Answer:
first negative and then positve
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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For a uniformly accelerated motion the graph of displacement versus time would be
Explanation:
hope this may help you
have a nice day
Which shows the weight of an atom?
atomic mass
atomic number
chemical symbol
none of the above
Answer:
Atomic mass
Explanation:
Answer:
atomic mass
Click thanks if thankc if this helped
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:
how to tell a girl you like her?
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.
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.
anything that has mass and occupies space is called
Answer:
Matter
Hope you could get an idea from here.
Doubt clarification - use comment section.
I change every second what am i
Explanation:
Ummm i think a timer or clock?
Answer: Timer or a clock
Explanation:
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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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
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✓
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!
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]
A 5kg ball is on top of the school building at a height of 40m above the ground.
A. calculate the potential energy of the ball at that height.
B. How much power would it take to get the ball from the ground to that height in 1 minute.
C. If it is dropped what type of energy will the ball have as it approaches the ground.
D. What speed will the ball hit the ground
Answer:
A-Caclcuate the potential energy of the ball at that height
Explanation:
(a). Mass of the Body = 10 kg.
Height = 10 m.
Acceleration due to gravity = 9.8 m/s².
Using the Formula,Potential Energy = mgh
= 10 × 9.8 × 10 = 980 J.
(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.
∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.
∴ Kinetic Energy = 980 J.
(c). Kinetic Energy = 980 J.
Mass of the ball = 10 kg.
∵ K.E. = 1/2 × mv²
∴ 980 = 1/2 × 10 × v²
∴ v² = 980/5
⇒ v² = 196
∴ v = 14 m/s.
Please help me! I don't understand how to do this!
Part 1: A 6 kg shopping cart goes from a velocity of 0.9 m/s to a velocity of 6 m/s over the course of 2.0 seconds. What is the cart's acceleration?
Part 2: What is the net force acting on the shopping cart during that time? You will need your answer from the previous part.
From the information in the question, the acceleration of the cart is 2.55 ms-2.
We can find the acceleration using the following information from the question;
u = 0.9 m/s
v = 6 m/s
a = 2.0 seconds
But;
a = v - u/t
v = final velocity
u = initial velocity
t = time
a = 6 - 0.9/2 = 2.55 ms-2
Force= ma
m = mass
a = acceleration
F = 6 kg × 2.55 ms-2
F = 15.3 N
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Starting with only the Balmer series light (visible light), how could we ensure that the solar panels generate a current that Mark can use for his power station? (It may help to look at the electromagnetic spectrum from week 3):.
We can ensure that the solar panels generate a current that Mark can use for his power station is by gradually increasing the brightness (amount) of light that we reflect on it.
Solar panel are photo-voltaic module or cells which means they uses sunlight as a source of energy to generate current.
when the sun rays hit a silicon crystal of the solar panel then the electron inside of the solar panel start to charge and travel rapidly. when such charged electron flow through the circuit then the solar panel generate electricity.
So, to optimize the electricity generation, the energy of light is to be maximized which is determined by the frequency of the source.
Hence, To ensure that the solar panels generate a current that Mark can use for his power station we need to gradually increase the brightness (amount) of light that we reflect on it.
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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
You are looking at a yellow flower growing outside in the sunshine. Why does it look yellow?
Answer:
It looks yellow because that is the only (major) color reflected.
Visible spectra is from about 4000-7000 Angstroms (10^-10 m).
Red are longer wavelengths and blue are the shorter wavelengths.
The Sodium doublet (yellow) occurs around 5900 Angstroms.
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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The blue, hotter stars tend to have shorter lifetimes than the red, cooler stars.
true
false
Answer:
A. True
Explanation:
edge 2022
The blue, hotter stars tend to have shorter lifetimes than the , cooler stars; this is a true fact as the lifespan of a star is determined by its mass. Blue, hotter stars have higher masses than red, cooler stars.
What is the lifespan of a star?The lifespan of a star is largely determined by its mass because a star's mass affects its internal pressure, temperature, and nuclear fusion rate, and more massive stars have stronger gravitational forces, which lead to higher pressures and temperatures in their cores. The rate of nuclear fusion reactions in a star determines how quickly it consumes its fuel, which in turn affects the lifespan of the star. Blue, hotter stars have a higher rate of nuclear fusion reactions and consume their fuel much more quickly than red, cooler stars.
Hence, the blue, hotter stars tend to have shorter lifetimes than the redder, cooler stars.
Learn more about the lifespan of a star here.
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#SPJ7
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.
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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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
The gravitational potential energy of the ball changes as it bounces. Where is the ball located when its gravitational potential energy is greatest? Where is the ball when its gravitational potential energy has the least value?
When the planet is closest to the Sun, speed v and kinetic energy are the highest, and gravitational potential energy is the lowest. When the planet moves farther away, the speed and kinetic energy decrease, and the gravitational potential energy increases.
according to newton, doubling the distance between two interacting objects:
Answer:
So as two objects are separated from each other, the force of gravitational attraction between them also decreases. If the separation distance between two objects is doubled (increased by a factor of 2), then the force of gravitational attraction is decreased by a factor of 4 (2 raised to the second power).
Explanation:
hope his helps
help please I really need this
Answer:
Explanation:
Height(h) = 4m; depth of the half filled tank 4m/2 =2m
Acceleration due to Gravity (g)=9.8m/s^2
Density (d)=1000kg/m^3
Pressure = hdg
Pressure = 2m × 9.8m/s^2 × 1,000kg/m^3
Pressure = 19,600Pa
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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can someone explain with steps please
Answer:
The speed of other projectile is [tex]3.1m/s[/tex]
Explanation:
Range of projectile is given by the equation
[tex]\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}[/tex]
Here we have same range
Hence
[tex]\frac{\mathrm{2.5}^{2} \cdot \sin (2 \times 65)}{\mathrm{g}}=\frac{\mathrm{u}^{2} \cdot \sin (2 \times 15)}{\mathrm{g}}\\\\u^2=\frac{2.5^2\sin130}{\sin30} \\\\u=3.10m/s[/tex]
Here
initial velocity=u
u1=2.5m/su2=?[tex]\theta_1=65°[/tex][tex]\theta_2=15°[/tex]Now
[tex]\\ \sf\longmapsto R_1=R_2[/tex]
[tex]\\ \sf\longmapsto \dfrac{u_1^2sin2\theta_1}{g}=\dfrac{u_2^2sin2\theta_2}{g}[/tex]
[tex]\\ \sf\longmapsto \dfrac{(2.5)^2.sin2(65)}{g}=\dfrac{u_2^2sin2(15)}{g}[/tex]
Cancel g[tex]\\ \sf\longmapsto 6.25sin130=u_2^2sin30[/tex]
[tex]\\ \sf\longmapsto 6.25(0.76)=u_2^2(0.5)[/tex]
[tex]\\ \sf\longmapsto 5.13=0.5u_2^2[/tex]
[tex]\\ \sf\longmapsto u_2^2\approx 10[/tex]
[tex]\\ \sf\longmapsto u_2\approx 3.1m/s[/tex]
Option B is correct
What is Newton's 1st Law?
Forces come in pairs that are equal in size and opposite in direction.
F=ma
An object in motion will stay in motion at the same speed and in the same direction (ie with constant velocity) unless a force makes it change.
Answer:
An object in motion will stay in motion at the same speed and in the same direction (ie with constant velocity) unless a force makes it change.
Explanation: