am i
Explanation:
Ummm i think a timer or clock?
Answer: Timer or a clock
Explanation:
What is the radius of a circular space station that spins with a linear speed of 90.0 m/s I’m order for its walls to supply a centripetal acceleration equal to the acceleration provided by gravity on the surface of the earth?
Answer:
a = v^2 / R centripetal acceleration
R = v^2 / a a at the surface of earth is 9.8 m/s^2
R = (90 m/s)^2 / 9.8 m/s^2 = 827 m
What chemical reactions support all life on earth?
Answer:
I’d say both anabolic and catabolic chemical reactions make life on Earth possible. Anabolic is the building up of molecules from atoms or small molecules and catabolic is the chemical reactions of breaking them down.
Explanation:
Describe the direction in which each object will accelerate. If the object won't accelerate, write "no acceleration." Explain what might be happening to each object in terms of its overall motion.
a water balloon is thrown a target at 18 m/s. if the water balloon has a mass of 0.4 kg, what is the momentum?
The momentum of the water balloon at the given speed is 7.2 kgm/s.
The given parameters:
Speed of the water balloon, v = 18 m/sMass of the balloon, m = 0.4 kgThe momentum of the water balloon is calculated as follows;
P = mv
Substitute the given values of mass and velocity as follows;
P = 0.4 x 18
P = 7.2 kg m/s.
Thus, the momentum of the water balloon at the given speed is 7.2 kgm/s.
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