Answer:
v = 1.74 m/s
Explanation:
Given that,
Diameter of a circular track, d = 100 m
Distance covered for the 4 laps,
[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]
Time, t = 12 minutes = 720 s
We need to find the velocity of the peter. It can be calculated as follows :
[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]
So, the speed is running with a velocity of 1.74 m/s.
Peter is running at 1.7453 m/sec.
Given to us,
Diameter of the circular track, D = 100 meters,
Number of laps Peter run, L = 4 laps,
Time taken by Peter, t = 12 minutes,
1 lap = circumference of the circle,
4 laps = 4 x circumference of the circle,
As we know, the circumference of a circle is given by πD.
So, 4 laps = 4 x circumference of the circle,
[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]
Also, we know that 1 minute has 60 sec.
so, 4 minutes = (4 x 60) seconds
Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]
Thus,
[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]
Hence, Peter is running at 1.7453 m/sec.
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How do I proton and and electron compared
what is the meaning of the word physics
Answer:
the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
Explanation:
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Bird A, with a mass of 2.2 kg, is stationary while Bird B, with a mass of 1.7 kg, is moving due north from Bird A at 3 m/s. What is the velocity of the center of mass for this system of two birds
Answer:
1.3 m/s
Explanation:
It is given that,
Mass of bird A, [tex]m_A=2.2\ kg[/tex]
Mass of bird B, [tex]m_B=1.7\ kg[/tex]
Initial speed of bird A is 0 as it was at rest
Initial speed of bird B is 3 m/s
We need to find the velocity of the center of mass for this system of two birds. Let it is V. so,
[tex]v_{cm}=\dfrac{m_Au_A+m_Bu_B}{m_A+m_B}\\\\v_{cm}=\dfrac{2.2\times 0+1.7\times 3}{2.2+1.7}\\\\v_{cm}=1.3\ m/s[/tex]
So, the center of mass for this system is 1.3 m/s.
A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.
Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.
Required:
a. Why does the person land in the moat if the rope's length is very short?
b. Why does the person land in the moat if the length is nearly the same as the height of the platform?
Answer:
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
Explanation:
For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.
Let's start with the projectile launch
as the body leaves the vertical its velocity must be horizontal
x = v₀ₓ t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero
t = √ 2 y₀ / g
we substitute
x = vox √2y₀ / g
v₀ₓ = √(g / 2y₀) x
In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)
v₀ₓ = √(g /(2 (H -L)) D
this is the minimum speed to cross the well.
Now let's use conservation of energy
starting point. On the platform
[tex]Em_{o}[/tex] = U = m g H
final point. At the bottom of the swing
Em_{f} = K + U = 1 / 2m v² + m g (H -L)
as there is no friction the mechanical energy is conserved
Em_{o} = Em_{f}
m g H = 1 / 2m v² + m g (H -L)
v = √ (2gL)
let's write our two equations
the minimum speed to cross the well
v₀ₓ = √ (g /(2 (H -L)) D
the speed at the bottom of the oscillatory motion
v = √ (2g L)
we analyze the extreme cases
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well
V₀ₓ = v
g / (2 (H -L) D² = 2g L
4 L (H- L) = D²
4 H L - 4 L2 - D² = 0
L² - H L - D² / 4 = 0
let's solve the quadratic equation
L = [H ± √ (H2-D2)] / 2
we assume that H> D
L = ½ H [1 + - RA (1 - (D / H) 2)]
The two values of La give the range of values for which the two speeds are equal
A) The person lands in the moat if the rope's length is very short because :
The speed of the platform is less than the required minimum speedB) The person lands in the moat if the rope length is similar to the height of the platform because :
The speed required to cross the moat approaches infinityFollowing the assumptions;
size of the person is much smaller than L and H
D = horizontal distance
The conditions that will cause the person to land on the moatThe person will land in the moat when the rope's length is very short because as the rope reduces in length the speed reduces as well such that the speed of the platform goes below the required minimum speed which will enable the person cross over. while As the magnitude of the length tends towards the same magnitude of the height the speed required to cross the moat increases towards infinity and this speed cannot be attained by the person hence he will land in the moat.Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed and The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.
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Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion is reversible but not isothermal and that the temperature decreases as the expansion proceeds. (a) Find an expression
Answer:
(a) The work done by the gas on the surroundings is, 17537.016 J
(b) The entropy change of the gas is, 73.0709 J/K
(c) The entropy change of the gas is equal to zero.
Explanation:
(a) The expression used for work done in reversible isothermal expansion will be,
where,
w = work done = ?
n = number of moles of gas = 4 mole
R = gas constant = 8.314 J/mole K
T = temperature of gas = 240 K
= initial volume of gas =
= final volume of gas =
Now put all the given values in the above formula, we get:
The work done by the gas on the surroundings is, 17537.016 J
(b) Now we have to calculate the entropy change of the gas.
As per first law of thermodynamic,
where,
= internal energy
q = heat
w = work done
As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.
So, at constant temperature the internal energy is equal to zero.
Thus, w = q = 17537.016 J
Formula used for entropy change:
The entropy change of the gas is, 73.0709 J/K
(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.
As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0
So, from this we conclude that the entropy change of the gas must also be equal to zero.
Explanation:
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.
Answer:
2 and 8
Explanation:
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A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east. what is the distance and displacement
Explanation:
Let east = E, and, west = opposite to east = - E.
Here, displacement:
=> 2m east + 4m west + 1m east
=> 2E + 4(-E) + 1E
=> 2E - 4E + 1E
=> - 1E
=> 1(-E)
=> 1m west
And, distance,
=> 2m + 4m + 1m = 7m
The distance of a person is 7 m and the displacement of the person is 1m west.
To find the distance and displacement, the given values are,
A person walks 2.00 m east, then turns and goes 4.00 m west, then turns and goes back 1.00 m east.
What is the distance and the displacement?Displacement:
The displacement is shortest distance between initial and final position or we can say it is the straight line distance between initial and final position.If object moves in a straight line path without any turn then the path length and the displacement is always same.Distance:
The distance is the total path length of the object while it will move from initial to final position.If the object move on curved path then displacement is smaller than the distance moved by the object.Let us consider East = E and west = opposite to east = - E.
Calculating the displacement:
= 2m east + 4m west + 1m east
= 2E + 4(-E) + 1E
= 2E - 4E + 1E
= - 1E
= 1(-E)
= 1m west.
The displacement is 1m west.
Now calculating the distance,
= 2m + 4m + 1m
= 7m
The distance is 7m.
Thus, the displacement and the distance is found as 1 m west and 7m.
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Thank you.
Answer:
(a) 102 cm/s
(b) 0.490 cm²
Explanation:
(a) Use Bernoulli equation.
P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂
0 + ½ ρ v₁² + ρgh₁ = 0 + ½ ρ v₂² + 0
½ ρ v₁² + ρgh₁ = ½ ρ v₂²
½ v₁² + gh₁ = ½ v₂²
½ (25.0 cm/s)² + (980 cm/s²) (5.00 cm) = ½ v²
v = 102 cm/s
(b) The flow rate is constant.
v₁ A₁ = v₂ A₂
(25.0 cm/s) (2.00 cm²) = (102 cm/s) A
A = 0.490 cm²
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?
Answer:
Option C.
Explanation:
To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:
1. Determination of the velocity.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Final velocity (v) =.?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 100)
v² = 0 + 1960
v² = 1960
Take the square root of both side.
v = √(1960)
v = 44.27 m/s
2. Determination of the time taken.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Time (t) =.?
h = ½gt²
100 = ½ × 9.8 × t²
100 = 4.9 × t²
Divide both side by 4.9
t² = 100 / 4.9
Take the square root of both side
t = √(100 / 4.9)
t = 4.52 s
From the above illustration,
Initial time (t1) = 0 s
Final time (t2) = 4.52 s
Initial velocity (u) = 0 m/s
Final velocity (v) = 44.27 m/s
Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.
A recipe gives the instructions below
After browning the meat pour off fat from the pan to further reduce fat use a strainer.
what type lf separation methods are described in the recipe
A decantation and screening
B distillation and screening
C decantation and centrifugation
D distillation and filtration
Answer:
A. decantation and screening
Explanation:
Decantation is the one of the process of separating the mixture. In this process the precipitated liquid is separated from the solid. According to the given instruction for the recipe, the fat which is in liquid state is separated from meat. In the process of screening, more liquid is separated by placing the mixture on the screen. Here, the gravity plays an important role for the process of separation.
Answer:
a
Explanation:
A car starts from rest and accelerates for 7.2 s with an acceleration of 1.4 m/s2. How far does it travel? Answer in units of m.
Answer:
xn = 36.28 [m]
Explanation:
To solve this problem we must use the following equation of kinematics, which is ideal for a body that moves with constant acceleration.
[tex]x=x_{0}+(v_{o} *t)+(\frac{1}{2} )*a*t^{2}[/tex]
where:
x - xo = displacement of the car [m]
Vo = initial velocity = 0
t = time = 7.2 [s]
a = acceleration = 1.4 [m/s^2]
The initial velocity is zero, as the car begins its movement from rest.
xn = (x - xo), Now replacing
xn = (0*7.2) + 0.5*1.4*(7.2^2)
xn = 36.28 [m]
During which process of the water cycle does water change from a gas to liquad
Magnets are usually made up of which material
A. plastic
B. iron ore
C. copper
D. gold
Answer:
B. iron ore
Explanation:
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What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
waht is science
wjwissbsskdldmndndnd
Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the information in the graph, it can be
that in the first second
Answer:speeding up constantly
Explanation:
The graph between the time and the speed of the car shows that the speed is increasing constantly, so, option C is correct.
What is speed?A moving object's rate of change in distance traveled is measured as speed. Speed is a scalar, which implies it is a measurement with a magnitude but no direction.
A thing that moves quickly and with high speed, covering a lot of ground in a short time. On the other hand, a slow-moving object traveling at a low speed covers a comparatively small distance in the same amount of time. An object with zero speed does not move at all.
Given:
The graph shows the speed of a car traveling east over a 12-second period,
As you can see from the graph, at time t = 0 sec the speed is 10 m/s,
At t = 3 sec, the speed = 15.3 m/s
At t = 6 sec, the speed = 20.3 m/s
Thus, speed is increasing constantly.
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One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.
Answer:
The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
Explanation:
From the question we are told that
The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]
The negative sign is because the direction is towards the south
The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]
The negative sign is because the direction is towards the west
The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]
The distance of the westbound plane is [tex]E = 15 \ km[/tex]
Generally the distance between the plane is mathematically represented using Pythagoras theorem as
[tex]R^2 = N^2 + E^2[/tex]
Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes
So
[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]
Here
[tex]R = \sqrt{N^2 + E^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R =33.54 \ m [/tex]
[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]
=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]
=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
The rate of change of the distance between the planes is 286.23 km/hr.
The given parameters;
speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 kmThe distance between the two planes is calculated by applying Pythagoras theorem as shown below;
[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]
The rate of change of the distance between the planes is calculated as follows;
[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]
Thus, the rate of change of the distance between the planes is 286.23 km/hr.
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1. What is Ohm"s law?
2. If you placed a negatively charged hairbrush near your hair, what charge would your hair be?
3. You must change a lightbulb and the new lightbulb has a larger resistance. If the voltage of the battery does not change, what happens to the current going through the flashlight?
HELLPPPP
1. Ohm's law shows the relationship between:
voltagecurrentresistanceFormula: voltage = current x resistance
2. The negative charge on the hairbrush will induce a positive charge on your hair. As a result, your hair is going to be attracted to the hairbrush (and repelled by other strands of hair.)
3. V = IR, so if the resistance of the current increases, and the voltage of the current stays the same, there is as a result, going to be less current.
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A tower crane has a hoist motor rated at 159 hp. If the crane is limited to using 72.0 % of its maximum hoisting power for safety reasons, what is the shortest time in which the crane can lift a 5550 kg load over a distance of 89.0 m
Answer:
The value is [tex]t = 56.68 \ s [/tex]
Explanation:
From the question we are told that
The rating of the hoist motor is [tex]k = 159hp = 159 *746 =118614 \ W[/tex]
The percentage of it power used is [tex]z = 0.72 * 118614=85402.08 \ W[/tex]
The mass of the load is m = 5550 kg
The distance is h = 89.0 m
The potential energy required to lift the load through that distance is
[tex]E = m * g * h[/tex]
=> [tex]E = 5550 * 9.8 * 89.0[/tex]
=> [tex]E = 4840710 \ J[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{E}{ z}[/tex]
=> [tex]t = \frac{4840710}{ 85402.08}[/tex]
=> [tex]t = 56.68 \ s [/tex]
True.or false A railroad track runs southwest to northeast.
Answer:
ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]
Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.
As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]
Contents
1 Definitions in American context
2 History
2.1 Faster inter-city trains: 1920–1941
2.2 Post-war period: 1945–1960
2.3 First attempts: 1960–1992
2.4 Renewed interest: 1993–2008
2.5 Plans for 2008–2013
3 Current state and regional efforts
3.1 The Northeast
3.1.1 Northeast Corridor: Next Generation High-Speed Rail
3.1.1.1 Proposed routes
3.1.2 Northeast Maglev proposal
3.1.3 New Jersey–New York City upgrades
3.1.4 New York
3.1.5 Pennsylvania
3.2 Western States
3.2.1 California
3.2.2 Pacific Northwest
3.2.3 Arizona
3.3 Mid-Atlantic and the South
3.3.1 Florida
3.3.2 Southeast
3.3.3 Texas
3.4 Midwest
3.4.1 Illinois and the Midwest
3.5 The Southwest
4 Federal high-speed rail initiatives
4.1 American Recovery and Reinvestment Act of 2009
4.1.1 Strategic plan
4.2 2009 federal grant funding
4.3 2010 allocation
4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida
4.4 2011 and 2012 proposals and rejections of funding
5 See also
6 Notes
7 Further reading
8 External links
Explanation:
A chef places an open sack of flour on a kitchen scale. The scale reading of
40 N indicates that the scale is exerting an upward force of 40 N on the sack. The magnitude of this force equals the magnitude of the force of Earth’s gravitational attraction on the sack. The chef then exerts an upward force of
10 N on the bag and the scale reading falls to 30 N.Draw a free-body diagram of the latter situation.
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
40N
Explanation:
trust
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
am I right? be honest
Answer:
I chose c because it is the greater slope at point c
block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force
Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s
Answer:
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
Explanation:
From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:
[tex]v = r\cdot \omega[/tex] (Eq. 1)
Where:
[tex]r[/tex] - Radius of rotation of the particle, measured in meters.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]v[/tex] - Linear velocity of the point, measured in meters per second.
But we know that angular velocity is also equal to:
[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)
Where:
[tex]\theta[/tex] - Angular displacement, measured in radians.
[tex]t[/tex] - Time, measured in seconds.
By applying (Eq. 2) in (Eq. 1) we get that:
[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)
From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:
[tex]s = r\cdot \theta[/tex]
[tex]v = \frac{s}{t}[/tex]
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]