Answer:
Velocity ratio of simple machine is the ratio of distance travelled by effort to the distance travelled by load in the machine. The efficiency of a simple machine is defined as the ratio of useful work done by machine (output work) to the total work put into the machine (input work).
steel wire 8m long and 4mm in diameter is fixed to two rigid b. [1] supports. Calculate the increase in tension when the temperature falls by 10°C. Given linear expansivity of steel, a 12x10 %/K, Young's modulus for steel, Y = 2x10¹¹ N/m².
Answer:
301.6 N
Explanation:
The length of the wire L₀ = 8 m and its diameter, d = 4 mm = 4 × 10⁻³ m. Since its temperature drops by 10°C, it will have a change in length ΔL = L₀αΔθ where α = linear expansivity of steel, a 12 × 10⁻⁶ /K, and Δθ = temperature change = -10°C = -10 K(negative since it is a drop)
So, the strain, ε = ΔL/L₀ = αΔθ = 12 × 10⁻⁶ /K × 10 K = 12 × 10⁻⁵
Now the Young's modulus of steel, Y = σ/ε where σ = stress = T/A where T = increase in tension in steel wire and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 4 × 10⁻³ m and ε = strain = 12 × 10⁻⁵
So, σ = Yε
Since Y = 2 × 10¹¹ N/m².
Substituting the values of the variables into the equation, we have
σ = Yε
σ = 2 × 10¹¹ N/m² × 12 × 10⁻⁵
σ = 24 × 10⁶ N/m²
Since σ = T/A
T = σA
T = σπd²/4
Substituting the values of the variables into the equation, we have
T = σπd²/4
T = 24 × 10⁶ N/m² × π × (4 × 10⁻³ m)²/4
T = 24 × 10⁶ N/m² × π × 16 × 10⁻⁶ m²/4
T = 24 × 10⁶ N/m² × π × 4 × 10⁻⁶ m²
T = 96 N × π
T = 301.59 N
T ≅ 301.6 N
So, the increase in tension in the steel wire is 301.6 N
A 43-kg child sits in a massless swing. With what horizontal force must the seat be pulled so that the ropes form an angle of 35o with respect to the vertical
Fx = 295.4N
180°-90°-35°= 55°
F= m(a)
Fy = 43kg(9.81m/s²)
Fy = 421.83N
tan(55°) = 421.83N/Fx
Fx = 421.83N/tan(55°)
Fx = 295.3685458N
What are horizontal and vertical forces?Each component describes the influence of that chain in the given direction. The vertical component describes the upward influence of the force upon Fido and the horizontal component describes the rightward influence of the force upon Fido.
What direction is friction force?The kinetic frictional force is always in the direction opposite to the velocity of the object.
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help with this physics question please :)
Answer:
is that rm in you profile pic
The energy required to change the state of a substance was determined to be E(H).
If the mass of the substance was DOUBLED, the value of E(H) will
(A) be halved
(B) be doubled
(C) be quadrupled
(D) remain constant
Answer:
ccccccccc or bbbbbbb or aaaaaaaa or ddddddd
to all the physicians please help this is for my assignment
Answer:
Q. 1. Newton's Law of gravitation states that all bodies in the universe exerts a force of attraction on all other bodies in the universe with a proportional force to both the product of the masses of the bodies and inversely proportional to the square of the distance between their centers
Mathematically, we have;
[tex]F = G \times \dfrac{m_1 \times m_2}{R^2}[/tex]
Where;
m₁, and m₂ are the masses of the bodies
R = The distance between their centers
G = The gravitational constant = 6.6743 × 10⁻¹¹ N·m²/kg²
The gravitational constant, G, is the Newton's law of gravitation's constant of proportionality between the force of attraction that exist two bodies and the product of their masses divided by the square of the distance between their centers
Q. 2. Newton's law of gravitation in vector form is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12}[/tex]
The above equation gives the gravitational force of attraction of body 1 on body 2, with the negative sign and unit vector indicating that the force of of gravity is towards body 1
The force of gravity of body 2 on 1 is presented as follows;
[tex]\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{12}^2} \cdot \hat R_{21}[/tex]
The gravitational force of attraction of body 2 on body 1 is therefore, equal in magnitude and opposite in direction of the gravitational force of body 1 on body 2 (towards body 2)
[tex]-\underset{F_{12}}{\rightarrow} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{12} = G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot -(\hat R_{21}) = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = -G \times \dfrac{m_1 \times m_2}{R_{21}^2} \cdot \hat R_{21} = \underset{F_{21}}{\rightarrow}[/tex]
[tex]-\underset{F_{12}}{\rightarrow} = \underset{F_{21}}{\rightarrow}[/tex]
Explanation:
Two football players run towards each other along a straight path in Penrith Park in the clash between the Melbourne storms and the Penrith Panthers a month ago. Melbourne's Justin Olam who is about 95kg and ran towards Viliame Kikau at 3.75m/s. Viliame Kikau is 111kg and moves towards Justin Olam at 4.10m/s. They end up in a head-on collision and are stuck together.
A) What is their velocity immediately after the collision?
B) What are the initial and final kinetic energies of the system?
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =[tex]\frac{m v_1 + M v_2}{m+M}[/tex]
let's calculate
v = [tex]\frac{ -95 \ 3.75 \ + 111 \ 4.10}{95+111}[/tex]
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J
A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radius of 40.0 m. The coefficient of static friction between the car tires and the road is 0.500. The car travels successfully around the turn at a constant speed of 10.0 m/s. Calculate the magnitude of the car's acceleration as it goes around the turn. _______ m/s^2
Answer:
2.5 m/s²
Explanation:
The given parameters are;
The mass of the car, m = 2,000 kg
The radius of the car, r = 40.0 m
The coefficient of friction between the car tires and the road, μ = 0.500
The constant speed with which the car moves, v = 10.0 m/s
The normal reaction of the road on the car, N = The weight of the car;
∴ N = m × g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
N ≈ 2,000 kg × 9.81 m/s² = 19,620 N
The frictional force, [tex]F_f[/tex] = μ × N
The centripetal force, [tex]F_c[/tex] = m·v²/r
The car moves without slipping when [tex]F_f[/tex] = [tex]F_c[/tex]
Therefore, [tex]F_f[/tex] = 0.500 × 19,620 N = 2,000 kg × [tex]v_{max}[/tex]²/40.0 m
∴ [tex]v_{max}[/tex] = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s
Therefore, the velocity with which the car moves, v < [tex]v_{max}[/tex]
The cars centripetal acceleration, [tex]a_c[/tex] = v²/r
∴ [tex]a_c[/tex] = (10.0 m/s)²/40.0 m = 2.5 m/s²
The cars centripetal acceleration as it goes round the turn, [tex]a_c[/tex] = 2.5 m/s².
I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work
This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.
The acceleration of the bird is "a = 26.13 m/s²".
First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:
[tex]h = v_it + \frac{1}{2}gt^2[/tex]
where,
h = height fall = 3 m
vi = initial velocity = 0 m/s
g = acceleration due to gravity = 9.8 m/s²
t = time taken = ?
Therefore,
[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]
The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:
[tex]s = v_it + \frac{1}{2}at^2[/tex]
where,
s = distance covered by the bird = 5 m + 3 m = 8 m
vi = initial velocity of the bird = 0 m/s
a = acceleration of the bird = ?
t = time taken = 0.78 s
Therefore, using these values we get:
[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]
a = 26.13 m/s²
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Why aren’t magazine photos a good representation of what a healthy person looks like
A simple machine has efficiency 90%. What does it mean? No machine has 109 efficiency,why?
Answer:
Efficiency of a machine is 90% this means that the total energy produced to the machine only 90% is useful and given as a output.
Machines are not 100% efficient because some of the work done by a machine is used to overcome friction
What do you mean by unit?
Unit is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature.
Answer:
The standard known quantity which is used to measure a physical quantities is known as unit.
at what angle should the circular road be banked so that a car running at 50 km per hour be safe to go round the circular from of 200m radius
Answer:
Mark brainliest please
answer is 5.5 degrees
Explanation:
Given,
Speed of car (v) = 50km/hr = 50×1000/60×60 m/s
Radius of circle (r) = 200m
Banking angle (θ) =?
we know ,
tanθ = v^2/rg
or , tanθ = (50×1000/60×60)^2/200×10 = 0.0965
∴ θ = tan-1(0.0965) = 5.5°
different between weight and mass
Answer:
So mass is the amount of matter in an object, but weight is the amount of force that is being applied to your body from gravity.
Mass does not change unless you change your person like your haircut, fingernail length, ect.
Weight changes depending on your mass, because the heavier you are, the stronger the gravitational pull on your person will be, which increases your weight.
That is why if you go onto the moon, for example, your weight is different, since the gravitational pull is weaker than on Earth, so you weigh less.
Explanation:
A jet airplane is in level flight. The mass of the airplane is m=8950 kg. The airplane travels at a constant speed around a circular path of radius R=9.33 mi and makes one revolution every T=0.123 h. Given that the lift force acts perpendicularly upward from the plane defined by the wings, what is the magnitude of the lift force acting on the airplane?
Answer:
The net force is 91780.8 N.
Explanation:
mass, m = 8950 kg
Radius, R = 9.33 miles = 15015.2 m
Time, T = 0.123 h = 442.8 s
There are two forces acting on the plane.
Horizontal force is the centripetal force and the vertical force is the weight.
[tex]Fx =m R w^2\\\\Fx = m R \frac{4\pi^2}{T^2}\\\\Fx = 8950\times 15015.2\times \frac{4\times 3.14\times 3.14}{442.8\times 442.8}\\\\Fx = 27030.8 N \\\\Fy = m g \\\\ Fy = 8950\times 9.8 \\\\Fy = 87710 N[/tex]
The net force is
[tex]F = \sqrt{Fx^2 + Fy^2}\\\\F = \sqrt {27030.8^2 + 87710^2}\\\\F = 91780.8 N[/tex]
An object is thrown with velocity 7.1 m s-1 vertically upwards on the Moon. The
acceleration due to gravity on the Moon is 1.62 ms? What is the time taken for the object to return to its starting point?
A 3.5 s
B 4.4 s
C 6.5 s
D 8.8 s
Answer:D
Explanation:
The spectra of most galaxies show redshifts. This means that their spectral lines Group of answer choices always are in the red part of the visible spectrum. have wavelengths that are longer than normal. have a higher intensity in the red part of the spectrum. have wavelengths that are shorter than normal.
have a higher intensity in the red part of the spectrum
Because of the Doppler effect, the spectra of most galaxies show redshifts. This means that their spectral lines Group of answer choices always are in the red part of the visible spectrum
Most galaxies' spectra have been redshifted, which means that their spectral lines are now pushed away from their normal or rest wavelengths and towards longer wavelengths. When an object travels away from an observer, the Doppler effect causes the shift phenomenon.
The Big Bang hypothesis and the concept of the expanding universe, show that galaxies are moving away from us, observed from the redshift lines.
The waves' apparent wavelengths shift as a result, changing the color of light waves or the pitch of sound waves.
Hence, because of the Doppler effect, the spectra of most galaxies show redshifts. This means that their spectral lines Group of answer choices always are in the red part of the visible spectrum.
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A van of mass 1200kg is moving with speed of 90km/hr. It is brought in 3 seconds by applying brakes. Calculate the force applied to bring van to rest.
Answer:
Explanation:
The main equation to solve this is F = ma, where F is the force applied to the brakes with respect to its acceleration. We have the mass that we need, but we do not have the acceleration. That's the first thing we have to find. However, our velocity needs to be stated in m/s and right now it's in km/h. Converting that:
[tex]90\frac{km}{hr}*\frac{1000m}{1km}*\frac{1hr}{3600s} =25\frac{m}{s}[/tex] Now we're ready to find the acceleration:
[tex]a=\frac{v_f-v_0}{t}[/tex] where the top line there translates to the final velocity minus the initial velocity.
[tex]a=\frac{0-25}{3.0}[/tex] so the acceleration is -8.3 m/s/s
We can use that now in the force equation above:
F = 1200(-8.3) and
F = -1.0 × 10⁴ N (that's 10,000 N to the correct number of sig dig's; the negative sign means that the force is being applied in the direction opposite to that which the van is currently moving)
When the system pressures are lower than the pressure in the refrigerant tank, such as when the system has just been evacuated or if the system is out of refrigerant, refrigerant can be added to both the high and low pressure sides of the system. A. True B. False
Answer: True
Explanation:
It should be noted that when refrigerant is added in vapor form to the operating refrigeration system, then the addition of the refrigerant should be to the low-pressure side of the refrigeration system.
Furthermore, when system pressures are lower than the pressure in the refrigerant tank as stated in the question, then the refrigerant can be added to both the high and low pressure sides of the system.
Therefore, the correct option is true.
a sphere of diameter 6.0
cm is moulded into a thin wire of diameter 0.2 mm calculate the length of the wire in metres
Answer:
V1 = 4/3 pi R^3 = pi D^3 / 6 D = 2 R volume of sphere
V2 = pi r^2 L = pi d^2 L / 4 volume of wire
V2 / V1 = 1 = 3/2 d^2 L / D^3 since volumes are equal
L = 2/3 D^3 / d^2 = 2/3 * 6^3 / .02^2 = 360,00 cm = 3600 m
What evidence is there that the use of fossil fuels, which produce carbon dioxide, is causing climate change?
Answer:
The carbon dioxide produced by fossil fuel into the air is a greenhouse gas just as is nitrous oxide, methane, water vapor, and synthetic gases that contain fluorine. Compared to the other greenhouse gases, carbon dioxide is the most abundantly present greenhouse gas.
The greenhouse gases cause greenhouse effect which is the observed trapping of heat in the atmosphere, similar to the glass roof of a greenhouse, thereby causing the average global temperature to rise, and changing the climate through global warming
There is presently a 1°C rise in the average global temperature as part of the Earth reflected heat from the Sun is returned back to Earth by the increasing greenhouse gas presence in the atmosphere which cause the greenhouse effect
Explanation:
Answer:
The carbon dioxide produced by fossil fuels in the air is a greenhouse gas, as is nitrous oxide, methane, water vapor, and synthetic gases containing fluorine. Compared to the other greenhouse gases, carbon dioxide is the most abundant greenhouse gas.
The greenhouse gases cause the greenhouse effect which is the observed trapping of heat in the atmosphere, similar to the glass roof of a greenhouse, thereby causing the average global temperature to rise, and changing the climate through global warming
There is presently a 1°C rise in the average global temperature as part of the Earth's reflected heat from the Sun is returned to Earth by the increasing greenhouse gas presence in the atmosphere which causes the greenhouse effect
Explanation:
I got a 100% on this assignment copy it and put it on your notes then go sentence by sentence copy but dont highlight any "." you will not be able to paste it into the box add the "." after you paste it, it one sentence.
ONE SENTENCE AT A TIME NO ". ' or 1°" AT ALLA car travels initially at 24 m/s, until it enters the highway. If the car accelerates at 4 m/s^2 for a 96 meters, what is the car's new velocity?
Using 3rd equation of kinematics
[tex]\boxed{\Large{\sf v^2-u^2=2as}}[/tex]
[tex]\\ \Large\sf\longmapsto v^2=u^2+2as[/tex]
[tex]\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)[/tex]
[tex]\\ \Large\sf\longmapsto v^2=576+768[/tex]
[tex]\\ \Large\sf\longmapsto v^2=1344[/tex]
[tex]\\ \Large\sf\longmapsto v=\sqrt{1344}[/tex]
[tex]\\ \Large\sf\longmapsto v=36.6m/s[/tex]
what is required for force to come into play
Answer:
An interaction of one object with another object results in a force between the two objects. Thus, at-least two objects must interact for a force to come into play.
the masses of your hand and your notebook are quite small, so the force of attraction between them is
Answer the following questions 1 What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars ? Why?
Answer:
difference between mass and weight is that mass is amount of matter in a material while weight is a measure of how the force of gravity acts upon that mass.the mass of an object will be same on earth and mars but its weight on both planets will be different.
A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find
i. it's period
ii. Angular Velocity
iii. Linear Velocity
iv. It's acceleration
PLEASE FORMULAE SHOULD BE WRITTEN CLEARLY.
YOU CAN SNAP AND SEND
Answer:
i) The period of the particle is 0.3 seconds
ii) The angular velocity is approximately 20.94 rad/s
iii) The linear velocity is approximately 1.047 m/s
iv) The centripetal acceleration is approximately 6.98 m/s²
Explanation:
The given parameters are;
The number of revolution of the particle, n = 800 revolution
The time it takes the particle to make 800 revolutions = 4 minutes
The dimension of the circle = 5 cm = 0.05 m
Given that the dimension of the circle is the radius of the circle, we have;
i) The period of the particle, T = The time to complete one revolution
T = 1/(The number of revolutions per second)
∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s
The period, T = 3/10 seconds = 0.3 seconds
ii) The angular velocity, ω = Angle covered/(Time)
800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes
∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3
The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s
iii) The linear velocity, v = r × ω
∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s
iv) The centripetal acceleration, [tex]a_c[/tex] = v²/r
∴ The centripetal acceleration, [tex]a_c[/tex] = (π/3)²/(0.05) = 20·π/9
The centripetal acceleration, [tex]a_c[/tex] = 20·π/9 m/s² ≈ 6.98 m/s²
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
Answer:
It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.
V^2=U^2+2*a*x
V=0 (as it is at max height)
U=30ms^-1 (initial speed)
a=-g /-9.8ms^-2 (as it is moving against gravity)
x is the variable you want to calculate (height)
0=30^2+2*(-9.8)*x
x=-30^2/2*-9.8
x=45.92m
Answer:
35
Explanation:
dfddffffffffffffdddfr
Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=4.00×103 kg and the second a mass of m2=7.50×103 kg. If the two satellites collide elastically rather than dock, what is their final relative velocity?
Answer:
Their final relative velocity is 0.190 m/s
Explanation:
The relative velocity of the satellites, v = 0.190 m/s
The mass of the first satellite, m₁ = 4.00 × 10³ kg
The mass of the second satellite, m₂ = 7.50 × 10³ kg
Given that the satellites have elastic collision, we have;
[tex]v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2[/tex]
[tex]v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2[/tex]
Given that the initial velocities are equal in magnitude, we have;
u₁ = u₂ = v/2
u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s
v₁ and v₂ = The final velocities of the satellites
We get;
[tex]v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
[tex]v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095[/tex]
The final relative velocity of the satellite, [tex]v_f[/tex] = v₁ + v₂
∴ [tex]v_f[/tex] = 0.095 + 0.095 = 0.190
The final relative velocity of the satellite, [tex]v_f[/tex] = 0.190 m/s
which statement accurately describes how the acceleration of an object in free fall changes
Answer:
A ball rolls off the table and free falls to the ground. Which statement accurately describes its acceleration? Its acceleration is downward but remains a constant value as it falls.
A object of mass 200kg is pushed from rest by a force of 500N along a horizontal plane for 5.0 seconds. Calculate the acceleration of the object
Answer:
force=mass×acceleration
hence
acceleration is given by force÷mass
(500÷200)*5=12.5
A car increase its speed steadily from 30km/hr to 60km/hr in 1 min A what is the average speed during this time
Explanation:
initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s
final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s
time taken(t) = 1 minutes
= 60 seconds
Now,
Average velocity = u+v/2
= 8.33 m/s + 16.67m/s÷2
=12.5 m/s
