Answer:
less than stating velocity due to friction and air resistance.
Explanation:
A plane starts from rest accelerates to 40 m/s in 10 seconds. How far did the plane travel during this time?
200 m
Explanation:We are given:
Initial velocity of the plane (u) = 0 m/s
Final velocity of the plane (v) = 40 m/s
Time interval (t) = 10 seconds
Displacement of the plane (s) = x m
Solving for x:
Acceleration of the plane
v = u + at [First equation of motion]
40 = 0 + a(10) [replacing known variables]
a = 4 m/s² [dividing both sides by 10]
Displacement of the Plane:
s = ut + 1/2 (at²) [Second equation of motion]
s = (0)(10) + 1/2(4)(10)² [replacing known variables]
s = 200 m
Hence, the Plane covers a distance of 200 m in the given time interval
You just tested the hypothesis a proposed model for the position of bright fringes produced by two slits. Explain what the model says and what you found in your investigations with the simulation.
Answer:
two-slit interference model was proposed by Young d sin θ = m λ
Explanation:
The two-slit interference model was proposed by Young, it establishes that if a coherent source of light passes through two slits, the shape of the given pattern is a consequence of the relative phase difference between the two rays; mathematically it can be expressed by
d sin θ = m λ
m= 0, 1, 2, 3, ...
for constructive interference, that is, the two rays arrive with a number between wavelengths.
D is the distance between the slits, tea the angle between the two rays, m an integer and m the wavelength used.
In a simulation a pattern of slits of equal intensity and equally spaced is observed.
Glycerin at 30°C has a density of 1,260 kg/m3 and a viscosity of 0.630 Pa s. In a laboratory experiment, some glycerin is forced through a horizontal tube that is 10.0 cm long and 1.00 cm in diameter. The high-pressure end of the tube is held at a gauge pressure of 618 Pa, while the other end is open to the atmosphere. What is the flow rate of the glycerin through the tube?
Answer:
Explanation:
Rate of flow of liquid through a tube can be expressed by the following expression
V = π P r⁴ / 8ηl
P is pressure difference between end of tube = 618 Pa
r , radius of tube = .5 x 10⁻²
η is viscosity of liquid flowing = .63
l is length of tube = .10 m
V = 3.14 x 618 x ( .5 x 10⁻² )⁴ / (8 x .63 x .10 )
= 240.64 x 10⁻⁸ m³ /s
mass = 240.64 x 1260 x 10⁻⁸ kg / s
= 3.03 x 10⁻³ kg /s
= 3.03 gram /s .
what body parts were scientists wanting to image that prompted the development of the CT scanner
Answer:
The head
Explanation:
At what speed do a bicycle and its rider, with a combined mass of 100 kg, have the same momentum as a 1500 kg car traveling at 1.0 m/s?
Answer:
15m/sExplanation:
Step one:
given data
mass of bicycle m=100kg
the velocity of bicyle v=?
mass of car M=1500kg
the velocity of car V=1.0m/s
Step two:
we know that the momentum is expressed as
P=mv
since the momentum of the bicycle must be equal to car then
mv=MV---------1
100*v=1500*1
divide both sides by 100
v=1500/100
v=15m/s
The velocity of the bicycle should be 15m/s
A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of sound in air a ordinary temperature is 343 m/s.
Answer:
480.2 m
Explanation:
The following data were obtained from the question:
Speed of sound (v) = 343 m/s.
Time (t) = 2.8 s
Distance (x) of the cliff =?
The distance of the cliff from the woman can be obtained as follow:
v = 2x /t
343 = 2x /2.8
Cross multiply
2x = 343 × 2.8
2x = 960.4
Divide both side by the coefficient of x i.e 2
x = 960.4/2
x = 480.2 m
Therefore, the cliff is 480.2 m away from the woman.
The distance should be 480.2 m
The calculation is as follows:Since A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. And, there is the velocity of 343 m/s
[tex]v = 2x \div t\\\\343 = 2x \div 2.8\\\\2x = 343 \times 2.8[/tex]
2x = 960.4
x = 480.2 m
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In contact forces, _____.
A.) objects do not touch each other
B.) objects must touch each other
C.) more work is done than in other forces
Answer:
B is the best answer for this
Calculate how long
in years it will take your radiation to reach Earth from Alpha Centauri (research this).
( MY RADIATION IS INFARED )
Alpha Centauri is 4.4 lightyears away. Traveling at 1/10 speed of light it would take you 44 years.
What is the value of the normal force of the coefficient of kinetic friction friction is 0.22 and kinetic friction force is 40 N
Answer: 1.8x10 sqaured newtons for all plato users
Explanation:
The value of the normal force of reaction is 181.8 N
Friction is defined as the resistance to motion caused by the contact between a surface and the object moving against it.
The normal reaction force is the weight of the body experiencing friction and normally acts vertically downwards.
The frictional force, the coefficient of kinetic friction and the normal force are related by the formula:
coefficient of friction = frictional force / normal reaction
Therefore, normal reaction = frictional force / coefficient of friction
Frictional force = 40 N; coefficient of kinetic friction friction is 0.22
Normal reaction = 40 N/ 0.22 = 181.8 N
Therefore, the value of the normal force of reaction is 181.8 N
Learn more at: https://brainly.com/question/18292235
What is the difference between renewable energy non-renewable energy? Use in your own words.
How does Hydroelectric Energy impact the environment? Use in your own words.
Explanation:
ok sooooooooooo Hydropower is fueled by water so it's a clean fuel source, meaning it won't pollute the air like power plants that burn fossil fuels.
Which statement is true about two isotopes of the same element?
Answer:
atoms that have the same # of protons but a DIFFERENT # of NEUTRONS
Explanation:
Answer:
D-They have different number of neutrons
Explanation:
brainliest? Plz
An object moves along the x-axis with its position x given as a function of time t by x(t)= Ht^2 - Ft + G
What is the object's velocity
Answer:
v(t) = 2Ht - F
Explanation:
Since, the position of the object is given in terms of time (t) as follows:
x(t) = Ht² - Ft + G
where,
H, F, G are constants.
Therefore, the velocity of the object can also be found in terms of the time (t), by simply taking the derivative of the given position equation with respect to time. So, the velocity can be found as follows:
(d/dt) x(t) = (d/dt)(Ht² - Ft + G)
v(t) = (d/dt)(Ht²) - (d/dt)(Ft) + (d/dt)(G)
v(t) = H (d/dt)(t²) - F (d/dt)(t) + (d/dt)(G)
v(t) = H(2t) - F(1) + 0
v(t) = 2Ht - F
At its maximum speed, a typical snail moves about 4.0 m in 5.0 min.
What is the average speed of the snail?
Answer:
Explanation:
Given
Distance = 4.0m
Time = 5.0 mins = 300secs
Required
Average speed
Average speed = Distance/Time
Average speed = 4.0/300
Average speed = 0.01333m/secs
Hence the average speed of the snail is 0.01333m/s
An object is allowed to fall freely near the surface of an unknown planet. The object falls 80 meters from rest in 5.0 seconds. The acceleration due to gravity on that planet is
Answer:
a=5 m/s^2
Explanation:
just took a quiz about it and got it right
Answer as soon as possible
Answer:
the velocity of the acorn
Explanation:
just do in in real life and see
Answer:
it is probably the velocity of the acorn
During takeoff, an airplane goes from 0 to 42 m/s is 5 s.
How far is it going after 4 s. Calculate answer to one decimal
Answer:
33.6 m
Explanation:
In order to solve the distance of the airplane traveled, we have to write each statement in the question:
1. The distance covered by the plane after 5s: 42 m ( This was calculated by getting the difference of the first and second speed recorded)(42 - 0 = 42)
2. Time taken for the plane to reach this speed: 5s
The question asks for:
Distance of the plane after 4s.
We then solve this by saying:
If 42 m are covered in 5s, how many are covered in 4s? or:
42 m = 5s
x m = 4s
We then cross multiply:
42 × 4 = 168 and 5 × (x) = 5x
After that we divide the "x" side on both sides:
168÷5 and 5x ÷ 5
= 33.6 m = x
Since its already rounded to one decimal place already, it remains unchanged.
According to the Law of Reflection, the angle of incidence the angle of reflection. O A. is greater than B. is less than C. equals D. is opposite from
Answer:
C. Equals
Explanation:
Law of reflection Equals the angle of incidence
What is the answer to the question ?
0.11m/s
Explanation:
avg. velocity = total displacement/ total time
total displacement = 2.55 - 1.09
= 1.46
total time taken = 12.8s
avg velocity = 1.46/12.8
=0.11
A student kicks a soccer ball upward at a 30º angle with an initial speed of 20 m∕s. What expression should the student use to calculate the magnitude of the ball’s initial velocity in the horizontal direction?
Answer:
[tex]\displaystyle x=10\sqrt{3}\ m/s[/tex]
[tex]y=10\ m/s[/tex]
Explanation:
Rectangular coordinates of vectors in 2D
Given a vector with a magnitude v and angle θ with respect to the positive horizontal direction, the x and y components of the vector are given by:
[tex]x=v\cos\theta[/tex]
[tex]y=v\sin\theta[/tex]
The soccer ball is kicked upward at an angle θ = 30° and at a speed v=20 m/s.
The rectangular components of the vector are:
[tex]x=20\cos 30^\circ[/tex]
[tex]\displaystyle x=20\cdot \frac{\sqrt{3}}{2}[/tex]
Operating:
[tex]\mathbf{\displaystyle x=10\sqrt{3}\ m/s}[/tex]
[tex]y=20\sin 30^\circ[/tex]
[tex]\displaystyle y=20\cdot \frac{1}{2}[/tex]
Operating:
[tex]\mathbf{y=10\ m/s}[/tex]
A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick together and continue moving at a new velocity. Assume that friction is negligible. What was the momentum of the shopping cart before the collision? a) 66.0 kg-m/s b) 0 kg.m/s c) 126 kg.m/s d) 378 kg.m/s What was the momentum of the box before the collision? a) 66.0 kg.m/s b) 0 kg.m/s c) 378 kg-m/s d) 126 kg-m/s What is the velocity of the combined shopping cart-box wreckage after the collision? a) 6.0 m/s b) 3.9 m/s c) 0 m/s d) 11.5 m/s
Answer:
a) 126 kgm/s
b) 0 kgm/s
c) 3.9 m/s
Explanation:
To solve this question, we will use the law of conservation of momentum.
Momentum before collision = momentum after collision
m1v1 + m2v2 = (m1 + m2)v, where
m1 = mass of the shopping cart, 21 kg
m2 = mass of the box, 11 kg
v1 = initial velocity of the shopping cart, 6 m/s
v2 = initial velocity of the box, 0 m/s
v = final velocity of the box+cart
a)
Momentum of the shopping cart before collision = P
P = mv
P = 21 * 6
P = 126 kgm/s = c
b)
Momentum of the box before collision
Like in question a above, the momentum of the box is P
P = mv
P = 11 * 0
P = 0 kgm/s = b
c)
Velocity of the combined shopping cart wreckage after collision is
m1v1 + m2v2 = (m1 + m2)v
(21 * 6) + (11 * 0) = (21 + 11)v
126 + 0 = 32v
32v = 126
v = 126/32
v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.
Please vote brainliest
According to the textbook, which indicator should Andrea use? Hint: Neutral substances have a pH of 7. Acids have a pH that is less than 7, and bases have a pH that is greater than 7. OA, phenol Red OB thymol Blue OC. phenolphthalein D. trinitrobenzoic acid Raclet Submit
Answer:
Thymol blue
Explanation:
Answer: c
Explanation:
A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant of the spring?
Answer:
Force constant of the spring (k) = 24.07 N/m
Concept/Theory:
The period [tex] \sf (T_s) [/tex] of a spring-mass system is proportional to the square root of the mass (m) and inversely proportional to the square root of the force constant of the spring (k).
Equation of period:
[tex] \boxed{ \bf{T_s = 2 \pi \sqrt{\dfrac{m}{k}}}}[/tex]
Explanation:
Mass = 0.250 kg
Period = 0.640 s
By substituting values in the equation, we get:
[tex] \rm \longrightarrow 0.640 = 2 \pi \sqrt{\dfrac{0.250}{k}} \\ \\ \rm \longrightarrow 2 \times 3.14 \sqrt{ \dfrac{0.25 0}{k} } = 0.640 \\ \\ \rm \longrightarrow 6.28 \sqrt{ \dfrac{0.250}{k} } = 0.640 \\ \\ \rm \longrightarrow \sqrt{ \dfrac{0.250}{k} } = \frac{0.640}{6.28} \\ \\ \rm \longrightarrow \frac{0.250}{k} = { \bigg(\frac{0.640}{6.28} \bigg) }^{2} \\ \\ \rm \longrightarrow \frac{k}{0.250} = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \\ \\ \rm \longrightarrow k = \bigg( { \frac{6.28}{0.640} \bigg) }^{2} \times 0.250 \\ \\ \rm \longrightarrow k = 24.07 \: N/m[/tex]
The force constant of the spring is approximately 24.038 newtons per meter.
As we are talking about Simple Harmonic Motion. In this exercise we need to determine the Spring Constant ([tex]k[/tex]), in newtons per meter, from the equation of the Period ([tex]T[/tex]), in seconds, which is described below:
[tex]T = 2\pi\cdot \sqrt{\frac{m}{k} }[/tex] (1)
Where [tex]m[/tex] is the mass of the moving element, in kilograms.
If we know that [tex]T = 0.640\,s[/tex] and [tex]m = 0.250\,kg[/tex], then the spring constant of the spring is:
[tex]0.640 = 2\pi\cdot \sqrt{\frac{0.250}{k} }[/tex]
[tex]\sqrt{\frac{0.250}{k} } \approx 0.102[/tex]
[tex]\frac{0.250}{k} \approx 0.0104[/tex]
[tex]k \approx 24.038\,\frac{N}{m}[/tex]
The force constant of the spring is approximately 24.038 newtons per meter.
Please see this question related to Simple Harmonic Motion for further details: https://brainly.com/question/17315536
A child and sled with a combined mass of 54.8
kg slide down a frictionless hill that is 11.5 m
high at an angle of 36 degrees
from horizontal.
The acceleration of gravity is 9.81 m/s^2.
If the sled starts from rest, what is its speed
at the bottom of the hill?
Answer in units of m/s.
Answer:
15.02 m/s.
Explanation:
Given that the height of the hill, h= 11.5 m.
Combined mass, m= 54.8 kg
The initial velocity of the combined mass, u=0
Acceleration due to gravity, [tex]g = 9.81 m/s^2[/tex].
Angle of the path the horizontal, [tex]\theta = 36[/tex] degree.
Let A be the initial position and B be the final position of the sled as shown in the figure.
The path is frictionless so the drag force =0
The gravitational force acting on the combined mass in the downward direction, [tex]F= mg\cdots(i)[/tex]
The component of force acting in the direction of motion = [tex]F\sin \theta.[/tex]
Let [tex]a[/tex] be the acceleration of the combined mass, m, So,
[tex]F\sin \theta= ma[/tex]
[tex]\Rightarrow mg \sin \theta= ma[/tex] [ from equation (i)]
[tex]\Rightarrow a = g \sin \theta \cdots(ii).[/tex]
Let v be the final velocity of the combined mass.
Now, by using the equation of motion,
[tex]v^2=u^2+2as\\\\\Rightarrow v^2=0^2+2as\\\\ \Rightarrow v^2=2as\cdots(iii)[/tex]
Here, s is the displacement in the direction of motion,
So, s= AB
Now, in the right-angled triangle ABO,
[tex]\sin\theta = OA/AB= h/AB\\\\\Rightarrow AB = h/ \sin\theta\\\\\Rightarrow s = h/ \sin\theta\cdots(iv)[/tex]
Now, from equations (ii), (iii) and (iv), we have
[tex]v^2= 2\times g \sin \theta \times \frac {h}{\sin\theta}\\\\\Rightarrow v^2= 2gh\\\\\Rightarrow v= \sqrt{2gh}[/tex]
By using the given values, we have
[tex]v= \sqrt{2\times 9.81\times 11.5}=\sqrt {225.63}\\\\\Rightarrow v = 15.02 m/s[/tex]
Hence, the speed of the combined mass at the bottom = 15.02 m/s.
The speed of the sled at the bottom of the hill is [tex]15.02m/s[/tex]
The speed of sled is calculated by using Newton's law of motion,
[tex]v^{2} =u^{2} +2gh[/tex]
where u is initial velocity, v is final velocity , g is acceleration due to gravity and h is height.
Given that, [tex]u = 0, g = 9.81m/s^{2}[/tex] and [tex]h = 11.5 m[/tex]
Substitute values in above equation.
[tex]v^{2}=0^{2}+2*9.81*11.5\\\\v^{2}=225.63\\\\v=\sqrt{225.63}=15.02m/s[/tex]
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How should the magnetic field lines be drawn for the magnets shown below?
Answer:
Magnetic field lines can be drawn by moving a small compass from point to point around a magnet. At each point, draw a short line in the direction of the compass needle.When opposite poles of two magnets are brought together, the magnetic field lines join together and become denser between the poles.
Explanation:
A horse trots away from its trainer in a straight line, moving away 8m away in 8.0s it then turns abruptly gallops halfway back in 2.0’s what is the magnitude of the horses average velocity for the entire trip
Answer:
v = 0.4 m/s
Explanation:
By definition, the average velocity is the rate of change of the position with respect to time, as follows:[tex]v_{avg} = \frac{x_{f}-x_{o}}{t_{f} -t_{o}} (1)[/tex]
Choosing arbitrarily x₀ = 0 and t₀ = 0, (1) reduces to:[tex]v_{avg} = \frac{x_{f}}{t_{f} }} (2)[/tex]
During the first part of the trip, the horse moves 8m away from its trainer, and during the the second part, gallops halfway back, which means that it finishes 4m away from its trainer.Since total time is 8.0 s + 2.0 s = 10s, replacing xf = 4m and tif = 10,0 s in (2) we get:[tex]v_{avg} =\frac{x_{f}}{t_{f} } = \frac{4.0m}{10s} = 0.4 m/s (3)[/tex]
Which of the following is an example of a mixture? A. calcium B. water C. carbon dioxide D. the atmosphere
Answer:
D
Explanation:
The atmosphere is made up of constituents not chemically combined unlike the rest of the options...
The atmosphere is made up of O2, N2, etc. and they are not chemically combined, hence the atmosphere is a mixture
If two balloons are charged and brought close to each other while hanging you observe that they move away from each other. What would the observation be if the balloons have larger charge?
A. There is no further effect. The repulsion magnitude is always the same.
B. There is no further effect. The magnitude of the force is determined by the charge signs not their magnitude
C. The balloons will lift in addition to separating as now they can start to overcome gravitational forces
D. The balloons will separate further as the repulsion magnitude increases
E. The balloons will come closer as the charges create larger polarization forces
F. The balloons will spin around each other since the electric force can produce rotational motion
Answer:
D. The balloons will separate further as the repulsion magnitude increases.
Big Bubba has a mass of 80 kg. What are his mass and weight on the moon respectively, if the acceleration due to gravity on the moon is 1.67 m/s2?
80 kg, 134 N
120 kg, 134 N
80 kg, 1180 N
120 kg, 704 N
Answer:
A
Explanation:
m = 80 kg
a = 1.67 m/s^2
The mass is the same anywhere in the universe. So Bubba will be 80 kg anywhere. That makes A and C the only possible answers.
F = Weight = m * a
F = 80 * 1.67
F = 134 N
The answer is A
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be the magnitude of the box's acceleration?
Acceleration of an object is the force divided by its mass. Hence, acceleration of the 100 Kg box with an applied force of 550 N is 5.5 m/s².
What is acceleration?Acceleration is a physical quantity having both magnitude and direction. This vector measures the rate of change of velocity of a moving body. Thus, acceleration has the unit of m/s².
According to Newton's second law of motion, force acting on a body is the product of its mass and acceleration. Thus, force is directly proportional to the mass and acceleration of the body.
Given the mass of the box = 100 Kg
Force applied on it = 550 N.
Acceleration = 550 N / 100 Kg
= 5.5 m/s² ( 1 N = 1 Kg m/s²)
Therefore, the acceleration of the box is 5.5 m/s².
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