Explanation:
For balanced forces, the severity of the 2 factors is equal, while the magnitude of the 2 factors is unequal in the situation of unbalanced forces. The three separate forces operate in opposing ways in balanced forces. In unbalanced forces, on the other hand, independent forces either behave in the same or reverse direction.
A weight lifter lifts a 250 Newton weight up 1.5 meters. How much work did they do?
Answer:
375 J
Explanation:
Work done is the product of force and distance moved by the object.
W=F*d
Given that ;
Force= F= 250 N
Distance= d =1.5 m
W= 250 * 1.5 =375 J
6/7 and 24/28 what is the answer proportion or no
Given that,
6/7 and 24/28
To find,
The above number are in proportion or not.
Solution,
First number is 6/7
Other number is 24/28
Multiply numerator and denominator by 4 to the first number.
[tex]\dfrac{6}{7}=\dfrac{6}{7}\times \dfrac{4}{4}\\\\=\dfrac{6\times 4}{7\times 4}\\\\=\dfrac{24}{28}[/tex]
So, 6/7 and 24/28 are in proportion.
Fraction A is proportional to fraction B because the value of fraction A is equal to fraction B.
Given the following data:
Fraction A = [tex]\frac{6}{7}[/tex]Fraction B = [tex]\frac{24}{28}[/tex]In this exercise, you're required to determine whether or not the two fractions are proportionate i.e whether fraction A is proportional to fraction B.
Two fractions are considered to be proportionate when the value of fraction A is equal to fraction B.
We would write fraction B in its simplest form by dividing both the numerator and denominator by their highest common factor (HCF).
HCF of 24 and 28 = 4.
Dividing by 4, we have:
[tex]Fraction\;B =\frac{24}{28}=\frac{6}{7}[/tex]
In conclusion, fraction A is proportional to fraction B because the value of fraction A is equal to fraction B.
Read more: https://brainly.com/question/18676621
What type of lever has the fulcrum between the resistance arm and the effort arm?
A) type 3 lever
B) type 2 lever
C) none of these
D) type 1 lever
E) type 4 lever
A generator delivers an AC voltage of the form Δv =(82 V) sin (75πt) to a capacitor. Themaximum current in the circuit is 1.00A. Find the following.
(a) rms voltage of the generator
1 V
(b) frequency of the generator
2 Hz
(c) rms current
3 A
(d) reactance
4 Ω
(e) value of the capacitance
5 F
Answer:
1. 57.99V
2. 37.5Hz
3. 0.7072A
4. 82 ohms
5. 5.18x10^-5F
Explanation:
In answer to this question, we have the Standard equation of AC emf to be
V = V0 x sin ωt
We have
V0 = 82V,
ω = 75π
1.
RMS Voltage =
V0/√2 = 82 /√2
= 82/1.414
= 57.99V
2.
ω = 2π* f
75π = 2πf
Frequency,f = 75π/2π
= 235.5/6.28
= 37.5 Hz
3.
RMS current
= Imax/√2
= 1.00/1.414
= 0.7072A
4.
Reactance
= Vrms/ Irms
= 57.99/0.7072
= 81.999
= 82.0 Ω
5.
Reactance = 1/ ω x C
Reactance = 82
ω = 75π
We put these values into the equation above and make c the subject of the formula
C = 1/82.0 x 75π
C = 1/ 82.0 x 75 x 3.14
C = 1/19311
Capacitance = 5.18x10^-5F
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
A brick of gold is 0.1 m wide, 0.1 m high, and 0.2 m long. The density of gold is 19,300 kg/m3. What pressure does the brick exert on the table if the brick is resting on its side?
Answer:
[tex]P=18,933.3Pa=18.9kPa[/tex]
Explanation:
Hello!
In this case, since we can compute the volume of the brick as shown below:
[tex]V=0.1m*0.1m*0.2m=0.002m^3[/tex]
Next, we can compute the mass of the brick given its density:
[tex]\rho =m/V\\\\m=V*\rho\\\\m=0.002m^3*19,300kg/m^3\\\\m=38.6kg[/tex]
Now, since the force exerted on the table corresponds to the weight of the brick, we use the gravity to obtain:
[tex]W=38.6kg*9.81m/s^2=378.7N[/tex]
Finally, since the surface of the brick in contact with the table corresponds to the 0.1x0.2 area (length and width), the area on which the weight force is exerted is:
[tex]A=0.1m*0.2m=0.02m^2[/tex]
Therefore, the pressure is:
[tex]P=\frac{F}{A}=\frac{W}{A}=\frac{378.7N}{0.02m^2}\\\\P=18,933.3Pa=18.9kPa[/tex]
Best regards!
A cat swats a ball on the coffee table. The ball rolls horizontally off the edge of a table. Which statement is true?
A. The balls horizontal velocity changes at the rate of 9.8 meters/second every second.
B. As the ball falls to the ground, it moves along a parabolic path.
C. No external vertical forces acts on the ball.
D. The ball had a constant vertical velocity.
Answer:
D
Explanation:
What is the first step in the event of a fire in a building?
A) Call 911.
B) Before opening a door, check to see if it is hot.
C) Stop, drop, and roll.
D) Leave the building quickly, if possible.
Answer:
B or C
Explanation:
i don't if I'm right or not
Order the layers, with the oldest at the bottom and most recent at the top.
someone plz help will give brainliest if correct
Answer:
adbce i think
Explanation:
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.
8 th Grade Physical Science Worksheet Part A: Read the scenario and answer the questions. David read that Fox brake pads and Best Brake pads were the best on the market. He always used NAPA pads and believed they performed the best. He decided to test all three pads and determine which was the best. David used the same car for each set of pads. He drove 25 mph and applied the brakes at the same point on the track. David then measured how many feet the car took to stop after the brakes were applied. 1. The hypothesis was: 2. The effects of the ________________(independent variable) on the __________________________________ (dependent variable). 3. List three constants:
Name at least two other viruses that are at all time lows due to vaccines.
The PE of an object is 2500 j. What is the mass of the object if it 4.0 m above the ground?
Answer:
The answer is 5.0
Explanation:
Hope that helped :D
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
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.
. 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 cannon shoots a 40 kg ball at a sailing ship and when it hits the
wall of the vessel it applies 2000N of force. How much was it
decelerated?
Answer:
ANSWER : 2000 / 40 = 50 m / s
Are vaccines 100% effective and safe for absolutely everyone?
What is the difference between Helium3, and Helium?
Explanation:
Helium is the type of atom, with 2 protons in the nucleus.
He-3 (which is the proper form), is an isotope of helium that has an atomic number of 3 (so 2 protons and 1 electron).
The most abundant helium isotope is He-4, just an extra fact.
HELP ASAP!
Everything on screenshot.
Answer:
I believe the answer is sea floor spreading
1. What is the value of the acceleration that the car experiences? 2. What is the value of the change in velocity that the car experiences? 3. What is the value of the impulse on the car? 4. What is the value of the change in momentum that the car experiences? 5. What is the final velocity of the car at the end of 10 seconds? The car continues at this speed for a while. 6. What is the value of the change in momentum the car experiences as it continues at this velocity? 7. What is the value of the impulse on the car as it continues at this velocity? The brakes are applied to the car, causing it to come to rest in 4 s. 8. What is the value of the change in momentum that the car experiences? 9. What is the value of the impulse on the car? 10. What is the value of the force (average) that causes the car to stop? 11. What is the acceleration of the car as it stops?
Answer:
All the answers are solved and explained below.
Explanation:
Note: This questions lacks the initial and most necessary data to answer these following questions. I have found a related question. I will be considering that question to carry out the answers.
Question: A car with a mass of 1000 kg is at rest at a spotlight. when the light turns green, it is pushed by a net force of 2000 N for 10 s. (This was the information missing in this question).
Data Given:
m = 1000 kg
F = 2000N
t = 10s
Q1 Solution:
Acceleration = a = ?
F = ma
a = F/m
a = 2000/ 1000
a = 2 [tex]m/s^{2}[/tex]
Q2: Solution:
Change in velocity = Δv = ?
acceleration = change in velocity / time
a = Δv/t
Δv = axt
Δv = 2 x 10
Δv = 20 m/s
Q3: Solution:
Impulse = I = ?
Impulse = Force x time
I = 2000 x 10
I = 20000 Ns
Q4: Solution:
Change in Momentum = Δp = ?
Δp = mΔv
Δp = 1000 x 20
Δp = 20000 Kgm/s
Q5: Solution:
Final velocity of the car at the end of 10 seconds = vf = ?
Δp = m x Δv
Δp = m x (vf-vi)
Δp = 1000 x (vf - 0 )
20000 = 1000 x vf
vf = 20000/1000
vf = 20 m/s
Q6: Solution:
Change in momentum the car experiences as it continues at this velocity?
Δp = ?
Δp = mΔv
Δp = m x (0)
Δp = 0
Q7: Solution:
Impulse = Change in momentum
Impulse = Δp
Implulse = 0
Q8: Solution:
Change in momentum = Δp = mΔv
Δp = m(vf-vi)
Δp = 1000 x (0-20)
Δp = -20000 kgm/s
Q9: Solution:
Impulse = Δp
Impulse = -20000 Ns
Q10: Solution:
Impulse = ?
Impulse = F x t
F = impulse/t
F = -20000/4s
F = -5000 N
Q11: Solution:
F = ma
a = ?
a = F/m
a = -5000/1000
a = -5[tex]m/s^{2}[/tex]
Water enters a shower head through a pipe of radius 0.0112 m at 3.25 m/s. What is it’s volume flow rate? (Unit= m^3/s)
Answer:
[tex]1.28 \times 10 ^{-3} m^3/s[/tex]
Explanation:
Given that the radius of the pipe, r =0.0112 m
So, the area of the cross-section, [tex]a= \pi (0.0112)^2 = 3.941\times 10^{-4} m^2[/tex]
Speed of water in the pipe, v=3.25 m/s
Volume flow rate =[tex]av= 3.941\times 10^{-4} \times 3.25 = 1.28\times 10 ^{-3} m^3/s[/tex]
Hence, the volume flow rate is [tex]1.28 \times 10 ^{-3} m^3/s[/tex]
Answer:
For Acellus the answer is 0.00128
Explanation:
Trust me
A 40.0-kg football player leaps through the air to collide with and tackle a 50.0-kg player heading toward him, also in the air. If the 40.0-kg player is heading to the right at 9.0 m/s and the 50.0-kg player is heading toward the left at 2.0 m/s, what is the speed and direction of the tangled players
Answer:
The speed and direction of the tangled players is 2.89 m/s towards right.
Explanation:
Given;
mass of the player heading right, m₁ = 40 kg
initial velocity of the player heading right, u₁ = 9.0 m/s
mass of the player heading left, m₂ = 50 kg
initial velocity of the player heading left, u₂ = -2.0 m/s
let the speed of the tangled players = v
Apply the law of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(40 x 9) + (50 x - 2) = v(40 + 50)
360 - 100 = v(90)
260 = v(90)
v = 260 / 90
v = 2.89 m/s (towards right, since it is in position direction)
Therefore, the speed and direction of the tangled players is 2.89 m/s towards right.
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]
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...
The world’s largest wind turbine has blades that are 80 m long and makes 1 revolution every 5.7 seconds.
What is the velocity for one of the blades?
(THIS IS PHYSICS, CIRCULAR MOTION)
Answer:
88.14 m/s
Explanation:
From the question given above, the following data were obtained:
Radius (r) = length of blade = 80 m
Revolution (rev) = 1
Time (t) = 5.7 s
Velocity (v) =?
The velocity of the blade can be obtained by using the following formula:
v = (rev × 2πr) / t
NOTE: Pi (π) = 3.14
v = (1 × 2 × 3.14 × 80) / 5.7
v = 502.4 / 5.7
v = 88.14 m/s
Therefore, the velocity of the blade is 88.14 m/s
what is the name of first magnet
Answer:
[tex]magnetite[/tex]
[tex]hope \: helps...[/tex]
The Sun does NOT rotate the way Earth does. Which statement best describes the difference in their rotations?
A
The Sun had different periods of rotation at different latitudes.
B
The Sun does NOT rotate.
C
The Sun rotated in different directions of rotation at different latitudes.
D
The Sun rotates in the opposite direction from Earth's rotation
Anyone can help?? I need it done before 9am please!!
Answer:
The answer is below
Explanation:
The equation for a linear line graph is given by:
y = mx + b, where y and x are variables, m is the slope of the graph and b is the y intercept (that is value of y when x is zero).
The slope (m) of a line passing through two points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given by:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
A) The first line passes through the point (0, 0) and (10, 60). It is represented as (time, velocity). Hence the slope is:
[tex]m=\frac{60 -0}{10-0}=6\ m/min^2[/tex]
B) The second line passes through the point (10, 60) and (15, 60). Hence the slope is:
[tex]m=\frac{60 -60}{15-10}=0\ m/min^2[/tex]
C) The third line passes through the point (15, 60) and (40, -40). Hence the slope is:
[tex]m=\frac{-40 -60}{40-15}=-4\ m/min^2[/tex]
D) The third line passes through the point (40, -40) and (55, 0). Hence the slope is:
[tex]m=\frac{0 -(-40)}{55-40}=2.67\ m/min^2[/tex]
If an object at the surface of the Earth has a weight Wt, what would be the weight of the object if it was transported to the surface of a planet that is one-sixth the mass of Earth and has a radius one third that of Earth
Answer:
We know that the gravitational acceleration in the surface of the Earth can be written as:
g = G*M/r^2
Where:
M = mass of the Earth
r = radius of the Earth.
G = gravitational constant.
The weight of an object of mass m, is written as:
W = m*g = m*(G*M/r^2)
Now, if we move our object to a place that has a mass equal to 1/6 times the mass of the Earth, and 1/3 the radius of the earth.
The gravitational acceleration on this planet is written as:
g' = G*(M/6)/(r/3)^2 = (1/6)*(G*M)/(r^2/9) = (9/6)*(G*M/r^2) = (3/2)*g
then the weight on this planet is:
W' = m*g' = m*(3/2)*g = (3/2)*(m*g)
and m*g was the weight on Earth, then:
W' = (3/2)*(m*g) = (3/2)*W
The new weight is 3/2 times the weight on Earth.