Answer:
B) 0.3Hz
Explanation:
I just took the test i hope i helped and i hope you pass the test
Answer:.5
Explanation:
I took the quiz
Acid rain is an example of which type of chemical weathering?
Neon has 3 naturally occurring isotopes, Neon-20, Neon-21, and Neon-22. Neon's average atomic mass on the periodic table is 20.179 amu. Based on this information which isotope is most abundant? Explain why abundance of an isotope matters when calculating the average atomic mass of an isotope.
IS BMI = Hight /Weight?
Yes or no ?
Answer:
Yes,BMI is a simple indicator of weight for height and can't differentiate between muscle mass and fat mass. So BMI tends to overestimate the health risk for adults with a high muscle mass, such as some athletes, and underestimate the risk for adults with a low muscle mass, as can occur with sedentary lifestyles.
Explanation:
Hope it is helpful...
what will create static electricity?
Answer:
Friction between two objects causes a transfer of electrons from one object to the other.
Explanation:
A 1020 kg car is pulling a 365 kg trailer. Together, the car and trailer have an acceleration of 2.21 m/s2 directly forward. (a) Determine the net force on the car. N forward (b) Determine the net force on the trailer. N forward
Answer:
a. 2936.2 N b. 773.8 N
Explanation:
Let m = mass of car = 1020 kg and m' = mass of trailer and a = acceleration = 2.12 m/s².
a. Determine the net force on the car. N forward
Since the car pulls both itself and the trailer, the combined mass is m + m' and the net force F on the car is F = (m + m')a
= (1020 kg + 365 kg)2.12 m/s²
= 1385 kg × 2.12 m/s²
= 2936.2 N
b. Determine the net force on the trailer. N forward
The net force F' on the trailer is F = m'a
= 365 kg × 2.12 m/s²
= 773.8 N
An object traveling a circular path of radius 5 m at constant speed experiences an acceleration of 3 m/s2. If the radius of its path is increased to 10 m, but its speed remains the same, what is its acceleration?
A. 0.3 m/s2
B. 1.5 m/s2
C. 6 m/s2
D. 12 m/s2
Answer:
1.5 m/s2
Explanation:
Gizmo Explanation: The acceleration of an object in uniform circular motion is inversely proportional to the radius of the motion. This means that, if the radius is multiplied by a number, the acceleration is divided by that same number. In this case, the radius of the path doubles, from 5 m to 10 m. From this you can conclude that the acceleration must be half of the original value, or 1.5 m/s2.
Car A is traveling at 18.0 m/s and car B at 25.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with an acceleration of 1.80 m/s2. How long does it take car A to overtake car B
Answer:
car A reaches and immediately overtakes the car B at 22.56 s.
Explanation:
After car A accelerate at 1.8 m/s2, it travels a distance x(A) and car B will have travels a distance x(B), let's recall that the initial distance between them is 300 m, so we have:
[tex]x_{A}=300+x_{B}[/tex]
Now, we can rewrite this equation in terms of speed and time
[tex]V_{iA}t+\frac{1}{2}at^{2}=300+V_{iB}t[/tex]
Where:
V(iA) is the initial speed of car A
V(iB) is the initial speed of car B
t is the time when car A reaches the car B
a is the acceleration
[tex]18t+\frac{1}{2}1.8t^{2}=300+25t[/tex]
[tex]0.9t^{2}-7t-300=0[/tex]
Solving this quadratic equation for t, and taking just the positive value, we will have:
t=22.56 s
Therefore, car A reaches and immediately overtakes the car B at 22.56 s.
I hope it helps you!
The age and crisis of the stage trust vs. mistrust
the orbit of an object is caused by what two factors *
1. The initial force of motion and an inward-acting force
2. 2 unbalanced forces
3, A push or pull on an object
The initial force of motion and inward acting force
Stoplight parrotfish:
The same car now travels on a straight track and goes over a hill with radius 146 m at the top. What is the maximum speed that the car can go over the hill without leaving
Answer:
v = 37.8 m/s
Explanation:
When at the top of the hill, the only force that keeps the car in the circular trajectory, is the centripetal force. This force is not a new force, is just the net force aiming to the center of the circle. In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward. So, we can write the following expression:[tex]F_{cent} = F_{g} - F_{n} (1)[/tex]
It can be showed that the centripetal force is related to the speed by the following expression: [tex]F_{cent} = m*\frac{v^{2}}{r} (2)[/tex]Replacing (2) in (1), and solving for Fn, we get: [tex]F_{n} = F_{g} - m*\frac{v^{2}}{r} (3)[/tex] Now, we need to find the value of v that makes Fn, exactly zero, because at a speed greater than this, the car will not be on track anymore.So we can write the following equation: [tex]m*\frac{v^{2}}{r} = F_{g} (4)[/tex] Replacing Fg by its value, simplifying, and solving for v, we get:[tex]v_{max} = \sqrt{g*r} =\sqrt{9.8 m/s2*146 m} = 37.8 m/s (5)[/tex]
So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, this picture is not quite complete! You are on the surface of the Earth, which is rotating. While answering the following questions you may ignore the Earth’s motion around the sun, galactic center, etc. (though that is another interesting question!) (a) What is the magnitude of the acceleration of a person sitting in a chair on the equator? (b) At the equator, is your mass times the gravitational acceleration of the Earth greater than, less than, or equal to the normal force exerted on you by the chair you are sitting on? Explain. (c) A classmate of yours asks you why we have ignored this acceleration for the whole first term of physics. "Is everything we’ve learned a lie?" they ask. Ease their fears by calculating the percentage difference between the normal force from the chair and your weight while sitting on equator. (d) The latitude of Corvallis is 44.4˚. What is your acceleration while sitting in your chair?
Answer:
a) [tex]a=33.73mm/s^{2}[/tex]
b) mg>N
c) [tex]\%_{change}=0.343\%[/tex]
d) [tex]a=24.07mm/s^{2}[/tex]
Explanation:
In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).
In this case, we know the centripetal acceleration is given by the following formula:
[tex]a_{c}=\omega ^{2}r[/tex]
where:
[tex]\omega=\frac{2\pi}{T}[/tex]
we know the period of rotation of the earth is about 24 hours, so:
[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]
so we can now find the angular speed:
[tex]\omega=\frac{2\pi}{86400s}[/tex]
[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]
So the centripetal acceleration will be:
[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]
which yields:
[tex]a_{c}=33.73mm/s^{2}[/tex]
b)
In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)
So we can do a sum of forces in equilibrium:
[tex]\sum F=0[/tex]
so we get that:
[tex]N-mg+ma_{c} = 0[/tex]
and solve for the normal force:
[tex]N=mg-ma_{c}[/tex]
In this case, we can clearly see that:
[tex]mg>mg-ma_{c}[/tex]
therefore mg>N
This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.
c)
So let's calculate our weight and normal force:
Let's say we weight a total of 60kg, so:
[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]
and let's calculate the normal force:
[tex]N=m(g-a_{c})[/tex]
[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]
N=586.58N
so now we can calculate the percentage change:
[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]
so we get:
[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]
[tex]\%_{change}=0.343\%[/tex]
which is a really small change.
d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).
There, we can see that the radius can be found by using the cos function:
[tex]cos \theta = \frac{AS}{h}[/tex]
In this case:
[tex]cos \theta = \frac{r}{R_{E}}[/tex]
so we can solve for r, so we get:
[tex]r= R_{E}cos \theta[/tex]
in this case we'll use the average radius of earch which is 6,371 km, so we get:
[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]
which yields:
r=4,551.91 km
and now we can calculate the acceleration at that point:
[tex]a=\omega ^{2}r[/tex]
[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]
[tex]a=24.07 mm/s^{2}[/tex]
In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:
a)[tex]a=33.73 mm/s^2[/tex]
b)[tex]mg>N[/tex]
c) [tex]Change=0.343%[/tex]
d) [tex]a=24.07 mm/s^2[/tex]
Then calculating from the information given in the text;
a)In this case, we know the centripetal acceleration is given by the following formula:
[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]
so we can now find the angular speed:
[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]
So the centripetal acceleration will be:
[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]
b)So we can do a sum of forces in equilibrium:
[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]
This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.
c) So let's calculate our weight and normal force:
[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]
So now we can calculate the percentage change:
[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]
d) There, we can see that the radius can be found by using the cos function:
[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]
And now we can calculate the acceleration at that point:
[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]
See more about acceleration at brainly.com/question/2437624
1. To get to school, a girl walks 1 km North in 15 minutes. She
then walks 200 m South-west in 160 seconds.
What is the girl's average velocity for her walk to school?
The girl's average velocity is 1.13 m/s
The formula used to determine average velocity is:
Velocity = distance/ timeV = D/tThis means to determine the girl's average velocity is necessary to find out the total distance and the total time.
Distance:
1 km north and 200 m SouthwestAs you can see there are two different distances and they are in different units (meters vs kilometers), let's convert the kilometers to meters so both distances are in the same unit.
1 x 1000 = 1000 (1 km = 1000 meters)Add the two distances:
1000 meters + 200 meters = 1200 meters
Time
15 minutes and 160 seconds
Let's convert the minutes into seconds:
15 x 60 = 900
Add the time:
900 + 160 = 1060
Finally, calculate the average velocity:
1200 meters / 1060 seconds = 1.13 meters per secondLearn more about velocity in: https://brainly.com/question/862972
Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign are placed near each other.
Answer:
As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
Explanation:
how are all the spheres similar
How much work is done by a person who
pushes a cant with a force of 200 newtons if
the cart moves 20 meters in the direction
of the fonce.
Wind is ________ rushing in to fill an area of low pressure.
Wind is the stream of air rushing in to fill an area of low pressure. Winds are generated by the differential distribution of temperature resulting in a pressure difference.
What are winds?The natural movement of air or other gases in relation to a planet's surface is referred to as wind. There are many different sizes of winds, ranging from thunderstorm flows that last only a few minutes to local breezes.
Large-scale atmospheric circulation is primarily brought on by the planet's rotation and the difference in temperature between the equator and the poles (Coriolis effect).
Thermal low circulations over topography and high plateaus can cause monsoon circulations in the tropics and subtropics. Local winds can be defined in coastal regions by the sea breeze/land breeze cycle; in regions with varying terrain, mountain and valley.
Find more on winds:
https://brainly.com/question/616790
#SPJ6
A 75kg encounters 10n of friction and slides down a hall with an acceleration of 3.60 find the coefficients of kinetic between the box and floor
Answer:
Explanation:
Given,
Mass of block = 75kg
Force of friction=10N
Acceleration of box = 3.60m/s^2
Acceleration due to gravity = 9.8m/s^2
Let tetha represent the angle of inclination.
Wherefore, we have mgsinθ - force of friction = ma........ 1
Substitute the values into equation 1
75×9.8×sinθ -10 = 75×3.60
735sinθ = 10+270
735sinθ = 280
Divide both sides by 735
Sinθ = 280/735
Sinθ = 0.3809
θ = sin^-1 0.3809
θ = 22.389°
We can now solve for the coefficient of friction by considering the formula:
Fs = us×R
Where R = mgcosθ
Fs = 10
10 = Us×75×9.8×cos22.389
10 = Us× 735×0.9246
10 = Us × 679.595
Divide both sides by 679.595
10/679.595 = Us
Us = 0.01471
Hence the coefficient of friction is 0.01471
Is the force exerted by the scale on the superball greater than, less than, or equal to the force exerted by the scale on the clay
Answer: A glob of very soft clay is dropped from above onto a digital scale. The clay sticks to the scale on impact. A graph of the clay's velocity vs. time, clay (t), is given, with the upward direction defined as positive.
The experiment is then repeated, but instead of using the clay glob, a superball with identical mass is dropped from the same height onto the scale.
Both the clay and the superball hit the scale 2.9s after they are dropped. Assume that the duration of the collision is the same in both cases and the force exerted by the scale on the clay and the force exerted by the scale on the superball are constant.
Explanation:
the actual ratio of weight to mass on earth is 9:8 N:1 kg what is the weight on earth of
an object whose mass is 2 kg
Answer:
The object weighs 19.6 N
Explanation:
Weight
It's the force exerted on a body by gravity.
This is often expressed in the formula
W = mg
Where
W = weight
m = mass of the object
g = gravitational acceleration
The ratio of weight to mass on Earth is 9.8 N: 1 kg. This gives us the value of [tex]g=9.8 \ m/s^2[/tex].
An object with a mass of m = 2 kg has a weight of:
[tex]W = 2\ Kg *9.8 \ m/s^2[/tex]
W = 19.6 N
The object weighs 19.6 N
A 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s what is the net force on the car
Answer: 20496N
Explanation:
The formula to calculate the net force will be given as:
Net force = Acceleration × Mass
Since 10248 kg car is pulled by a low tow truck that has an acceleration of 2.0m/s, the net force would be:
= 10248 × 2
= 20496N
An object is released from a rocket moving upwards at a speed of 8.9 m / s.
What is the downward acceleration of this object?
Answer:
9.81 m/s²
Explanation:
It is given that the upward speed of the object is 8.9 m/s.
An object is released.
Now, we know that if the object is released then it will move under the control of gravity which means that the initial velocity of the rocket will not affect the acceleration of the object.
Hence, the downward acceleration of the object will be equal to g i.e. 9.81 m/s².
HELP ASAP!
Everything on screenshot.
Answer:
I believe the answer is sea floor spreading
Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance
Answer: [tex]T=26.36^{\circ}C[/tex]
Explanation:
Given
mass of the first object [tex]m_1=5\ kg[/tex]
mass of the second object [tex]m_2=0.5\ kg[/tex]
specific heat of the two is [tex]c_1=c_2=4200\ J/kg/^{\circ}C[/tex]
The temperature of the first [tex]T_1=20^{\circ}[/tex]
The temperature of the second [tex]T_2=90^{\circ}[/tex]
Heat flows from high temperature to low temperature
Suppose T is the common temperature
[tex]m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5\times 4200\times (T-20)=0.5\times 4200\times (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^{\circ}C[/tex]
. A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20.0o C. (a) Consider the flute to be a pipe, open at both ends, and find its length, assuming that the middle-C frequency is the fundamental. (b) A second player, nearby in a colder room, also attempts to play middle C on an identical flute. A beat frequency of 3.00 Hz is heard. What is the temperature of the room
Answer:
Explanation:
The first case relates to open end pipe with fundamental frequency . Let the length of flute be l . The length vibrating column is l .
λ = v / n where n is frequency , v is velocity of light and λ is wavelength of sound produced .
λ = 343 / 261.6
= 1.31 m .
For fundamental frequency
λ / 2 = l
l = λ / 2
= 1.31 / 2
= .655 m
= 65.5 cm .
b )
Nearby colder room will create lower frequency because velocity of sound will be smaller .
frequency of note produced = 261.6 - 3 = 258.6 Hz .
velocity of sound v = n λ
= 258.6 x 1.31 m
= 338.76 m /s
decrease of velocity = 343 - 338.76 = 4.24 m /s
1 degree change in temperature produces a change of .61 m/s change in velocity .
decrease in temperature = 4.24 / .61
= 7⁰C
Temperature of colder room = 20 - 7 = 13⁰C .
A railroad car of mass M moving at a speed v1 collides and couples with two coupled railroad cars, each of the same mass M and moving in the same direction at a speed v2. (a) What is the speed vf of the three coupled cars after the collision in terms of v1 and v2
Answer:
vf = v₁/3 + 2v₂/3
Explanation:
Using the law of conservation of linear momentum,
momentum before impact = momentum after impact
So, Mv₁ + 2Mv₂ = 3Mv (since the railroad cars combine) where v₁ = initial velocity of first railroad car, v₂ = initial velocity of the other two coupled railroad cars, and vf = final velocity of the three railroad cars after impact.
Mv₁ + 2Mv₂ = 3Mvf
dividing through by 3M, we have
v₁/3 + 2v₂/3 = vf
vf = v₁/3 + 2v₂/3
A car is traveling to the right with a speed of 2.0 m/s on an icy road when the brakes are applied. The car begins sliding with constant acceleration for 3.0 m until it comes to a stop. How long does it take the car to slide to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.
Answer:
t = 3.0s
Explanation:
U = 2.0m/s , V = 0 (stop) , S = 3m , t =?
From V^2 - U^2 = 2aS
=) a = -4/6 = -0.667m/s^2
Now again by V-U = at
We have t = -U/a = 2/0.667 = 3s
Required time is t = 3.0s
A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T
Answer:
The force acting on the shaft is 1324.75 N
Explanation:
Given that,
Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]
Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]
Total mass M=24.5+0.5=25 kg
Diameter of piston, d=10 cm
We need to calculate the cross section area of piston
Using formula of area
[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]
Put the value into the formula
[tex]A_{p}=\pi\times\dfrac{0.1^2}{4}[/tex]
[tex]A_{p}=0.00785\ m^{2}[/tex]
We need to calculate the weight of the piston and shaft
Using formula of weight
[tex]W=mg[/tex]
Put the value into the formula
[tex]W=25\times9.81[/tex]
[tex]W=245.25\ N[/tex]
We need to calculate the force due to gas pressure
Using formula of force
[tex]F_{gas}=P_{gas}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{gas}=3\times10^{5}\times0.00785[/tex]
[tex]F_{gas}=2355\ N[/tex]
We need to calculate the force due to atmospheric pressure,
Using formula of force
[tex]F_{atm}=P_{atm}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{atm}=1\times10^5\times0.00785[/tex]
[tex]F_{atm}=785\ N[/tex]
We need to calculate the force acting on the shaft
from free body diagram
[tex]F_{gas}=F+F_{atm}+W[/tex]
Put the value into the formula
[tex]2355=F+785+245.25[/tex]
[tex]F=2355-785-245.25[/tex]
[tex]F=1324.75\ N[/tex]
Hence, The force acting on the shaft is 1324.75 N
The complete question is :
The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.
Solution :
Given
The cross sectional area of the shaft, [tex]$A_{s} = 0.8 \ cm^2$[/tex]
Gas pressure, [tex]$P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$[/tex]
The total mass, m = 24.5 + 0.5
= 25 kg
Diameter of the piston, d = 10 cm
The cross sectional area of the piston, [tex]$A_{p} = \frac{\pi}{4} \times (0.1)^2$[/tex]
Weight of the piston and shaft, W = mg
= 25 x 9.81
= 245.25 N
Force due to the gas pressure,
[tex]$F_{gas} = P_{gas} \times A_{p}$[/tex]
[tex]$F_{gas} = 3 \times 10^5 \times 0.00785 $[/tex]
= 2355 N
Force due to atmospheric pressure,
[tex]$F_{atm} = P_{atm} \times A_{p}$[/tex]
[tex]$= 1 \times 10^5 \times 0.00785$[/tex]
= 785 N
Now [tex]$F_{gas} = F + F_{atm} + W$[/tex]
⇒ 2355 = F + 785 + 245.25
⇒ F = 1324.75 N
a crane lifts a 200 kilogram weight to a height of 50 meters. what would be the gravitational potential energy ?
Answer:
98,000 JExplanation:
The gravitational potential energy of a body can be found by using the formula
GPE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 9.8 m/s²
From the question we have
GPE = 200 × 9.8 × 50
We have the final answer as
98,000 JHope this helps you
a body is undergoing non uniform circular motion work done by tangential force on body is
Answer:
non-zero
Explanation:
The tangential force acts as a circular moment on the body and increases its velocity, and also increases its kinetic energy.
In unequal circular movement work done on the object is non-zero. If the object's velocity increases in the tangential direction, the force is acting in the same direction.